Algebraic. techniques1

Transcription

1 techniques Algebraic An electrician, a bank worker, a plumber and so on all have tools of their trade. Without these tools, and a good working knowledge of how to use them, it would be impossible for them to perform their jobs successfully. In the same way, mathematics practitioners have various tools of their trade. The most vital of these tools is algebra, since it underpins so much of mathematics and its study. In this chapter, techniques of algebra (both old and new) are developed, so that you will be equipped with this vital tool for your further study of mathematics.

2 TI-Nspire 0., 0. ClassPad 0., 0. MathsWorld Mathematical Methods CAS Units &. Algebra of polynomials Polynomials A polynomial is an algebraic expression of the form Px ( ) = a n x n + a n x n + + a x+ a 0 where n is a positive integer or zero and the coefficients a n, a n,, a 0 are real numbers. The term a n x n, a n 0, is called the leading term of the polynomial. The degree of the polynomial is n. The remainder and factor theorems The remainder theorem states that if a polynomial P(x) is divided by the linear expression (x a) then the remainder is given by P(a). If the remainder is 0 then (x a) is a factor of P(x). The factor theorem states that if (x a) is a factor of P(x) then P(a) = 0 and, conversely, if P(a) = 0 then (x a) is a factor of P(x). Dividing by (bx c) is the same as dividing by b c x --. In this case the remainder is b given by P c. b -- Example If (x ) is a factor of P(x) = x + kx + bx 5 and P( ) = 5, find the values of k and b. Hence find the remainder when P(x) is divided by (x + ). (x ) is a factor of P(x) so P() = 0. + k + b 5 = 0 k + b = () P( ) = 5 + k b 5 = 5 k b = 6 k b = 8 () k = 6 () + () k = Substitute k = into equation (). + b = b = Hence P(x) = x x + x 5. The required remainder is given by P( ). P( ) = 5 = Tip The Solve command of a CAS can be used to solve for b and k. Define P(x) then solve P() = 0 and P( ) = 5 for b and k.

3 Algebraic techniques Long division of polynomials and factorisation Long division is a useful technique, particularly for factorising polynomials by hand.. Example Factorise: a x + x 7x 0 b x 7x + 6 a Let P(x) = x + x 7x 0. Use the factor theorem to find a factor. P( ) = ( ) + ( ) 7( ) 0 = 0 So (x + ) is a factor. Now use long division to find the other factor. x + x 0 x + ) x + x 7x 0 x + x x 7x x + x 0x 0 0x 0 So, P(x) = x + x x 0 = (x + )(x + x 0) = (x + )(x + 5)(x ) 0 Tip If the coefficient of the leading term of P(x) is, when using the factor theorem to find a factor of P(x) (i.e. finding a value of a for which P(a) = 0), the value of a must be a factor of the constant term. In this case you would only test the values,,,, 5, 5 and 0, 0. b Let P(x) = x 7x + 6. Use the factor theorem to find a factor. P() = () 7() + 6 = 0 So (x ) is a factor. Now use long division to find the other factor. x + x 6 x ) x + 0x 7x + 6 x x x 7x x x 6x + 6 6x + 6 So, P(x) = x 7x + 6 = (x )(x + x 6) = (x )(x + )(x ) 0 Tip When performing long division it is recommended to keep the same powers of x in vertical columns. In some cases it may be necessary to include zero terms to make this easier.

4 TI-Nspire.9 ClassPad.9 MathsWorld Mathematical Methods CAS Units & Other techniques that do not involve the use of long division can also be used to factorise cubic polynomials. Example Factorise: a x + 5x + 6x b x + x x 6 a x + 5x + 6x = x(x + 5x + 6) (taking out a common factor first) = x(x + )(x + ) b x + x x 6 = x (x + ) (x + ) (grouping terms) = (x + )(x ) = (x + )(x + )(x ) Tip The Factor command of a CAS can be used to factorise polynomials such as those in example. Equality of polynomials Two polynomials P (x) and P (x) such that P (x) = a n x n + a n x n + a n x n + + a x + a 0 P (x) = b n x n + b n x n + b n x n + + b x + b 0 are equal for all values of x if and only if a n = b n, a n = b n,, a = b, a 0 = b 0. Example Find the values of a, b and k for which 8x + (a 6)x + (b a)x = kx for all values of x. 8x + (a 6)x + (b a)x = kx 8x + (a 6)x + (b a)x = kx + 0x + 0x Equating coefficients gives k = 8 (x ) a 6 = 0 (x ) a = 6 b a = 0 (x) b = a = 6

5 Algebraic techniques Example 5 Factorise x 5x x + without using long division.. Use the factor theorem to find one factor. P( ) = ( ) 5( ) ( ) + = 0 So (x + ) is a factor. The other factor must be quadratic (i.e. of the form ax + bx + c). (x + )(ax + bx + c) = x 5x x + Expanding gives ax + bx + cx + ax + bx + c = x 5x x + ax + (a + b)x + (b + c)x + c = x 5x x + Equating coefficients of like powers of x: a =, c = a + b = 5 + b = 5 b = 7 So, x 5x x + = (x + )(x 7x + ) = (x + )(x )(x ) Tip a and c can be found by inspection before expanding. Then it is only necessary to determine either the x or x term in the expansion of the left hand side of the equation in order to find b. exercise. Find the remainder when x + x x + is divided by (x + ). Use the remainder theorem to find the remainder when x x + x is divided by (x ). When x + ax + bx is divided by (x ) the remainder is, and when it is divided by (x + ) the remainder is 6. Find the values of a and b. x + ax + x 5 and x + x 8 both have the same remainder when divided by (x ). Find the value of a. 5 The polynomial x + px + qx + 6 is exactly divisible by (x + ) and (x ). Find the values of p and q. 6 Factorise: a x x x + b x + x x c x + x 0x d x x 9x + 8 e 6x + x 5x f x x 5x + 8 g x 8x + 5x 5 h x + 9x 8x 7 If x + x x + = (x + )(ax + bx + c) + d for all values of x, find the values of a, b, c and d. 8 If x + x x + = x + (mx + n) + 5 for all values of x, find the values of m and n. 9 a If x = a(x + ) + b(x + ) + c for all values of x, find the values of a, b and c. b If x x + = A(x h) + k, find the values of A, h, and k. 5

6 TI-Nspire 0. ClassPad 0. MathsWorld Mathematical Methods CAS Units & of polynomial equations Solving an equation means finding the value(s) of the variable that makes the equation true. Linear equations A linear equation in one variable is an equation involving polynomials whose degree is. Thus x + 5 = x is a linear equation in the variable x. Some equations reduce to linear equations when suitably transformed. For example, the x equation = 7 can be expressed as x = 7(x ). x Linear equations are usually solved by isolating the variable on one side of the equation. Example 6 Solve for x the equation (x ) 5 = (x + ) + 5. Expand the bracketed terms and simplify. x = x + Isolate the variable by subtracting x from both sides and adding to both sides. x = Check that x = is the correct solution. LHS = ( ) 5 = 77; RHS = ( + ) + 5 = 77 = LHS Example 7 Solve = for x. x x + Clear the fractions by multiplying both sides by (x )(x + ). Thus: (x + ) = (x ) x + 6 = x 8 = x Hence x = 8. Tip The Solve command of a CAS can be used as shown. Quadratic equations The standard form of a quadratic equation is ax + bx + c = 0, where x is the variable and a, b and c are constants with a 0. Methods for solving quadratic equations include factorisation, completing the square, using the quadratic formula and numerical or graphical approximation. When quadratic equations are expressed as the product of linear factors, the null factor law can be used to find solutions. Null factor law If ab = 0, then either a = 0 or b = 0 or both a and b = 0. 6

