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1 Higher Mathematics UNIT Mathematics HSN000 This document was produced speciall for the HSN.uk.net website, and we require that an copies or derivative works attribute the work to Higher Still Notes. For more details about the copright on these notes, please see

2 Higher Mathematics Unit Mathematics Contents Polnomials and Quadratics Quadratics The Discriminant Completing the Square Sketching Parabolas 7 Determining the Equation of a Parabola 9 6 Solving Quadratic Inequalities 60 7 Intersections of Lines and Parabolas 6 8 Polnomials 6 9 Snthetic Division 6 0 Finding Unknown Coefficients 68 Determining the Equation of a Graph 70 Iteration 7 Integration 7 Indefinite Integrals 7 Preparing to Integrate 76 Differential Equations 77 Definite Integrals 78 Geometric Interpretation of Integration 80 6 Areas between Curves 8 7 Integrating along the -ais 89 Trigonometr 90 Solving Trigonometric Equations 90 Trigonometr in Three Dimensions 9 Compound Angles 96 Double-Angle Formulae 98 Further Trigonometric Equations 99 Circles 0 Representing a Circle 0 Testing a Point 0 The General Equation of a Circle 0 Intersection of a Line and a Circle 0 Tangents to Circles 0 6 Equations of Tangents to Circles 06 7 Intersection of Circles 07 - ii - HSN000

3 Higher Mathematics Unit Mathematics OUTCOME Polnomials and Quadratics Quadratics A quadratic has the form provided a 0. You should alread be familiar with the following. a + b + c where a, b, and c are an numbers, The graph of a quadratic is called a parabola. There are two possible shapes: concave up (if a > 0 ) concave down (if a < 0 ) This has a minimum turning point This has a maimum turning point To find the roots of the quadratic factorisation a b c + + = 0, we can use: b ± b ac the quadratic formula: = (this is not given in the eam) a EXAMPLES. Find the roots of = 0 = 0. ( + )( ) = 0 + = 0 or = 0 = =. Solve = = 0 ( )( ) + + = 0 + = 0 or + = 0 = = Page HSN000

4 Higher Mathematics Unit Mathematics. Find the roots of + = 0. We cannot factorise +, so we use the quadratic formula: ± ( ) = ± 6 + = ± 0 = = ± = ± Note If there are two distinct solutions, the curve intersects the -ais twice. If there is one repeated solution, the turning point lies on the -ais. If b ac < 0 when using the quadratic formula, there are no points where the curve intersects the -ais. Page HSN000

5 Higher Mathematics Unit Mathematics The Discriminant Given a + b + c, we call b ac the discriminant. It is the part of the quadratic formula which determines how man real roots a + b + c = 0 has. If b If b If b EXAMPLE ac > 0, the roots are real and unequal (distinct). ac = 0, the roots are real and equal (i.e. a repeated root). ac < 0, the roots are not real; the do not eist.. Find the nature of the roots of a = 9 b = c = 6 Since b Using the Discriminant b = 0. ac = 9 6 = = 0 ac = 0, the roots are real and equal. two roots one root no real roots The discriminant has man uses, usuall involving an unknown term in a quadratic. EXAMPLES. Find the values of q such that Since a = 6 b = c = q 6 q 0 6 q = has real roots, b b ac 0 6 q 0 q 0 q q q = has real roots. ac 0 Page HSN000

6 Higher Mathematics Unit Mathematics. Show that ( k ) ( k ) ( k ) a = k + b = k + c = k = 0 alwas has real roots. b ac ( k ) ( k )( k ) 9k k ( k )( k 8) = + + = = k k 8k = k + k + ( k 6) = + Since b ac ( k ) 6 = + 6 0, the roots will alwas be real. Completing the Square Completing the square involves epressing = a( + p) + q. = a + b + c in the form In this form, we can determine the turning point of an parabola, including those with no real roots. The ais of smmetr is = p and the turning point is ( p, q). The following process can be used to complete the square:. Make sure the equation is in the form = a + b + c.. Separate the constant term ( c ) b bracketing everthing else on the RHS. Take out an constant common factor from the brackets.. Add on half of the coefficient squared and subtract half of the coefficient squared (note that adding on a number and taking it awa keeps the equation balanced).. Factorise the first three terms in the bracket. 6. State our answer in the form = a( + p) + q. EXAMPLES. Write = + 6 in the form = ( + p) + q. ( 6 ) ( ) = + = = ( + ) 9 = ( + ) (Step s -) (Step ) (Step ) (Step 6) Note You can alwas check our answer b epanding the brackets Page HSN000

7 Higher Mathematics Unit Mathematics. Write = + in the form = ( + p) + q. ( ) = + ( ( ) ( ) ) ( 9 ) ( ) = + + = + = +. Write = + 8 in the form = ( + a) + b and then state: (i) the ais of smmetr, and (ii) the minimum turning point of the parabola with this equation. ( 8 ) ( ) = + = ( ) ( ) = + 6 = + 9 (i) The ais of smmetr is =. (ii) The minimum turning point is (, 9).. A parabola has equation = + 7. Epress the equation in the form = ( + a) + b, and state the turning point of the parabola. ( ) ( ) = + 7 = + 7 ( ( ) ( ) ) = ( ( 9 ) ) + 7 = ( ) = ( ) The turning point is ( ) = + + 7, Page 6 HSN000