7 TI-Nspire 0. ClassPad 0. Algebraic techniques Example 8 Solve each of the following for x. a x 8 = 0 b x = 5 x c x + 6x = 0 d x + x 6 = 0 a x 8 = 0 (x 6) = 0 (common factor) (x )(x + ) = 0 (difference of two squares) x =, b x = 5 x x + x 5 = 0 (using the null factor law) (taking all terms to one side) (x 5)(x + ) = 0 (factorising by inspection) 5 x = --, (using the null factor law) c x + 6x cannot be factorised by inspection so complete the square. x + 6x = 0 x + 6x = 0 (x + ) = 0 ( x + ) ( x + + ) = 0 (since = ) So x = +,. d x + x 6 cannot be factorised by inspection so use the quadratic formula. If ax b ± b ac + bx + c = 0 then x = a Here, a =, b = and c = 6. ± ( ) ( 6) x = ± 76 = ± 9 = ± 9 = Tip To avoid errors when cancelling, taking out a common factor in the numerator may be advantageous. ± 9 = ( ± 9) = ± 9 = Tip The Solve command of a CAS can be used to find exact or approximate solutions to a quadratic equation such as that in example 8 part d. Press the appropriate key combination to obtain the numerical approximations. 7

8 MathsWorld Mathematical Methods CAS Units & Equations reducible to quadratics Some equations are reducible to quadratics, as shown in the following examples. Example 9 Solve for x: x -- 5 x = 0. x x x -- 5 x = 0 x x Let a = x -- so the equation becomes a 5a + 6 = 0. x Now factorise and solve: (a )(a ) = 0 a = 0 or a = 0 So, x -- = 0 or x -- = 0. x x x x = 0 or x x = 0 Using the quadratic formula: ± ( ) ( )( ) x = or x = x = ± or x = ± ± ( ) ( )( ) Example 0 Solve for x: x = x 5. Isolate the square root term to one side and then square both sides. x = x 5 (subtract from both sides) x 6x + 9 = x 5 (square both sides; note ( ) = ) x 9x + = 0 (x )(x 7) = 0 x =, 7 Check: x = : LHS = = ; RHS = = = LHS Tip When solving equations involving surds, it is necessary x = 7: LHS = 7 = 6; RHS = 6 = LHS to always check solutions in the original equation. So the only solution is x =. 8

9 TI-Nspire.,., 0. ClassPad.,., 0. Algebraic techniques Equations involving higher degree polynomials A polynomial equation has the form a n x n + a n x n + + a x + a 0 = 0. Techniques for solving linear (degree ) and quadratic (degree ) equations have already been considered. With cubic (degree ) and higher degree polynomial equations it is necessary to use either factorisation or technology to find solutions.. Example Solve each of the following for x, giving exact solutions where possible. a x + x 5x 5 = 0 b x 5 x + x = 0 c x = x a x + x 5x 5 = 0 Factorising the left-hand side by grouping gives: x (x + ) 5(x + ) = 0 (x + )(x 5) = 0 ( x + ) ( x 5) ( x + 5) = 0 x =, 5, 5 b x 5 x + x = 0 Factorising gives: x(x x + ) = 0 x(x )(x + ) = 0 x(x )(x + )(x + ) = 0 x = 0,, c x = x x x + = 0 However, x x + does not have a rational linear factor, so the equation is solved using technology. A graphical approach can be used. Plot the graph of y = x x + and find any zeros or x-intercepts using the appropriate jump-to feature of a CAS. From the screenshot, there is exactly one solution, x =.5, correct to decimal places. (An alternative graphical method is to plot the graphs of y = x and y = x and find any points of intersection.) Alternatively, the Solve command of a CAS can be used as shown. Tip The factor theorem could also be used to identify the factor (x + ). 9

10 MathsWorld Mathematical Methods CAS Units & exercise. 0 Solve each of the following for x. a (x + ) = (x + ) b 9(x + ) 55x = 6 5x x 6 5 c = d -- + = x x e (x + ) = (x ) + 6 f (x + )(x ) (x ) = (x + ) Find all the exact solution(s) to each of the following equations. Check the solution(s) with your CAS. a (x + )(x ) + 5 = 0 b x + 6x = 0 c x + 6x + 7 = 0 d 5x 6x = 0 e (x + 5) = 5 0x f x = x g x + 5x 6 = 0 h 6 x + -- x = 0 i x x 7x = 0 j x x x + 6 = 0 k x x + x = 0 l x 6 = 8 7x m x + = x + n + x + = x o 5 x x = x (Hint: In part o, square both sides, taking care with the left-hand side, then isolate the remaining square root term and square both sides a second time.) Use your CAS to solve each of the following equations, giving answers correct to decimal places. a x x 5 = 0 b x = 5 x c x x + x = 0 d x x = 0 e x = + x f x 6 = 5 + x Binomial expansions Pascal s triangle Pascal s triangle is named after the French mathematician Blaise Pascal. Although Pascal was not responsible for its discovery (it was discovered by the Chinese about 500 years before his birth), he was the first mathematician to perform any extensive study of it. Constructing Pascal s triangle continued. At the top we place the number. This is referred to as row zero.. At each end of every subsequent row, place the number.. Each other element is found by adding the two numbers diagonally above it in the previous row. 0

11 Algebraic techniques Rows 0 to 6 of Pascal s triangle are shown below For example, in row, the second element is obtained by adding the numbers and diagonally above it. Similarly, the third element 6 is found by adding the numbers and. Pascal s triangle and binomial expansions Consider the expansion of (a + b) n for n = 0,,,, and 5. Since the bracketed expression consists of two terms, it is said to be binomial and hence its expansion is known as a binomial expansion. (a + b) 0 = (a + b) = a + b (a + b) = a + ab + b (a + b) = a + a b + ab + b (a + b) = a + a b + 6a b + ab + b (a + b) 5 = a 5 + 5a b + 0a b + 0a b + 5ab + b 5 It is important to note the following features of each expansion.. There are n + terms in each expansion.. In each term the sum of the powers of a and b is n.. In successive terms, the powers of a decrease from n to 0 and the powers of b increase from 0 to n.. The coefficients of the terms in each expansion relate to the rows of Pascal s triangle as shown below. (a + b) 0 (a + b) (a + b) (a + b) (a + b) 6 (a + b)

12 TI-Nspire.9 ClassPad.9 MathsWorld Mathematical Methods CAS Units & Example Use Pascal s triangle to expand: a (x ) b (x ) 6 c a To expand (x ) we obtain the coefficients from the th row of Pascal s triangle. So the expansion of (a + b) is (a + b) = a + a b + 6a b + ab + b. Here a = x, b =. (x ) = x + x ( ) + 6x ( ) + x( ) + ( ) = x x + 5x 08x + 8 b To expand (x ) 6 we obtain the coefficients from the 6th row of Pascal s triangle. So the expansion of (a + b) 6 is (a + b) 6 = a 6 + 6a 5 b + 5a b + 0a b + 5a b + 6ab 5 + b 6 Here a = x, b =. (x ) 6 = (x) 6 + 6(x) 5 ( ) + 5(x) ( ) + 0(x) ( ) + 5(x) ( ) + 6(x)( ) 5 + ( ) 6 = 6x 6 576x x 0x + 860x 96x + 79 c To expand x -- we obtain the coefficients from the rd row of Pascal s triangle. x So the expansion of (a + b) is (a + b) = a + a b + ab + b Here a = x, b = --. x x -- = ( x) x + x ( ) -- x ( ) -- x + x + -- x = 8x x x x x -- x Tip The Expand command of a CAS can be used to obtain binomial expansions. For example the screenshot on the right shows the results for example parts a and b.

13 TI-Nspire.8 ClassPad.8 Algebraic techniques Expansion of (a + b) n Recall from Year Combinatorics that n n n! = C. The elements in Pascal s r r = r! ( n r)! triangle can be expressed as combinations, as shown below For example, 5 5 5! 5! = C and. = = = ! 5! = C!!! = =!! =! 0 These agree with the values in row 5 of Pascal s triangle (see page ). Binomial expansion In general: (a + b) n = n C 0 a n b 0 + n C a n b + + n C r a n r b r + + n C n a b n + n C n a 0 b n = a n + n C a n b + + n C r a n r b r + + n C n ab n + b n where n N. The (r + )th term, or general term, is given by n C r a n r b r, r = 0,,,, n. Tip A CAS can be used to calculate a combination as shown in the following screenshot. Using combinations, the expansion for a high power of (a + b) n can be found without having to write out all the previous rows of Pascal s triangle.

14 The greatest coefficient SAC analysis task MathsWorld Mathematical Methods CAS Units & Example Expand (x + y) 5 using (a + b) n = a n + n C a n b + n C a n b + + b n (x + y) 5 = (x) C (x) y + 5 C (x) y + 5 C (x) y + 5 C (x)y + y 5 = x 5 + 5(8x )y + 0(7x )y + 0(9x )y + 5(x)y + y 5 = x x y + 70x y + 90x y + 5xy + y 5 exercise. State the number of terms in the expanded form of: a (x + ) 7 b (x 5) 0 c ( z) Use Pascal s triangle to expand: a (x + ) 5 b (x ) c (x ) 5 d (x + ) 6 e (x ) 5 f ( x) g (t + y) h 5 Use (a + b) n = a n + n C a n b + n C a n b + + b n to expand: a (y x) 6 b (x y) c (x + x) d (x y ) 5 e x -- + f x g ---- x 7 h ( x ) 6 x 6 Find the th term in the expansion of each of the following. a (x ) 8 b (x t) Analysis task the greatest coefficient. x continued x -- 6 x SAC In this analysis task, you will be asked to determine a general expression for the greatest coefficient in the expansion of ( + x) n. Note that the binomial coefficients are all positive. Part Gathering the tools for analysis Let T r + be the (r + )th term and T r be the rth term in the expansion of ( + x) n for r n. a Write down general expressions for T r + and T r. b Find a general simplified expression for the ratio of the coefficients of T r + and T r. c Under what condition relating n and r must the coefficient of T r + always be greater than or equal to the coefficient of T r? Part d By using Pascal s triangle and the results from the previous part, determine a general expression for the greatest coefficient in the expansion of ( + x) n. Explain your reasoning carefully and illustrate your findings with some specific examples.

15 Approximations using binomial expansions SAC analysis task Algebraic techniques Analysis task approximations using binomial expansions. Expansions of binomial expressions are not limited to positive integer powers. The expansion of ( + x) can be used to approximate reciprocals of numbers close to ; the expansion of ( + x) can be used to approximate square roots of numbers close to. Part ( + x) = x + x x +, where the nth term is ( x) n a Write down the next four terms in the series. b i What is the value of x in this expansion to approximate ?. ii Use technology to evaluate and write down the result correct to decimal places.. iii Use the binomial expansion above with terms to approximate To how many. decimal places is this accurate? iv How many terms are required to approximate correct to decimal places?. Part ( + x) --x = x , 6 where the nth term is x (, n )! > (n ) terms c Simplify this series expansion. d What are the next three terms in this series in simplest form? e Write down the value of x if this expansion is used to approximate: i. ii 0.6 iii. iv 0.7 v. vi 0.8 vii. viii 0.9 f Complete the following table. + x i ii x Value from technology Approximation using SAC term terms terms terms 5 terms 6 terms 7 terms In each case, how many terms are required for decimal place accuracy? As each extra term is added the approximation approaches the actual value. This is referred to as convergence. How does the value of x affect the rate of convergence?. 5

16 MathsWorld Mathematical Methods CAS Units &. Systems of simultaneous linear equations A set of two or more equations in the same variables is referred to as a system of equations. In this section we will consider systems of linear equations. Systems of linear equations in two unknowns Recall that solving simultaneous linear equations in two unknowns algebraically involves obtaining a single linear equation in one variable by eliminating the other variable. Example Solve these simultaneous linear equations using an algebraic method: y = x and x + y = 8 y = x () x + y = 8 () Substitute x for y in equation (). x + (x ) = 8 x + x 6 = 8 7x = x = Substitute x = in equation (). y = () y = So the solution is x = and y =. Tip It is a good idea to check your answer. Substitute x = and y = into the left-hand side of equation (). LHS = () + () = 8 = RHS. Example Solve these simultaneous linear equations: x + 5y = and x y = 7 These equations can be solved algebraically. x + 5y = () x y = 7 () x + 5y = () 0x 5y = 5 () = () 5 x = () + () x = Substitute x = in equation (). y = 7 y = So the solution is x = and y =. 6

17 TI-Nspire. ClassPad. TI-Nspire 8., 8. ClassPad 8., 8. TI-Nspire 8., 8., 0. ClassPad 8., 8., 0. Algebraic techniques Simultaneous linear equations can also be solved graphically by finding the point of intersection of the straight lines corresponding to each equation.. [ 0, 0] by [ 6.7, 6.7] They can also be solved using matrix methods on a CAS. The command rref uses elimination methods to solve the equations. To use this method, first write each equation in the form ax + by = c. Then create a matrix whose elements are the coefficients of x and y and the constant terms in the equations. x + 5y = 5 x y = 7 7 To solve the equations, apply the command rref to the matrix above. The result when rref is applied is shown in the screenshot at right. The rref command has transformed the original pair of equations into: x + 0y = 0x + y = So the solution is x =, y =. (Further details and examples of the use of matrices in the solution of simultaneous linear equations are found on pages.) Tip The matrix used with the rref command can be stored and then used as shown. Alternatively, the Solve command can be used to solve simultaneous linear equations as shown. 7

18 MMCASSB_ed_0_pp.fm Page 8 Tuesday, June, 009 : PM MathsWorld Mathematical Methods CAS Units & Parallel and coincident lines The graphs of the two equations are straight lines. If the graphs are of parallel lines, which do not intersect, there will be no solutions to the simultaneous equations. If the graphs are coincident, that is, the same straight line, there will be an infinite number of solutions to the simultaneous equations. Example Solve the simultaneous equations x y = and 6x y = k if: a k=6 b k=0 a If k = 6, then: () x y = 6x y = 6 () 6x y = 6 () = () 6x y = 6 () The graphs of these two equations are the same straight line, that is, the two lines are coincident. Any point on the first line will also lie on the second. There is an infinite number of solutions to the simultaneous equations consisting of the points lying on the line. Let y = t. Then: 6x t = 6 6x = 6 + t x = t So, for any real value of t, x = t and y = t satisfy the equations. b If k = 0, then: x y = () 6x y = 0 () 6x y = 6 () = () 6x y = 0 () Subtracting these equations gives 0 = 6. This is impossible, so these simultaneous equations have no solution. This is because the graphs of two original equations are parallel lines that never intersect. y 6x y = x y = x