8 Higher Mathematics Unit Mathematics Sketching Parabolas The method used to sketch the curve with equation depends on how man times the curve intersects the -ais. = a + b + c We have met curve sketching before. However, when sketching parabolas, we do not need to use calculus. We know there is onl one turning point, and we have methods for finding it. Parabolas with one or two roots Find the -ais intercepts b factorising or using the quadratic formula. Find the -ais intercept (i.e. where = 0). The turning point is on the ais of smmetr: O The ais of smmetr is halfwa between two distinct roots O A repeated root lies on the ais of smmetr Parabolas with no real roots There are no -ais intercepts. Find the -ais intercept (i.e. where = 0). Find the turning point b completing the square. EXAMPLES. Sketch the graph of = Since b ac = ( 8 ) 7 > 0, the parabola crosses the -ais twice. The -ais intercept ( = 0) : = ( 0) 8( 0) + 7 = 7 ( 0, 7 ) The -ais intercepts ( = 0) : = 0 ( )( 7) = 0 = 0 or 7 = 0 = = 7 (, 0 ) ( 7, 0 ) The ais of smmetr lies halfwa between = and = 7, i.e. =, so the -coordinate of the turning point is. Page 7 HSN000

9 Higher Mathematics Unit Mathematics We can now find the -coordinate: ( ) ( ) = = = 9 So the turning point is (, 9).. Sketch the parabola with equation = ( 0) 6( 0) 9 = 9 ( 0, 9) = 6 9 Since b ac = ( 6) ( ) ( 9) = 0, there is a repeated root. The -ais intercept ( = 0) : The -ais intercept ( = 0) : Since there is a repeated root, (, 0) is the turning point.. Sketch the curve with equation = ( ) = 0 ( + )( + ) = 0 + = 0 = = 8 +. (, 0) Since b ac = ( 8) < 0, there are no real roots. The -ais intercept ( = 0) : = ( 0) 8( 0) + = ( 0, ) So the turning point is (, ). Complete the square: = O 9 = 6 9 ( ) = + = ( ) 8 + O = ( ) + = + = (, 9) 8 O (, ) Page 8 HSN000

10 Higher Mathematics Unit Mathematics Determining the Equation of a Parabola In order to determine the equation of a parabola, we need two pieces of information: The -ais intercept(s) or turning point Another point on the parabola EXAMPLES. A parabola passes through the points (, 0 ), (, 0 ) and ( 0, ). Find the equation of the parabola. Since the parabola cuts the -ais where = and =, the equation is of the form: = k ( )( ) To find k, we use the point ( 0, ): = k ( )( ) = k ( 0 )( 0 ) = k k = So the equation of the parabola is: = ( )( ) = ( + ) = 8 +. Find the equation of the parabola shown below. (, 6) Since there is a repeated root, the equation is of the form: = k ( + ) 6 Hence = ( + ). O To find k, we use (, 6 ) : = k ( + ) 6 = k ( + ) 6 6 k = 6 = Page 9 HSN000

11 Higher Mathematics Unit Mathematics. Find the equation of the parabola shown below. O (, ) 7 Since the turning point is (, ), the equation is of the form: = a( ) = 6. Hence ( ) To find a, we use ( 0, 7) : a = 6 ( ) = a ( ) 7 = a 0 6a = 6 Solving Quadratic Inequalities The most efficient wa of solving a quadratic inequalit is b making a rough sketch of the parabola. To do this we need to know: the shape concave up or concave down the -ais intercepts. We can then solve the quadratic inequalit b inspection of the sketch. EXAMPLES. Solve + < 0. The parabola with equation The -ais intercepts are given b: + = 0 ( + )( ) = 0 + = 0 = or = 0 = = + is concave up. Make a sketch: = + So + < 0 for < <. Page 60 HSN000

12 Higher Mathematics Unit Mathematics. Find the values of for which The parabola with equation The -ais intercepts are given b: = 0 ( ) 7 6 = 0 ( + )( ) = 0 + = 0 or = 0 = = = is concave down. Make a sketch: = So for.. Solve 0 >. The parabola with equation The -ais intercepts are given b: = 0 ( )( ) = 0 = 0 or = 0 = = = is concave up. Make a sketch: = So 0 > for < and >.. Find the values of q for which ( ) Since ( ) a = b = q c = + q + q = 0 has non-real roots, b q + q + q = 0 has non-real roots. ( ) ( )( ) = b ac q q ( )( ) q 8q 6 q = + = q q q = + q 0q 6 ac < 0: We now need to solve the inequalit q 0q + 6 < 0 : The parabola with equation q q = is concave up. Page 6 HSN000

13 Higher Mathematics Unit Mathematics The -ais intercepts are given b: q 0q + 6 = 0 ( q )( q ) 8 = 0 q = 0 or q 8 = 0 Therefore b q = q = 8 ac < 0 for q 8 non-real roots when < q < 8. 7 Intersections of Lines and Parabolas Make a sketch: = q 0q < <, and so ( ) + q + q = 0 has To determine how man times a line intersects a parabola, we substitute the equation of the line into the equation of the parabola. We can then use the discriminant or factorisation to find the number of intersections. If b ac > 0, the line and curve intersect twice If b ac = 0, the line and curve intersect once (i.e. the line is a tangent to the curve) If b ac < 0, the line and the parabola do not intersect EXAMPLES. Show that the line = is a tangent to the parabola and find the point of contact. Substitute = into: + = 0 ( )( ) = 0 = + = + + = 0 Since there is a repeated root, the line is a tangent at =. = + To find the -coordinate, substitute = into the equation of the line: = ( ) = So the point of contact is (, ). Page 6 HSN000

14 Higher Mathematics Unit Mathematics. Find the equation of the tangent to = + that has gradient. The equation of the tangent is of the form = m + c, with m =, i.e. = + c. Substitute this into: c = + + = + c + = 0 Since the line is a tangent: b ac = 0 ( ) ( c ) = c = 0 c = c = Therefore the equation of the tangent is: = = 0 8 Polnomials Polnomials are epressions with one or more terms added together. Each term has a number (called the coefficient) followed b a variable (such as ) raised to a whole number power. For eample: or The degree of the polnomial is the value of its highest power, for eample: has degree has degree 8 Note that quadratics are polnomials of degree two. Also, constants are 0 polnomials of degree zero (e.g. 6 is a polnomial, since 6 = 6 ). Page 6 HSN000