19 Algebraic techniques exercise. Solve each of the following pairs of simultaneous equations by hand then check your answers using technology. a x + y = b --x y = c y = x x y = 5 x + y = 8 --x + --y = 0 y x y d x + -- = e = f x = y 5x + y = 7 y x -- = --x --y = 8 Find the value(s) of k for which the simultaneous equations have: i no solution ii an infinite number of solutions a x + y = b x y = k c x + 9y = k x + y = k x + y = 6 0.8x +.8y = A straight line has equation y = mx + c. Use simultaneous equations to find the equation of the straight line that passes through the points (, ) and (, ). A parabola that passes through the points (, ) and (, 5) has equation y = x + bx + c. Use simultaneous equations to find the values of b and c. Simultaneous linear equations in more than two unknowns Situations often arise in which there are three or more linear equations with corresponding numbers of unknowns. Such systems of equations can be solved by hand, eliminating one variable at a time until a single variable remains and solving the resulting linear equation. The values of the other variables can then be found by substitution. Systems of equations are more easily solved using technology.. Example Solve the following system of equations, using: a algebra (by hand) b technology. x y + z = x + y + z = x y + z = a x y + z = () x + y + z = () x y + z = () x y + 6z = 6 () = () x + y + z = () x y + z = 8 () = () continued 9

20 TI-Nspire 8., 8. ClassPad 8., 8. MathsWorld Mathematical Methods CAS Units & Now y can be eliminated. x + 8z = 0 (5) = () + () 5x + z = (6) = () + () x + 8z = 0 (5) 0x + 8z = (7) = (6) 7x = x = Substitute x = in equation (6) to find z. 0 + z = z = z = -- Substitute x = and z = -- in equation () to find y. + y + = y = y = -- The solution is x =, y = -- and z = --. b First create a matrix whose elements are the coefficients of the variables and the constant terms in the equations. Now use the rref command. This has transformed the equations into x + 0y + 0z = 0x + y + 0z = -- 0x + 0y + z = -- Read off the solution: x =, y = -- and z = --. Tip The matrix can be stored first, then rref applied to the stored matrix. 0

21 Algebraic techniques Matrix representation of systems of linear equations The system of equations in example can be written in the form of a matrix equation as follows. x y + z = x + y + z = x y + z = A X B The matrix form on the right is written as AX = B, where A is the matrix formed by the coefficients of the variables, X is a column matrix containing the variables and B is a column matrix formed by the constants on the right hand side of the equations. To recover the original equations from the matrix equation, start by multiplying the first row of A by the column matrix X and equate the result to the first entry in B: ()(x) + ( )(y) + ()(z) =, i.e. x y + z =. The remaining equations can be recovered in the same way. When written in this form the equations can be solved on a TI-Nspire using the Simult command, accessible from the Matrix menu or the CATALOG. The screenshot at right shows how, with the matrices A and B above stored as a and b. Of course, the system of equations can also be solved using the rref or Solve commands. The role of parameters x y z = A system of linear equations in which there are fewer equations than unknowns can have infinitely many solutions. The solutions are then given in terms of a parameter, another variable. Then substitution of a particular value for the parameter gives one particular solution for the other variables. In some cases, a parametric solution occurs for an obvious geometrical reason, as in example part a, where the two equations represent coincident lines.. Example 5 Solve the system of equations: x + z = x y 7z = 9 x y + z = 7 First create a matrix whose elements are the coefficients of x, y and z and the constant terms in the equations continued

22 TI-Nspire 8., 8. ClassPad 8., 8. TI-Nspire 0., 0. ClassPad 0., 0. MathsWorld Mathematical Methods CAS Units & Now use the rref command, as shown on the right. 5 5 The resulting matrix shows x + z = and y z = -----, so there is an infinite set of solutions. This set of solutions can be written in terms of a parameter. Let z = t. 5 5 x + t = and y t = x = t 5 y = ( t) 5 The solutions are x = t, y = ( t), z = t, where t R. 5 If t = 0, the solution is x =, y = -----, z = 0. If t =, then x = 9, y = 5, z =. Similarly, further specific solutions can be found by assigning t any real number value. Tip Using the Solve command, the same set of solutions as found above is identified. Note that the TI-Nspire expresses these in terms of a parameter c; the ClassPad is interpreted as z can take any real value, and then x and y are determined in terms of that value (so z is taken to be a parameter). Inconsistent systems Sometimes there are no solutions to a system of linear equations. In such cases, the system is said to be inconsistent. Example 6 Solve the system of equations: x y z = x + y + z = 0 x + y z = First create a matrix whose elements are the coefficients of x, y and z and the constant terms in the equations. 0 continued

23 Algebraic techniques Now use the rref command as shown in the screenshot. The last row of this matrix reads 0x + 0y + 0z =. This cannot be true, so the system of equations has no solution. Alternatively, using the Solve command returns the value false on a TI-Nspire or No on a ClassPad, which again means that the system is inconsistent.. Warning Interpreting error messages The alternative matrix method of solving systems of equations using the TI-Nspire command Simult only works if there is a single (unique) solution to the simultaneous system. The screenshot below shows what happens when this method is applied in examples 5 and 6. The error message means that there is a problem with the matrix of coefficients. So this method is unable to deal with examples in which there are infinitely many solutions, or there are inconsistent equations. In such cases the rref or Solve commands are more informative. exercise. continued 5 Solve the following systems of linear simultaneous equations by hand. Check your answers using technology. a x + y + z = b x + y z = c x + y + z = 0 x + y + z = 6 x y + z = 5 x + 5y z = 7 9x + y + z = x + y z = x + y + z = d 5x + z = e x y + z = 8 f x + y + z = y + z = 5 x y + z = 8 x y z = 6 x + y = x + y z = x + y z = 6 A parabola with equation y = ax + bx + c passes through the points (, 0), (, ) and (, 6). Find the values of a, b and c.

24 MathsWorld Mathematical Methods CAS Units & 7 For each of the following: i determine whether the system has a unique solution or an infinite number of solutions or no solution ii find the solutions to those systems of simultaneous equations for which solutions exist. a x + y + z = b x + y + z = 0 c x y + z = 6 x + y + z = 5 x + y z = x y + 5z = 7 x y z = x y z = x + y z = 5 d x + y 5z = e x y + z = f x y z = 5x + y 8z = 6 x + y z = 0 x y + z = x + y 6z = 5 x y + z = x z = 0 8 Express the following systems of equations as matrix equations and use technology to solve them. a x y 5z w = 0 b x y + 6w = x 5y z w = x 5y + z w = x 8y z 0w = y + z + w = x 5y + z 5w = x + y + z = 0 9 A curve passes through the points (, 9), (, 5) (, ) and (, 7). If the curve has equation y = ax + bx + cx + d, find the values of a, b, c and d. 0 Five numbers add to zero. The first is equal to the sum of the second and the fourth. The third is equal to the sum of the fourth and the fifth. The sum of the first two numbers is more than the fifth number. The fifth number is three times the sum of the third and fourth numbers. Find the five numbers. Consider the pair of linear equations below, where k is a real constant. 5x + ky = k kx + 0y = k a Find the solution if k = 5. b Show that the equations are inconsistent if k = 0. c Solve the pair of equations if k = 0. d Solve the pair of equations for x and y in terms of k for all k R\ { 0, 0}. Consider the following system of linear equations, where n is a real constant. x + (n + )y (n + )z = y + (n + )z = x y + z = a Find the solution if n =. b Are there any values of n for which the equations are inconsistent? c Are there any values of n for which the equations have an infinite set of solutions? State the solutions in terms of a parameter t in this case.