15 Higher Mathematics Unit Mathematics 9 Snthetic Division Snthetic division provides a quick wa of evaluating functions defined b polnomials. For eample, consider f ( ) = Using a calculator, we find f ( 6) =. We can also evaluate this using snthetic division with detached coefficients. Step Detach the coefficients, and write them across the top row of the table. Note that the must be in order of decreasing degree. If there is no term of a specific degree, then zero is its coefficient. Step Write the number for which ou want to evaluate the polnomial (the input number) Step Bring down the first coefficient. 6 9 Step Multipl this b the input number, writing the result underneath the net coefficient. Step Add the numbers in this column. Repeat Steps and until the last column has been completed. The number in the lower-right cell is the value of the polnomial for the input value, often referred to as the remainder Page 6 HSN000

16 Higher Mathematics Unit Mathematics EXAMPLE. Given f ( ) = + 0, evaluate f ( ) using snthetic division. 0 So f ( ) = 0. using the above process Note In this eample, the remainder is zero, so f ( ) = 0. This means + 0 = 0 when =, which means that = is a root of the equation. So + is a factor of the cubic. We can use this to help with factorisation: ( ) f ( ) = ( + ) q ( ) where q ( ) is a quadratic Is it possible to find the quadratic q ( ) using the table? Tring the numbers from the bottom row as coefficients, we find: ( + )( 0) = = = f ( ) 0 So using the numbers from the bottom row as coefficients does give the correct quadratic EXAMPLES. Show that is a factor of is a factor = is a root Since the remainder is zero, = is a root, so is a factor. Page 6 HSN000

17 Higher Mathematics Unit Mathematics. Given f ( ) = (i) show that = 7 is a root of f ( ) = 0, and (ii) hence full factorise f ( ) Since the remainder is zero, = 7 is a root. Hence we have f ( ) = Show that = is a root of factorise the cubic ( 7)( 7 ) = + + ( 7)( )( ) = + Using snthetic division to factorise =, and hence full Since = is a root, + is a factor. ( )( ) = This does not factorise an further since b ac < 0. In the eamples above, we have been given a root or factor to help factorise polnomials. However, we can still use snthetic division if we do not know a factor or root. The roots of the polnomial f ( ) are numbers which divide the constant term eactl. So b tring snthetic division with these numbers, we will (eventuall) find a root.. Full factorise 8 +. Numbers which divide : ±, ±, ±, ±. Tr = : ( ) + ( ) 8( ) = Tr = : ( ) + ( ) 8( ) = Note For ±, it is simpler just to evaluate the polnomial directl, to see if this is a root Page 66 HSN000

18 Higher Mathematics Unit Mathematics Tr = : Since = is a root, is a factor. So + 8 = ( )( + + ) = ( )( + )( + ) Using snthetic division to solve equations We can also use snthetic division to help solve equations. EXAMPLE 6. Find the roots of =. Numbers which divide : ±, ±, ±, ±, ± 6, ±. Tr = : ( ) ( ) + 6( ) + = Tr = : ( ) ( ) + 6( ) + Tr = : In general = For a polnomial f ( ): If f ( ) is divided b h Since = is a root, is a factor: = 0 ( )( ) 6 = 0 ( )( + )( 6) = 0 = 0 = or + = 0 = then the remainder is f ( ) h, and f ( h) = 0 h is a factor of f ( ). or 6 = 0 = 6 As we saw, snthetic division helps us to write f ( ) in the form + where q ( ) is called the quotient. ( h) q ( ) f ( h) Page 67 HSN000

19 Higher Mathematics Unit Mathematics EXAMPLE 7. Find the quotient and remainder when f ( ) = + is divided b +, and epress ( ) The quotient is ( ) ( )( ) f = Finding Unknown Coefficients f as ( ) q( ) f ( h) and the remainder is, so Consider a polnomial with some unknown coefficients, such as + p p +, where p is a constant. If we divide the polnomial b h, then we will obtain an epression for the remainder in terms of the unknown constants. If we alread know the value of the remainder, we can solve for the unknown constants. EXAMPLES. Given that is a factor of is a factor = is a root. p p 6 + p + p p Since is a factor, the remainder is zero: + p = 0 p = p = + +, find the value of p. Page 68 HSN000

20 Higher Mathematics Unit Mathematics. When f ( ) = p + q 7 + q is divided b, the remainder is 6, and is a factor of f ( ). Find the values of p and q. When f ( ) is divided b, the remainder is f ( ) : p q 7 q p p + q 8 p + q p p + q p + q 7 8 p + 8q Since f ( ) = 6, we have: 8 p + 8q = 6 8p + 8q = 0 p + q = () Since is a factor, f ( ) = 0 : f ( ) = p( ) + q ( ) 7( ) + q = p + q 7 + q = p + q 7 i.e. p + q 7 = 0 () Note There is no need to use snthetic division here Solving () and () simultaneousl, we obtain: p + q = 7 () p q = () q = q = Giving p = q = = from (). Hence p = and q =. Page 69 HSN000

21 Higher Mathematics Unit Mathematics Determining the Equation of a Graph Given the roots, and at least one other point ling on the graph, we can establish the graph s equation. EXAMPLE. Find the equation of the cubic shown in the diagram below. 6 O 6 Step Write out the roots, then rearrange to get the factors. Step The equation is then these factors multiplied together with a constant, k. Step Substitute the coordinates of a known point into this equation to find the value of k. Step Replace k with this value in the equation. = 6 = = + 6 = 0 + = 0 = 0 = k ( + 6)( + )( ) Using ( 0, 6) : k ( 0 + 6)( 0 + )( 0 ) = 6 k ( )( )( 6) = 6 8k = 6 = ( + 6)( + )( ) k = ( )( ) ( ) = = = Page 70 HSN000

22 Higher Mathematics Unit Mathematics Repeated Roots If a repeated root eists, then a stationar point lies on the -ais. Recall that a repeated root eists when two roots, and hence two factors, are equal. EXAMPLE. Find the equation of the cubic shown in the diagram below. 9 O = = = + = 0 = 0 = 0 So = k ( + )( ). Use ( 0, 9 ) to find k : 9 = k ( 0 + )( 0 ) 9 = k 9 k = = + + = = + 9 So = ( + )( ) ( )( 6 9) ( 6 9 8) Page 7 HSN000

23 Higher Mathematics Unit Mathematics Iteration When a root of a polnomial is not rational, an estimate of the root can be determined using an iterative process. EXAMPLE If f ( ) = + 7, show that there is a real root between = and =. Find this root correct to two decimal places. Evaluate f ( ) at = and = : f ( ) = ( ) ( ) + 7 = f ( ) = ( ) ( ) + 7 = Since f ( ) > 0 and f ( ) < 0, f ( ) has a root between = and =. Tr halfwa between = and = : f (. ) =. (. ) (. ) + 7 =. 6 < 0 f ( ) > 0 and f (. ) < 0 so the root is between = and =.. Tr roughl halfwa between = and =. : f (. ) =. (. ) (. ) + 7 = 0. 6 < 0 f (. ) =. (. ) (. ) + 7 = > 0 f (. ) > 0 and f (. ) < 0 so the root is between =. and =.. Tr halfwa between =. and =. : f (. ) =. (. ) (. ) + 7 = 0. 0 > 0 f (. ) > 0 and f (. ) < 0 so the root is between =. and =.. Tr roughl halfwa between =. and =. : f (. 7) =. 7 (. 7) (. 7) + 7 = > 0 f (. 8) =. 8 (. 8) (. 8) + 7 = < 0 f (. 7) > 0 and f (. 8) < 0 so the root is between =. 7 and =. 8. Tr halfwa between =. 7 and =. 8: f (.7) =. 7 (. 7) (. 7) + 7 = f (. 7) > 0 and f (. 8) < 0 so the root is between =. 7 and =. 8. Therefore the root is =. 8 to d.p. Page 7 HSN000

24 Higher Mathematics Unit Mathematics OUTCOME Integration Indefinite Integrals Integration is another branch of calculus. It is the reverse of differentiation. The basic technique for integrating is: n+ n d = + c n, c is the constant of integration n + Stated simpl: raise the power (n) b one ( n + ), divide b the new power ( n + ), and add the constant of integration ( c ). EXAMPLES d.. Find. Find. Find d = + c = + c d. d = + c = + c d. 9 9 d = 9 + c = 9 + c The smbol we use for integration is The must be used with d in the eamples above, to indicate that we are integrating with respect to The constant of integration is included to represent an constant term in the original epression, since this would have been zeroed b differentiation Integrals with a constant of integration are called indefinite integrals Page 7 HSN000

25 Higher Mathematics Unit Mathematics Checking the answer Since integration and differentiation are reverse processes, if we differentiate our answer we should get back to what we started with. For eample, if we differentiate our answer to Eample above, we do get back to the epression we started with: d d Integrating terms with coefficients ( ) + c = The above technique can be etended to: n+ n n a a d = a d = + c n, a is a constant n + Stated simpl: raise the power (n) b one ( n + ), divide b the new power ( n + ), and add on c. EXAMPLES 6 d.. Find 6 6 d = + c = + c. Find d. d = + c = 8 + c 8 = + c Note It can be eas to confuse integration and differentiation, so remember: d = + c k d = k + c Page 7 HSN000

26 Higher Mathematics Unit Mathematics Other variables Just as with differentiation, we can integrate with respect to an variable. EXAMPLES p dp. 6. Find p = + c p p dp = + c 7. Find p d. p d = p + c Integrating several terms The following rule is used to integrate an epression with several terms: f ( ) + g ( ) d = f ( ) d + g ( ) d Stated simpl: integrate each term separatel. EXAMPLES d. 8. Find 9. Find d = + c 8 = + c = + c d d = c = c 8 8 c = Note dp tells us to integrate with respect to p Note Since we are integrating with respect to, we treat p as a constant Page 7 HSN000

27 Higher Mathematics Unit Mathematics Preparing to Integrate As with differentiation, it is important that before integrating, all brackets are multiplied out, and there are no fractions with an term in the denominator (bottom line), for eample: = EXAMPLES = = d. Find for 0. d can be written as d, so:. Find d = d d = = + c = + c d d for 0. = d = d = + c = + c = = 7. Find p dp where p 0. 7 p 7 dp = p dp 7 p = + c 7 = + c p Page 76 HSN000

28 Higher Mathematics Unit Mathematics. Find d. d = d 6 = + c 6 6 = 8 + c 6 = c Differential Equations As we have previousl stated, integration is the reverse of differentiation. So b integrating a function which has been differentiated, we obtain the original function. For eample: f ( ) d = f ( ) + c or d d = + c d If we have additional information about the function (such as a point its graph passes through), we can find the value of the constant of integration ( c ) and obtain a particular solution. EXAMPLES. The graph of = f ( ) passes through the point (, ). d If =, epress in terms of. d d = d d = d = + c We know that when =, = so we can find c: So = + c = ( ) ( ) + c = 9 + c c = = +. Page 77 HSN000

29 Higher Mathematics Unit Mathematics. The function f is defined on a suitable domain such that f ( ) = + +. Given that f ( ) =, find a f ( ) in terms of. f ( ) = f ( ) d = + + = + + d d = + + c = + + c We know that f ( ) =, so we can find c: ( ) f = + + c = + + c c = f = + +. So ( ) Definite Integrals = ( ) + ( ) + c If F( ) is the integral of f ( ), then we define: b a ( ) = ( ) ( ) f d F b F a where a and b are called the limits of the integral. Stated simpl: Work out the integral as normal, leaving out the constant of integration Calculate the integral for the value Calculate the integral for the value = b (the upper limit value) = a (the lower limit value) Subtract the lower limit value from the upper limit value Since there is no constant of integration and we are calculating a numerical value, this is called a definite integral. Page 78 HSN000