25 Algebraic techniques. Algebra of exponential and logarithmic functions Definition of a logarithm Recall that if a > 0 and a x = y then log a y = x. This means that x is the power to which a is raised in order to obtain y. For example, log 9 =, since = 9. Laws of logarithms The laws of logarithms can be deduced from the index laws. Logarithm of a product (first logarithm law) log a (xy) = log a x + log a y, where x, y > 0. (In words, the logarithm of a product is equal to the sum of the logarithms.) Let m = log a x and n = log a y. x = a m and y = a n (definition of a logarithm) xy = a m a n = a m + n (st index law) log a (xy) = m + n (logarithm form) = log a x + log a y Logarithm of a quotient (second logarithm law) x log a -- = log a x log a y, where x, y > 0. y (In words, the logarithm of a quotient is equal to the difference of the logarithms.) Let m = log a x and n = log a y. x = a m and y = a n (definition of a logarithm) x a -- = m y a n = a m n (nd index law) x log a -- = m n (logarithm form) y = log a x log a y Logarithm of a power (third logarithm law) log a (x n ) = n log a x, where x > 0. (In words, the logarithm of a power is equal to the product of the power and the logarithm.) Let m = log a x. x = a m (definition of a logarithm) x n = (a m ) n = a mn (rd index law) log a (x n ) = mn (logarithm form) = n log a x. 5

26 MathsWorld Mathematical Methods CAS Units & Two special cases of the third law are important. If n = 0, log a (x 0 ) = 0, i.e. log a = 0. Of course, this also follows from the fact that a 0 =. If n =, log a (x ) = log a x, i.e. log a -- = log a x. x Comparison of index and logarithm laws Index form Logarithm form. First law: a m a n = a m + n log a (xy) = log a x + log a y. Second law: a m = a m n x log a -- = log a x log a y a n y. Third law: (a m ) n = a mn log a (x n ) = n log a x. Special value: a 0 = log a = 0 Changing the base of logarithms When a decimal approximation to a logarithm is required, the change of base rule is useful, together with a calculator that will evaluate logarithms with base 0. Suppose we need an approximation for log a x. Let y = log a x, so that x = a y. Take logarithms (base b) of each side: log b x = log b a y = y log b a = (log a x)(log b a) log Therefore log a x = b x log b a In the special case where b = 0, we have: log log a x 0 x = log 0 a Example Correct to decimal places, log 5 =.65 and log = 0.6. Use these to find an approximate value for the following. a log 0 b log 0 c log -- 5 d log 5 a log 0 = log (5 ) b log 0 = log (8 5) = log 5 + log = log 8 + log 5 = = log + log 5 =.096 = log + log 5 = =.58 c log -- = log log 5 d log 5 = log (9 5) 5 = log log 5 = log 9 + log 5 = log log 5 = +.65 = =.65 = 0.0 6

27 TI-Nspire. ClassPad. Algebraic techniques Example log Evaluate log 5 log log 5 log = log 5 log = log 5 = Warning It is of course completely incorrect to cancel the logs as in: log = -- =. log 5. Example Express a in terms of b if: a log 0 a = log 0 b + b log a = 0.5b + a log 0 a = log 0 b + b log a = 0.5b + = log 0 b + log 0 0 a = 0.5b + = log 0 0b = 0.5b a = 0b = ( 0.5 ) b 6 = 6 b Example Find the value of log, correct to decimal places, using common (base 0) logarithms. Use the change of base rule to change the logarithm base to base 0. log log a x = 0 x log 0 a log log = log 0 =.070 (correct to decimal places) Tip The calculation using base 0 is shown in the screenshot opposite. However, note that a CAS can find the value directly as shown. (Remember to press the appropriate key combination to obtain the numerical approximation.) 7

28 MathsWorld Mathematical Methods CAS Units & exercise. Evaluate each of the following without the use of a calculator. a log 6 b log 9 c log log d 5 8 log e f log 8 log log 5 log 6 7 g log log 6 h log 6 log 9 i log 0 log Express each of the following as a single logarithm. a log 5 x log 5 (x + ) b log log 0 x c log (x ) log log x d log x y log xy x + log ---- y e log x -- log f ( x 6 y + ) log x -- log 0 x + log 0 x If log a x = 0. and log a y = 0., find: x x a log a (xy) b log c d log a xy a ---- log a y Given that log 5 = 0. and log 5 = 0.68 correct to decimal places, find an approximate value for each of the following. a log 5 b log 5 5 c log 5 60 d log 5 e log 5 0 f 0 log Find an expression for a in terms of b, if: a log 0 a = b b log 0 a = b c log 0 a = log 0 b d y log 0 a = --log 0 b 6 If x = log a and y = log b, express log a b in terms of x and y. b 7 Use the change of base rule and common logarithms (base 0) to evaluate the following, correct to decimal places. a log 7 b log 5 9 c log 0 Solving exponential equations and inequations Consider the exponential equation x = 8. The solution can be obtained as follows: x = 8 = x = It is not always possible to express the term on the right-hand side in a simple way involving indices as in the example above. In such a case, logarithms can be used. 8

29 TI-Nspire. ClassPad. TI-Nspire., 0. ClassPad..0. Algebraic techniques Example 5 Solve for x, giving answers correct to decimal places. a x = 5 b 5 0.x =.7 c x + = 5 x d x + = x a One solution method is to use common logarithms as follows. log 0 x = log 0 5 (taking the common logarithm of both sides) (x ) log 0 = log 0 5 log x = log 0 log x = = (correct to decimal places) log 0 Alternatively, x = 5 is equivalent to x = log 5, i.e. x = log 5 + (note how the change of base rule can be used here to reveal the answer as expressed in the solution above). Hence, correct to decimal places, the answer is x = (refer to the screenshot). log b 5 0.x =.7 is equivalent to 0.x = log 5.7, i.e. x = Hence, correct to decimal places, the answer is x = 0.8 (refer to the screenshot). c x + = 5 x log 0 x + = log 0 5 x (x + ) log 0 = x log 0 5 log 0 = x log 0 5 x log 0 log x = =.5 (correct to decimal places) log 0 5 log 0 d x + = x Technology is required in this case. One method is to plot the graphs of y = x + and y = x with a CAS and find the x-coordinates of the points of intersection as shown. Alternatively, the Solve command can be used. Hence, correct to decimal places, the solution is x =.96 or x =.7.. 9

30 TI-Nspire 0. ClassPad 0. MathsWorld Mathematical Methods CAS Units & Tip The Solve command can be used to find the exact or approximate solutions to exponential equations. For example, consider example 5 part a. The screenshot ln0 opposite gives the exact answer in the form , ln where the symbol ln stands for log natural, which uses a new base, e, as the default base for logarithms (ln can also be written as log e ). This new base e is explored on page of this chapter. The numerical approximation is as before. The change of base rule can be used to convert answers to different bases. Thus: ln0 x = ln = log 0 log00 = (and so on) log 0 Note that the TI-Nspire has a command Convert to logbase (found in the Algebra menu) that can be used as shown in the screenshot. Example 6 Find, correct to decimal places where necessary: a {x : x + < } b {x : (0.5) x } a {x : x + < } x + < log 0 x + < log 0 (x + ) log 0 < log 0 x log 0 < log 0 log 0 log x < 0 log log 0 x < 0.6 The solution set is {x : x < 0.6}. b {x : (0.5) x } (0.5) x (0.5) x -- We could take the logarithm of both sides and solve as in part a, taking care to note that log is negative. However, in this case it is easier to use indices. Since 0.5 = --, the inequality becomes -- x -- x (taking the reciprocal of both sides) x x The solution set is {x : x }. 0