30 Higher Mathematics Unit Mathematics EXAMPLES d.. Find. Find d = ( ) ( ) = = = = + d. 0 d 0 + = +. Find d. = = = = + 8 = d = d = = = ( ) = = 8 0 Page 79 HSN000

31 Higher Mathematics Unit Mathematics Geometric Interpretation of Integration We will now consider the meaning of integration in the contet of areas. Consider 0 d = = = 0 Looking at the graph of = : = The shaded area is given b 0 d Therefore the shaded area is square units In general, the area enclosed b the graph = f ( ) and the -ais, between = a and = b, is given b: EXAMPLE O b a f ( ) d. The graph of = is shown below. Calculate the shaded area. = O d = = ( ) ( ) = = 6 8 = So the shaded area is square units. Page 80 HSN000

32 Higher Mathematics Unit Mathematics Areas below the -ais Care needs to be taken if part or all of the area lies below the -ais. For eample if we look at the graph of = : The shaded area is given b d = O = ( ) = 6 + = 6 0 = 0 = 9 In this case, the negative indicates that the area is below the -ais, as can be seen from the diagram. The area is therefore 9 square units. Areas above and below the -ais Consider the graph from the eample above, with a different shaded area: O = Area From the eamples above, the total shaded area is: Area + Area = + 9 = square units Using the method from above, we might tr to calculate the shaded area as follows: d = = ( ) = 0 + = 8 = 6 Clearl this shaded area is not 6 square units since we alread found it to be square units. This problem arises because Area is above the -ais, while Area is below. = Area To find the true area, we needed to evaluate two integrals: d and d We then found the total shaded area b adding the two areas together. Page 8 HSN000

33 Higher Mathematics Unit Mathematics EXAMPLES. Calculate the shaded area shown in the diagram below. O = To calculate the area from = to = : d = = = ( ) ( ) ( ) ( ) ( ) ( ) = [ ] = + 9 = 0 So the area is 0 square units. We have alread integrated the equation of the curve, so we can just substitute in new limits to work out the area from = to = : d = = ( ) ( ) ( ) ( ) ( ) ( ) = 6 8 = 8 + = So the area is square units. So the total shaded area is 0 + = square units. Page 8 HSN000

34 Higher Mathematics Unit Mathematics. Calculate the shaded area shown in the diagram below. = + O First, we need to calculate the root between = and = : + = 0 ( )( + 6) = 0 = or = 6 So the root is = To calculate the area from = to = : + d = + = + ( ) ( ) ( ) ( ) ( ) ( ) = + + = = 6 6 = 7 So the area is 7 square units. To calculate the area from = to = : + d = + = + + = = 6 = So the area is square units. ( ) ( ) ( ) ( ) ( ) ( ) So the total shaded area is 7 + = square units. Page 8 HSN000

35 Higher Mathematics Unit Mathematics 6 Areas between Curves The area between two curves between b = a and = b is calculated as: upper curve lower curve d square units a So for the shaded area shown below: = f ( ) a O = g ( ) b The area is b a f ( ) g ( ) d square units When dealing with areas between curves, areas above and below the -ais do not need to be calculated separatel. However, care must be taken with more complicated curves, as these ma give rise to more than one closed area. These areas must be evaluated separatel. For eample: = g ( ) O a b c = f ( ) b In this case we appl upper curve lower curve d to each area. So the shaded area is given b: b a a g ( ) f ( ) d + f ( ) g ( ) d c b Page 8 HSN000

36 Higher Mathematics Unit Mathematics EXAMPLES. Calculate the shaded area enclosed b the curves with equations = 6 and =. = 6 O = To work out the points of intersection, equate the curves: 6 = = 0 9 = 0 ( + )( ) = 0 = or = Set up the integral and simplif: upper curve lower curve d ( 6 ) ( ) = d = = 9 d d Carr out integration: 9 d = 9 ( ) ( ) = 9( ) 9( ) = = = 6 Therefore the shaded area is 6 square units. Page 8 HSN000

37 Higher Mathematics Unit Mathematics. Two functions are defined for R b f ( ) = and g ( ) = +. The graphs of = f ( ) and = g ( ) are shown below. = g ( ) O 6 Calculate the shaded area. Since the shaded area is in two parts, we appl upper lower d twice. Area from = to = : upper lower d ( ) = d = d 7 = ( ) 7( ) = ( ) + ( ) = = = 9 So the area is 9 square units. b a Note The curve is at the top of this area Page 86 HSN000

38 Higher Mathematics Unit Mathematics Area from = to = 6 : 6 upper lower d 6 ( ) = d 6 = + 7 d 7 = = = [ ] = 96 = 6 So the area is 6 square units. Note The straight line is at the top of this area So the total shaded area is = 9 square units. Page 87 HSN000

39 Higher Mathematics Unit Mathematics. A trough is metres long. A cross-section of the trough is shown below. = + = The cross-section is part of the parabola with equation = +. Find the volume of the trough. To work out the points of intersection, equate the curve and the line: + = + = 0 ( )( ) = 0 so = or = Set up the integral and integrate: O ( ) upper lower d = + d = + d = + ( ) ( ) = + ( ) ( ) + ( ) ( ) = [ ] + = = = Therefore the shaded area is square units. Volume = cross-sectional area length = = 8 = Therefore the volume of the trough is cubic units. Page 88 HSN000

40 Higher Mathematics Unit Mathematics 7 Integrating along the -ais For some problems, it ma be easier to find a shaded area b integrating with respect to rather than. EXAMPLE The curve with equation 9 = 9 = is shown in the diagram below. =6 = O Calculate the shaded area, which lies between = and = 6. We have = 9 9 = = ± 9 = ± = The shaded area in the diagram to the right is given b: 6 6 d = d 6 = = 6 = 6 = 6 8 = = Since this is half of the required area, the total shaded area is square units. O =6 Page 89 HSN000