31 Algebraic techniques Example 7 Solve for x, giving exact answers. a 5 x 5 x 5 = 0 b x = 0 x + 9 a 5 x 5 x 5 = 0 (5 x ) 5 x 5 = 0 Factorising gives: (5 x + )(5 x 5) = 0 Using the null factor law: 5 x = (which is impossible as 5 x is positive for all x), or 5 x = 5 x = b x = 0 x + 9 Multiply both sides by x : ( x ) = x ( x ) 9 x 0 = 0 Factorising gives: ( x + )( x 0) = 0 x = (which is impossible), or x = 0 x = log 0 The screenshot at right shows that the answer using the ln0 Solve command is expressed as x = , which by ln the change of base rule is equivalent to x = log 0. If required, the numerical approximation can be obtained by pressing the appropriate key combination.. Solving logarithmic equations The definition of a logarithm and the laws of logarithms are used in solving logarithmic equations. Example 8 Solve for x. a log (x + ) = 5 b log (x + ) log (x ) = c log x + log (x 8) = a log (x + ) = 5 x + = 5 x + = x = 9 Check: LHS = log (9 + ) = log = 5 = RHS continued

32 MathsWorld Mathematical Methods CAS Units & b log (x + ) log (x ) = log x = x x = x x + = 8(x ) x + = 8x 6 9 = 7x 9 x = Check: LHS = log log = log log = log = log 8 = = RHS Tip c log x + log (x 8) = log (x(x 8)) = x(x 8) = x 8x 9 = 0 (x 9)(x + ) = 0 x = 9, Check: x = 9: LHS = log 9 + log (9 8) = + log = = RHS x = : LHS = log ( ) + log ( 9) It is not possible to evaluate the logarithm of a negative number. So, x = 9 is the only solution. When solving logarithmic equations it is necessary to always check solutions. Be sure to check the original equation. exercise. 8 Solve for x. a 5 x = 5 b x = -- c 6 x = 9 d x 5 = e 9 x = 7 f x = Solve for x. a x + = x b 5 x = c x 5x 6 = 5 x + d ( 6) x = 6 5 x e 7 x = ( 7 x ) f 7 x = 0 Solve for x, giving answers correct to decimal places. a x = 0 b -- c 5 x = 0.5 x = 8 d 0 0.x = 5 e 0.0x = 0. f 0 x = 5 Solve for x, giving answers correct to decimal places. a x = x + b x = 5 0.x c x x = x Working correct to decimal places where necessary, find a {x : x > 5} b {x : x < 0.} c {x :5 x } d x : e {x : x + > 0.} f -- x 0.5 x : + x -- x continued

33 Algebraic techniques Solve for x. a x 6 x + 8 = 0 b x x + 7 = 0 c 5 x 0 5 x + 5 = 0 d x 5 x 6 = 0 e x + x 5 = 0 f x 8 = 9 x g x + 7 x + 8 = 0 h + x = x Solve for x. a log (x + ) = b log x 5 = c log x = d log x = e log (x + ) log (x + ) = f log (x + ) log x = log (x + ) g log 0 5x = log 0 x h log x (x + 6x) = i log x = (log x) j log (x + ) + log (x ) = k (log x) 8 log x + 8 = 0 l log 0 x log 0 = log 0 (x ). The base e Consider the binomial expansion of + -- n + -- n = n C n -- n n C -- n n C -- n , where n is a natural number. n! = ( n )! n -- n! ! ( n )! n -- n!! ( n )! n nn ( )! = ( n )! n -- nn ( ) ( n )! ! ( n )! n -- nn ( ) ( n ) ( n )! ! ( n )! n nn ( ) = ! n -- nn ( ) ( n ) ! n = ! n! n -- + n As n, n -- 0, n -- 0, n So, + -- n lim = n n!! The irrational number e is defined as: e = + -- n lim n n = !! =.788 In many applications of exponential and logarithmic functions, the base number e is used. It is referred to as the natural base. The function y = e x, x R, is referred to as the natural exponential function and the function y = log e x, x > 0 is referred to as the natural logarithmic function. Note that log e x is also denoted by ln x. Natural logarithms can be evaluated directly using the ln function of a CAS.

34 MathsWorld Mathematical Methods CAS Units & Example 9 Evaluate each of the following correct to decimal places. a e b -- e c log e 5 d ln The screenshot at right shows the relevant calculations, obtained by pressing the appropriate key combination to give numerical approximations. Hence the answers are: a 0.7 b c.89 d 0.66 The techniques for solving equations, already covered in this chapter, are also valid for equations involving the natural base. Example 0 Solve for x, giving exact answers in terms of the natural logarithm function. a e x = 7 b e x 5e x + = 0 a e x = 7 e x 7 = -- 7 x = log e -- 7 x = --log e -- b e x 5e x + = 0 (e x ) 5e x + = 0 (e x )(e x ) = 0 e x = or e x = x = ln or x = ln x = 0 or x = ln A decimal approximation can also be obtained, if necessary. Thus, correct to decimal places, the answers are 0. in part a, and 0 or.86 in part b.

35 Algebraic techniques Example Solve for x giving exact answers. a log e x = -- b ln(x) = 6 a log e x = -- x = = e e b ln(x) = 6 ln (x) = x = e x = --e Tip Always give an exact answer unless a decimal approximation is required or necessary.. exercise. 5 Solve for x in each of the following giving i exact answers. ii answers correct to decimal places. a e x = 5 b e x = c e x = 8 d e x + = e e x = 5 f 0e 0.x + 5 = 65 g = 9 h e x 9e x 0 = 0 i e x = 6e x Solve for x in each of the following, giving exact answers. a ln x = b log e x = 9 c ln(x ) = 0 d ln(x ) = e ln(x ) = ln(x + ) f log e (5 x ) = log e (x x) g (ln x)(ln x ) = 0 h log e -- x = i ln(x + ) ln(x + ) = ln x 7 Let a and b be positive real numbers. a If a x + = b x, express x in terms of a and b using natural logarithms. p log b Hence show that the solution to x + = 0 x can be written in the form e q , log where p, q and r are positive integers with p >. e r 8 Let a and b be positive real numbers. a Prove that (log a b) (log b a) = b Prove that = log a b + log b a log a b log b a c e x Simplify: log e -- e log e ē - continued 5