41 Higher Mathematics Unit Mathematics OUTCOME Trigonometr Solving Trigonometric Equations You should alread be familiar with solving some trigonometric equations. EXAMPLES. Solve sin = for 0 < < 60. sin = 80 = 0 or 80 0 = 0 or 0 S A T C First quadrant solution: = sin. Solve cos = for 0 < < 60. cos = = 0 S A T C ( ) = 80 6 or = 6 6 or. Solve sin = for 0 < < ( ) = cos = 6 (to d.p.) There are no solutions since sin. Since sin is positive Since cos is negative Note that cos, so cos = also has no solutions Page 90 HSN000

42 Higher Mathematics Unit Mathematics. Solve tan = for 0 < < 60. Note tan = 80 S A T C = tan ( ) = (to d.p.) = or = 0 0 or 8 0 Since tan is negative All trigonometric equations in Higher can be reduced to problems like those above. The onl differences are: the solutions could be required in radians in this case, the question will not have a degree smbol, e.g. Solve tan = rather than tan = eact value solutions could be required in the non-calculator paper Questions can be worked through in degrees or radians, but make sure the final answer is given in the units asked for in the question. EXAMPLES. Solve sin = 0 where sin = 80 S A 0 60 sin = T C 0 70 = 0 or = sin = 0 or or or = 0 or 0 or 90 or 0 = or 7 or 9 or ( ) Note There are more solutions ever 60, since sin(0 ) = sin(0 +60 ) = So keep adding 60 until > 70 Page 9 HSN000

43 Higher Mathematics Unit Mathematics 6. Solve cos = where 0 π. cos = π S A T C = π or π π or π + π = π or 7π = π 7 8 or π 8 7. Solve cos = where 0 < < π. ( cos ) = cos = ± cos = ± π + π = cos = π ( ) 0 π 0 π S A Since cos can be positive or negative T C = cos = π 6 ( ) = π 6 or π π 6 or π + π 6 or π π 6 or π + π 6 = π 7 6 or π 6 or π 6 or π 6 8. Solve tan( 0 ) = where tan( 0 ) = tan( 0 ) = S A T C ( ) 0 = tan = 9 06 (to d.p.) 0 = 9 06 or or or or or or π π 6 π π Page 9 HSN000

44 Higher Mathematics Unit Mathematics 0 = 9 06 or 9 06 or 9 06 or or or = or 9 06 or 9 06 or or or = 6 or 86 or 6 or 06 or 66 or 6 9. Solve ( π ) ( π ) cos + = 0 8 for 0 < < π. cos + = 0 8 S A 0 < < π T C 0 < < π π < + π < π + π 07 < + π < 6 (to d.p.) π + = cos ( 0 8 ) Remember Make sure our = 0 6 calculator uses radians + π = 0 6 or π 0 6 or π or π + π 0 6 or π + π or π + π + π π.... = 660 or or 9 or 89 =. 6 or. 89 or or. =. 07 or. 90 or. 8 or Trigonometr in Three Dimensions Solving problems in three dimensions uses the same techniques as in two dimensions. The use of sketches in solutions to these problems is advised. The angle between a line and a plane The angle a between the plane P and the line ST is calculated b drawing a line perpendicular to the plane and then using basic trigonometr. T P a S Page 9 HSN000

45 Higher Mathematics Unit Mathematics EXAMPLE. The triangular prism ABCDEF is shown below. E A 0 cm Calculate the acute angle between: (a) The line AF and the plane ABCD. (b) AE and ABCD. (a) Start with a sketch: Opposite tana = Adjacent = 0 ( ) A a = tan 0 cm D 0 = (or 0. 9 radians) (to d. p.) Note Since the angle is in a right-angled triangle, it must be acute (i.e. lie in the first quadrant) and therefore we do not use a CAST diagram. (b) Again, make a sketch: b angle to be A calculated We need to calculate the length of AC first using Pthagoras s Theorem: C AC = cm A = 6 0 cm D Therefore: E Opposite tanb = cm Adjacent = C 6 A a b F 6 cm B cm E cm C F D C 6 cm cm ( ) b = tan 6 =. 6 (or 0. radians) (to d. p.) Page 9 HSN000

46 Higher Mathematics Unit Mathematics The angle between two planes The angle a between planes P and Q is calculated b drawing a line perpendicular to Q and then using basic trigonometr. P Q a EXAMPLE. ABCDEFGH is a cuboid with dimensions 8 8 cm as shown below. H G E F 8 cm 8 cm A cm B (a) Calculate the size of the angle between the planes AFGD and ABCD. (b) Calculate the size of the acute angle between the diagonal planes AFGD and BCHE. (a) Start with a sketch: F A a cm 8 cm B D Opposite tana = Adjacent = 8 ( ) C Note Angle GDC is the same size as angle FAB a = tan =. 690 (or radians) (to d. p.) (b) Again, make a sketch: E F T A B Let AF and BE intersect at T ATB is isosceles, so TAB = TBA ɵ =. 690 ATB = 80 ( ) =. 60 Note The angle could also have been calculated using rectangle DCGH So the acute angle is: BTF = ATE = = (or. 76 radians) (to d. p.) Page 9 HSN000