36 MathsWorld Mathematical Methods CAS Units &. Algebra of circular functions The unit circle, symmetry properties and exact values The unit circle is a circle of radius unit with centre at the origin, O. Let P(x, y) be a point on the unit circle such that OP makes an angle θ with the positive direction of the x-axis. A tangent is drawn to this circle through the point A(, 0). When the line OP is produced it meets this tangent at Q. In general, for any point P on the unit circle, we define the functions cosine and sine by: cosine θ = cos θ = x and sine θ = sin θ = y where x is the x-coordinate of P and y is the y-coordinate of P. Using Pythagoras theorem on triangle OMP: OM + MP = OP cos θ + sin θ = This is known as an identity, since it is true for all values of θ. We define the tangent function by: tangent θ = tan θ = AQ, the y-coordinate of Q AQ y sin θ tan θ = = -- = , cos θ 0 x cos θ (, 0) The angle θ is positive if the angle of rotation from the positive direction of the x-axis is in an anticlockwise direction and negative if the angle of rotation from the positive direction of the x-axis is in a clockwise direction. The four quadrants in the Cartesian plane are numbered according to the system shown on the right. Recall the following properties of symmetry of the trigonometric functions. y (0, ) O (0, ) θ x Quadrant Quadrant y P y M Q A (, 0) x Quadrant Quadrant x Quadrant Quadrant Quadrant Radians sin (π θ ) = sin θ cos (π θ ) = cos θ tan (π θ ) = tan θ sin (π + θ ) = sin θ cos (π + θ ) = cos θ tan (π + θ ) = tan θ sin (π θ ) = sin θ cos (π θ ) = cos θ tan (π θ ) = tan θ Degrees sin (80 θ ) = sin θ cos (80 θ ) = cos θ tan (80 θ ) = tan θ sin (80 + θ ) = sin θ cos (80 + θ ) = cos θ tan (80 + θ ) = tan θ sin (60 θ ) = sin θ cos (60 θ ) = cos θ tan (60 θ ) = tan θ 6

37 Algebraic techniques For negative angles: sin ( θ) = sin θ cos ( θ) = cos θ tan ( θ) = tan θ The signs of the circular functions can be remembered using the following diagrams. SIN is positive y ALL functions are positive S y A. TAN is positive COS is positive x T C x This is sometimes known as the CAST rule. π The exact values of sin, cos and tan for --, π 6 --, π -- come from the triangles drawn below. π π 6 π Other exact values are obtained directly from the unit circle. π y π π π π π θ π π 6 (0, ) sin θ cos θ (, 0) O (, 0) 0, x tan θ undefined 0 undefined 0 Example (0, ) Find the exact value of: a sin 0 b cos π c tan 5π a sin 0 = sin (80 60 ) b cos π = c = 6 π -- π 6 tan 5π π tan = sin 60 = cos -- π 6 = tan π + -- π = = = tan -- π = 7

38 TI-Nspire.6 ClassPad.6 MathsWorld Mathematical Methods CAS Units & Tip A CAS can be used to find exact values of sin θ, cos θ and tan θ. With the mode set to Radian, you can still find the values for angles in degrees by using the symbol. However, it is important that you learn the exact values for the special angles, so that you can answer questions about these without the use of technology. Solving equations involving circular functions Due to the periodic nature of circular functions, a trigonometric equation will have an infinite number of solutions. However, in most situations, only particular solutions are of interest and therefore we usually solve these equations over a finite interval. In some instances, the equations we wish to solve have recognizable exact solutions and we can solve these by hand, i.e. without the use of technology. Other equations require the use of technology to find solutions. Example π If cos θ = 0.6 and π< θ< -----, find the exact values of a sin θ b tan θ a cos θ + sin sinθ θ = b tan θ = ( 0.6) + sin cosθ θ = sin θ = 0.8 = sin θ = sin θ = 0.8 (θ in rd quadrant) = -- Note that tan θ is positive as expected (θ in rd quadrant) Example Solve for x: cos x = , 0 x 60 Identify the quadrants in which the solutions lie. The value of cos x is positive so there is a solution in the first quadrant and a solution in the fourth quadrant (CAST rule). Find the reference angle in the first quadrant: cos 0 = So the solution in the first quadrant is x = 0. By symmetry, the solution in the fourth quadrant is x = 60 0 = 0. So x = 0 or x = 0. S T A C 8

39 TI-Nspire.5,.6, 0. ClassPad.5,.6, 0. Algebraic techniques Example Solve for x: sin x = 0., 0 x π for x. Give your answers correct to decimal places. Method : using symmetry properties Identify the quadrants in which the solutions lie. The value of sin x is positive so there is a solution in the first quadrant and a solution in the second quadrant. Find the reference angle in the first quadrant. With your CAS in Radian mode, the reference angle is So the solution in the first quadrant is x = By symmetry the solution in the second quadrant is x = π 0.57 =.859. Hence x = 0.57 or x =.859. Method : using the Solve command of a CAS Solve the equation as shown in the screenshot opposite. (Note that it is important to include the restriction on x.) Hence x = 0.57 or x =.859. S T A C. Warning Check the mode! When solving an equation where the variable is in degrees, make sure your CAS is set to Degree mode. When solving an equation where the variable is in radians, make sure your CAS is set to Radian mode. Example 5 Solve for θ : sin θ = cos θ, 0 θ π sin θ = cos θ sin θ = cos θ sin θ tan θ =, since = tan θ cos θ Identify the quadrants in which the solutions lie. The value of tan θ is negative, so there is a solution in the second quadrant and a solution in the fourth quadrant (CAST rule). Find the reference angle in the first quadrant: tan -- π =. By symmetry, the solution in the second quadrant is x π -- π π = = -----, and the solution in the fourth quadrant is x π -- π 5π = = π 5π Hence x = or x = S T A C 9

40 TI-Nspire 0. ClassPad 0. MathsWorld Mathematical Methods CAS Units & Tip The Solve command can be used to find the exact solutions to trigonometric equations where possible. For example, consider the equation in example. With the mode set to Degree, the solution is x = 0 or x = 0, agreeing with the solution to example. By changing the mode to Radian, the solution to the π 5π equation in example 5 is shown as x = or x = -----, agreeing with the solution to example 5. However, it is important that you are able to solve equations like these without the aid of technology. Example 6 Solve each of the following equations for x if 0 x π. a sin x sin x = 0 b tan x = 0 c sin x + cos x = 0 a sin x sin x = 0 Take out the common factor sin x. sin x ( sin x ) = 0 sin x = 0 or sin x = -- (null factor law) If sin x = 0 then x = 0 or x = π or x = π. If sin x = π 5π S A -- then x = -- or x = Hence the solutions are 0 π 5π, --, -----, π, π. T C 6 6 b tan x = 0 ( tan x )( tan x + ) = 0 (difference of two squares) tan x = or tan x = If tan x = then x = π π π 5π -- or x = If tan x = then x = or x = S A S A T C T C π Hence the solutions are --, π, π, π. continued 0

41 Algebraic techniques c sin x + cos x = 0 ( cos x) + cos x = 0 cos x + cos x = 0 cos x cos x + = 0 ( cos x )(cos x ) = 0 cos x = -- or cos x = π 5π If cos x = -- then x = -- or x = If cos x = then x = 0 or x = π. Hence the solutions are 0 π 5π, --, -----, π. S T A C Tip The identity sin x + cos x = can be useful when solving equations. If sin x is made the subject, then sin x = cos x and if cos x is made the subject, then cos x = sin x.. exercise. Find the exact value of: a sin 0 b cos 00 c tan 5 d sin ( 50 ) e cos ( 80 ) f tan ( 0 ) g sin π h cos π 6 i tan π j sin π k cos π l tan π 6 m sin 5π n cos π o tan π a If sinθ = and θ is in the second quadrant, find the exact values of i cosθ ii tanθ b If cosθ = k, < k < 0, and θ is in the third quadrant, express in terms of k: i sinθ ii tanθ iii cos θ sin θ Solve each of the following for x if 0 x 60. Give exact values for x. a cos x = b sin x = -- c tan x = d sin x = 0 e tan x = f sin x cos x = 0 Solve each of the following for x if 0 x 60, correct to the nearest tenth of a degree. a sin x = 0.6 b cos x = c tan x = 5 Solve each of the following for x if 0 x π. Give exact values for x. a tan x = b cos x = c sin x = d cos x = 0 e sin x + cos x = 0 f cos x = g cos x = h sin x = i tan x = 0 6 Solve each of the following for x if 0 x π, correct to decimal places. a cos x = 0.6 b sin x = 0.5 c tan x = 7 Solve each of the following for x if 0 x π. Give exact values for x. a sin x = 0 b tan x tan x = 0 c cos x = 0 d cos x + cos x + = 0 e tan x = tan x

42 MathsWorld Mathematical Methods CAS Units & Equations on other intervals In all of the previous examples, solutions were found on the interval [0, 60 ] or [0, π]. It is sometimes necessary to obtain solutions of equations on other intervals. Example 7 Find the exact values of x for which cos x = , 0 x π. Find the solutions between 0 and π first. Identify the quadrants in which the solutions lie. S A The value of cos x is negative so there is a solution in the second quadrant and a solution in the third quadrant. T C Find the reference angle in the first quadrant: cos -- π = By symmetry, the solution in the second quadrant is x π -- π π = = Similarly, the solution in the third quadrant is x π -- π 5π = + = The solutions between π and π can be found by going round the unit circle again, that is, by adding π to those already found. π 5π x = π x = π π π = = π Hence the solutions are -----, π, π, π. Equations of the form Af(bx) = d, where f is one of sine, cosine or tangent Example 8 If 0 x π, find the values of x for which sin x + = 0. sin x + = 0 sin x = If 0 x π, then 0 x π. So we need to solve sin x = for x, with 0 x π.

43 Algebraic techniques Find the solutions for x between 0 and π first. Identify the quadrants in which the solutions lie. The value of sin x is negative so there is a solution in the third quadrant and a solution in the fourth quadrant. Find the reference angle in the first quadrant: sin -- π = By symmetry, the solution in the third quadrant is x π -- π π = + = Similarly, the solution in the fourth quadrant is x π -- π 5π = = Additional solutions can be found by adding π to those already found. S T A C x = π π x = π π 0π π Warning Order! = = The addition of π must be carried The solutions for x are: out before the division by. π x = -----, π, π, π The solutions for x are found by dividing by : π x = -----, π, π, π 6 6. exercise. continued 8 Solve each of the following equations on the interval specified. Give exact answers where possible. Otherwise, give answers correct to decimal places. a cos θ = --, 0 θ π b sin x = , 0 x π c tan θ =, 0 θ π d sin x = , π x π e sin θ = 0., 0 θ π f tan x = --, π x π 9 Solve for x, giving exact values. a tan x = , 0 x π b sin x =, 0 x π c cos x = , 0 x π d sin x =, π x π e cos----- πx =, x f tan x = 0, 80 x 80 0 Solve for x: cos ( x 50) = for 0 x 60. Give exact values for x. Solve for x, 0 x π. Give exact values for x. π a cos x -- b c π = -- sin x + -- π = -- tan x -- = Solve for θ : sin θ -- π cos θ π = 00, θ π 6

44 MathsWorld Mathematical Methods CAS Units & General solutions Due to the periodic nature of circular functions, a trigonometric equation on the open interval R will have an infinite number of solutions. In this case, solving the equation gives a general solution. Example 9 Find the general solution to each of the following equations. a sin x = b cos x = c tan x = a First solve sinx = for x on [0, π]. Identify the quadrants in which the solutions lie. The value of sin x is positive so there is a solution in the first quadrant and a solution in the second quadrant. S A T C Find the reference angle in the first quadrant: sin -- π = π So the solution in the first quadrant is x = --. By symmetry the solution in the second quadrant is x π -- π π = = Additional solutions can be obtained by adding or subtracting integer multiples of π. x = π --, ± π, -- π ± π, -- π ± 6π, or x = π -----, π ± π, π ± π, π ± 6π, So the general solution is x = π π -- + nπ or x = nπ, where n is an integer, i.e. n Z. b First solve cos x = for x on [0, π]. Identify the quadrants in which the solutions lie. The value of cos x is positive so there is a solution in the first quadrant and a solution in the fourth quadrant. S A T C Find the reference angle in the first quadrant: cos -- π = π So the solution in the first quadrant is x = By symmetry, the solution in the fourth quadrant is x π -- π π = = Additional solutions can be obtained by adding or subtracting integer multiples of π. So the general solution is x = π π -- + nπ or x = nπ, where n Z. 6 6 Dividing by gives x = π nπ π nπ or x = , n Z. 8 8 continued

45 TI-Nspire. ClassPad. Algebraic techniques However, the general solution can be written in a simpler, equivalent way in this case. Rather than giving the solution in the fourth quadrant as x π -- π π = = , we can just π 6 6 as easily give the solution as x = --, since the negative angle is also in the fourth 6 quadrant. π π The general solution then becomes x = -- + nπ or x = -- + nπ, which can be more π 6 6 simply written as x = ± -- + nπ. 6 π nπ Dividing by then gives x = ± , n Z, and this is equivalent to the solution 8 above. c First solve tan x = for x on [0, π]. Identify the quadrants in which the solutions lie. The value of tan x is negative, so there is a solution in the second quadrant and a solution in the fourth quadrant. Note that these solutions are exactly π apart, so we can S A use a slightly different method of solution from that in parts a and b. T C Find the reference angle in the first quadrant: tan -- π = π 5π By symmetry, the solution in the second quadrant is x π -- = Additional solutions can be obtained by adding or subtracting integer multiples of π. This will automatically give the solution in the fourth quadrant and all other solutions to the equation over R. 5π So the general solution is x = nπ, n Z. 6 Tip When solving equations involving tan x, the solutions are always π apart. Recall that the graph of y = tan x has period π. A general solution can be obtained by initially finding one solution to the equation and then adding integer multiples of π to this. For equations involving tan ax, add integer multiples of -- π. a. Finding general solutions with a CAS Consider the equation in example 9 part a, sinx = , where x R. Using the Solve command of a CAS gives the general solution. The TI-Nspire (see screenshot) expresses the solution as x = nlπ π or x = nlπ +. The symbol n is used to -- π denote a general integer. 5

46 MathsWorld Mathematical Methods CAS Units & The ClassPad (see screenshot) expresses the solution as π π x = πconstn( ) + --, x = πconstn( ) The symbols constn() and constn() are used to denote general integers. Although the symbols and style are different, the solution in either case is equivalent to the one found by hand in example 9 part a, where we used the symbol n to denote a general integer. Example 0 Find the first 6 positive solutions to the equation sin x = π π The general solution is x = -- + nπ or x = nπ, n Z, found in example 8 part a and confirmed above by the CAS solution. The first 6 positive solutions can easily be obtained without technology by successively substituting the values 0, and for n and simplifying. Alternatively, if CAS has been used to find the general solution, the substitution can be done directly or by copying, pasting and editing as shown in the screenshots. π Hence the first 6 positive solutions are π π π 7π,,, , , π. Example Find the first positive solutions to the equation cos x = 0.. Using the Solve command of a CAS gives the general solution x = ± nπ, n Z.In this case, substituting 0 and for n gives one negative and three positive solutions. To get the first positive solutions, substitute and for n in the second case (refer to the screenshot opposite). Hence the first positive solutions are.8755,.077, 8.589, , correct to decimal places. 6