47 Higher Mathematics Unit Mathematics Compound Angles When we add or subtract angles, we call this a compound angle. For eample, + 0 is a compound angle. Using a calculator, we find: sin( + 0 ) = sin( 7 ) = sin( ) + sin( 0 ) =. 07 (both to d.p.). This shows that sin ( A + B ) is not equal to sin A + sin B. Instead, we can use the following identities: sin( A + B ) = sin Acos B + cos AsinB sin( A B ) = sin Acos B cos Asin B These are given in the eam in a condensed format: sin( A ± B ) = sin Acos B ± cos AsinB EXAMPLES cos( A ± B ) = cos Acos B sin Asin B. Epand and simplif cos( + 60 ). cos( + 60 ) = cos cos 60 sin sin 60. Show that ( ) cos( A + B ) = cos AcosB sin Asin B cos( A B ) = cos AcosB + sin Asin B = cos sin sin a + b = sina cosb + cos a sinb for a = π 6 and b = π. LHS = sin( a + b) RHS = sin a cosb + cosa sinb = sin π π ( 6 + = sinπ ) 6 cosπ + cosπ 6 sinπ = sin π ( ) = ( ) + ( ) = = + = Since LHS = RHS, the claim is true for a = π 6 and b = π.. Find the eact value of sin7. ( ) sin 7 = sin + 0 = sin cos0 + cos sin0 ( ) ( ) = + = + = + = 6+ Page 96 HSN000

48 Higher Mathematics Unit Mathematics Finding Trigonometric Ratios You should alread be familiar with the following formulae (SOH CAH TOA). Opposite sin a = Hpotenuse Adjacent cosa = Hpotenuse Opposite tana = Adjacent p If we have sin a = q where 0 < a < π, then we can form a right-angled triangle to represent this ratio. Opposite p Since sin a = = q then: q Hpotenuse p the side opposite a has length p a the hpotenuse has length q The length of the unknown side can be found using Pthagoras s Theorem. Once the length of each side is known, we can find cosa and tana using SOH CAH TOA. The method is similar if we know cosa and want to find sin a or tana. EXAMPLE Hpotenuse a Adjacent Opposite. Acute angles p and q are such that sin p = and sinq =. Show that sin 6 ( p + q) = 6. p sin p = cos p = q sinq = cosq = ( ) sin p + q = sin pcosq + cos psinq ( ) ( ) = + = = 6 6 (as required) Note Since Show that is used in the question, all of this working is required This is a common eam question. Page 97 HSN000

49 Higher Mathematics Unit Mathematics Using compound angle formulae to confirm identities EXAMPLES. Show that sin( π ) sin π ( ) = cos. = sin cosπ π cos sin = sin 0 cos = cos (as required) sin( s + t ) 6. Show that = tan s + tant for cos s 0 and cost 0. cos s cost sin( s + t ) sin s cost + cos s sint = cos s cost cos s cost sin s cost cos s sint = + cos s cost cos s cost sin s sint = + cos s cost = tan s + tan t (as required) Remember sin = tan cos Double-Angle Formulae If we use the compound angle identities with A = B, then we obtain epressions for sina and cosa. These are called double-angle formulae. sin A = sin A cos A cos cos sin A = A A = cos A = sin It is important to remember that these are given in the eam. EXAMPLE Given tanθ = for acute angle θ, find the eact value of sinθ and cosθ. A θ sinθ = cosθ = sin θ = sinθ cosθ = = θ = θ θ cos cos sin ( ) ( ) = = 9 6 = 7 Note An of the cosa formulae can be used in this eample Page 98 HSN000

50 Higher Mathematics Unit Mathematics Further Trigonometric Equations We will now consider trigonometric equations where the double-angle formulae are used to find solutions. These equations will involve: sin and either sin or cos cos and cos cos and sin Solving equations involving sin and either sin or cos EXAMPLE. Solve sin = sin for 0 < 60. sin cos = sin sin cos + sin = 0 sin = 0 sin ( cos + ) = 0 cos + = 0 Replace sin using the double angle formula Take all terms to one side, making the equation equal to zero Factorise the epression and solve = 0 or 80 or 60 cos = = or So = 0 or 0 or 80 or 0. Solving equations involving cos and cos EXAMPLE. Solve cos = cos for 0 π. = 0 or 0 Remember The double-angle formulae are given in the eam S A T C = cos = 60 ( ) cos = cos cos = cos cos cos 0 = ( cos + )( cos ) = 0 cos + = 0 cos = = π π or π + π = π or π So = 0 or π or π or π. Replace cos b cos Take all terms to one side, making a quadratic equation in cos Solve the quadratic equation (using factorisation or the quadratic formula) S A T C = cos = π ( ) cos = 0 cos = = 0 or π Page 99 HSN000

51 Higher Mathematics Unit Mathematics Solving equations involving cos and sin EXAMPLE. Solve cos = sin for 0 < < π. cos = sin sin = sin + = sin sin 0 ( sin )( sin + ) = 0 Replace cos b sin Take all terms to one side, making a quadratic equation in sin Solve the quadratic equation (using factorisation or the quadratic formula) sin = 0 sin = = π 6 or π π 6 = π 6 or π 6 So = π 6 or π 6 or π. S A T C = sin = π 6 ( ) sin + = 0 sin = = π Page 00 HSN000

52 Higher Mathematics Unit Mathematics OUTCOME Circles Representing a Circle The equation of a circle with centre ( a, b ) and radius r units is: This is given in the eam. ( ) ( ) a + b = r For eample, the equation of a circle with centre (, ) and radius is: ( ) ( ) ( ) ( ) + + = + + = 6 Note that the equation of a circle with centre ( 0, 0 ) is of the form + = r. EXAMPLE Find the equation of the circle with centre (, ) and radius units. ( ) ( ) a + b = r ( ) ( ) ( ) + ( ) = ( ) ( ) + + = Testing a Point Given a circle with centre ( a, b ) and radius r units, we can determine whether a point ( p, q ) lies within, outwith or on the circumference using the following rules: ( ) ( ) ( ) ( ) ( ) ( ) p a + q b < r the point lies within the circle p a + q b = r the point lies on the circumference of the circle p a + q b > r the point lies outwith the circle Page 0 HSN000

53 Higher Mathematics Unit Mathematics EXAMPLE A circle has the equation ( ) ( ) + + = 9. Determine whether the points (, ), ( 7, ) and (, ) lie within, outwith or on the circumference of the circle. Point (, ) : ( ) + ( + ) = ( ) + ( + ) = = 6 > 9 So outwith the circle. Point ( 7, ) : ( ) + ( + ) = ( 7 ) + ( + ) = + = 9 So on the circumference. Point (, ) : ( ) + ( + ) ( ) ( ) = + + = + = < 9 So within the circle. The General Equation of a Circle The equation of a circle can be written in the form: where the centre is ( g, f ) This is given in the eam. g f c = 0 and the radius is g + f c units. Note that the above equation onl represents a circle if since: if if g f c g f c + < 0 then we cannot obtain a real value for the radius g f c EXAMPLE + > 0, + = 0 then the radius is zero the equation represents a point.. Find the radius and centre of the circle with equation = 0. Comparing with g = so g = f = 8 so f = c = 7 g f c = 0: Centre is, ( g f ) = (, ) Radius is g + f c ( ) 7 = + = = units Page 0 HSN000

54 Higher Mathematics Unit Mathematics. Find the radius and centre of the circle with equation = 0. The equation must be in the form each term b : + + = 0 Now compare with g = so g = f = so f = c =. Eplain wh g f c g f c = 0: Centre is, ( g f ) (, ) = = 0, so divide Radius is g + f c ( ) ( ) = + + = = = 8 8 units = 0 is not the equation of a circle. Comparing with + + g + f + c = 0 : g = so g = g + f c = + ( ) 9 f = 8 so f = = 9 < 0 c = 9 The equation does not represent a circle since satisfied. g f c + > 0 is not Intersection of a Line and a Circle A straight line and circle can have two, one or no points of intersection: two intersections one intersection no intersections If a line and a circle onl touch at one point, then the line is a tangent to the circle at that point. To find out how man times a line and circle meet, we can use substitution. Page 0 HSN000

55 Higher Mathematics Unit Mathematics EXAMPLES. Find the points where the line with equation = intersects the circle with equation + = 0. + = 0 + ( ) = = 0 0 = 0 = = ± = = ( ) = ( ) So the circle and the line meet at (, ) and (, ) = = =.. Find the points where the line with equation = + 6 and circle with equation = 0 intersect. Substitute = + 6 into the equation of the circle: + ( + 6) + + ( + 6) 8 = 0 + ( + 6)( + 6) = = = 0 ( ) = 0 ( )( ) + + = 0 + = 0 + = 0 = = ( ) + 6 = = = ( ) + 6 = So the line and circle meet at (, ) and (, ). Remember ( ab) = m m m a b Page 0 HSN000

56 Higher Mathematics Unit Mathematics Tangents to Circles As we have seen a line is a tangent if it intersects the circle at onl one point. To show that a line is a tangent to a circle, the equation of the line can be substituted into the equation of the circle, and solved there should onl be one solution. EXAMPLE Show that the line with equation + = is a tangent to the circle with equation = 0. Substitute for equation of straight line: = 0 ( ) ( ) = 0 ( )( ) ( ) = = 0 + = 0 ( ) + = 0 + = 0 Then (i) factorise or (ii) use the discriminant + = 0 ( )( ) = 0 = 0 = = 0 = Since the solutions are equal, the line is a tangent to the circle. a = b = c = + = 0 b ac = ( ) ( ) = = 0 Since b ac = 0, the line is a tangent to the circle. Note If the point of contact is required then method (i) is more efficient. To find the point, substitute the value found for into the equation of the line (or circle) to calculate the corresponding -coordinate. Page 0 HSN000

57 Higher Mathematics Unit Mathematics 6 Equations of Tangents to Circles If the point of contact between a circle and a tangent and is known, then the equation of the tangent can be calculated. If a line is a tangent to a circle, then the radius will meet at the tangent at right angles. The gradient of this radius can be calculated, since the centre and point of contact are known. EXAMPLE Using mradius mtangent =, the gradient of the tangent can be found. Show that A (, ) lies on the circle equation of the tangent at A. Substitute point into equation of circle: = + + 6( ) + ( ) = The equation can then be found using b = m( a), since the point is known, and the gradient has just been calculated = 0 and find the = 0 Since this satisfies the equation of the circle, the point must lie on the circle. Find gradient between the centre (, ) and point (, ): mradius = + = + = so m = since m m = tangent radius tangent Find equation of tangent using point (, ) and gradient m = : b = m( a) = ( ) = + = + + = 0 Therefore the equation of the tangent to the circle at A is + = 0. Page 06 HSN000

58 Higher Mathematics Unit Mathematics 7 Intersection of Circles Consider two circles with radii r and r with r > r. Let d be the distance between the centres of the two circles. r d r There are three cases to consider. The circles do not touch d The meet at one point onl Eternall d The meet at two distinct points d d > r + r d = r + r r r < d < r + r Internall d d EXAMPLES d < r r d = r r. Circle P has centre (, ) and radius units, circle Q has equation = 0. Show that the circles P and Q do not touch. To find the centre and radius of Q: Compare with + + g + f + c = 0: g = so g = Centre is ( g, f ) Radius rq = g + f c f = 6 so f = = (, ) = + 9 c = = 9 = units r =. We know P has centre (, ) and radius P units So the distance between the centres d = ( + ) + ( + ) = + ( ) = 9. 9 units Since r P + r Q = + = < d, the circles P and Q do not touch. Page 07 HSN000

59 Higher Mathematics Unit Mathematics. Circle R has equation equation ( ) ( ) eternall. + = 0, and circle S has + 6 =. Show that the circles R and S touch To find the centre and radius of R: Compare with g = so g = f = so f = g f c c = = 0: Centre is, To find the centre and radius of S: Compare with ( ) ( ) r a = b = 6 = so r = ( g f ) = (, ) a b r + = : Centre is, = ( a b) (, 6) Radius rr = g + f c S = ( ) + ( ) + = 9 = units Radius r = units So the distance between the centres d = ( ) + ( 6) ( ) ( ) = + = units Since r R + r S = + = = d, the circles R and S touch eternall. = Page 08 HSN000