CHAPTER 4. APPLICATIONS AND REVIEW IN TRIGONOMETRY

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1 CHAPTER 4. APPLICATIONS AND REVIEW IN TRIGONOMETRY In the present chapter we apply the vector algebra and the basic properties of the dot product described in the last chapter to planar geometry and trigonometry. The reader may already know some trigonometric identities presented in the present chapter, largely learned by memorizing. Here we give a careful derivation of these identities, to consolidate the your knowledge in trigonometry, as well as what you have learned in the last chapter. Recall the following important identity from the last chapter u v = u v cos θ. (4.0) where the dot product u v of vectors u = (u 1,u ) and v = (v 1,v ) is given by u v = u 1 v 1 +u v. It is not an exaggeration to say that almost all crucial and nontrivial identities in trigonometry are consequences of this identity! Application 1: The addition formulae for sine and cosine. Take arbitrary angles α and β and draw the unit vectors u = (cosα, sin α) and v = (cosβ, sin in ) the Cartesian plane: Figure 1. For convenience, let us assume that α > β. Then, from the picture you draw, we can see that the angle between u and v is given by θ = α β. From u = (cosα, sin α) and v = (cosβ, sin β) we have u v = cosαcosβ + sin α sin β (4.1) On the other hand, from (4.0) we have u v = u v cos θ = cos(α β), (4.) 1

2 in view of the obvious facts that u = 1, v = 1 and θ = α β. Putting (4.1) and (4.) together, we have cos(α β) = u v = cos α cos β + sin α sin β. We have arrived at one of the basic addition formulae in trigonometry: cos(α β) = cosαcos β + sinαsin β. (4.3) From this, we derive other addition formulae as follows. Replace β by β in (4.3) to get cos(α ( β)) = cosαcos( β) + sinαsin( β). Using the facts that cos( β) = cosβ, sin( β) = sin β, (4.4) we obtain cos(α + β) = cosαcos β sin α sin β. (4.5) To get the addition formula for the sine function, we make use of the following basic identities ( π ) sin θ = cos θ ( π ) cos θ = sin θ. (4.6) Why are these identities valid? Well, draw a right triangle ABC with B = π/ and A = θ so that C = π θ. Put BC = a, AB = c and AC = b. Figure. Then sin A = cos C = a/b and cos A = sin C = c/b. Now (4.6) is more and less clear. We use (4.3), (4.5), (4.6) to get formulae for sin(α + β) and sin(α β) as follows: ( π ) (( π ) ) sin(α + β) = cos (α + β) = cos α β

3 ( π ) ( π ) = cos α cos β + sin α sin β = sinαcos β + cos α sin β. Hence we have arrived at sin(α + β) = sinαcos β + cos α sin β. (4.7) In the above identity, replace β by β. Using (4.5), we obtain sin(α β) = sinαcos β cosαsin β. (4.8) Identities (4.3), (4.5), (4.7), (4.8) constitutes the addition formulae for the sine and the cosine functions. Most students in natural sciences or engineering memorize them. In the future when we study complex numbers, you will learn a way to survive in trigonometry if you cannot remember these formulae. Review Exercise 4.1. What do you get from (4.3), (4.5), (4.7), (4.8) if you let β = α? Review Exercise 4.. From the last exercise you get cos α = cos α sin α. Explain how to derive from it the following identities. cos α = cos α 1, cos α = 1 sin α Then explain how to derive from the last two identities the following so called half angle formulae. cos θ = 1 + cosθ, sin θ = 1 cos θ. (4.9) (Most students in natural sciences and engineering memorize them in the format cos x = 1 + cos x, sin x = 1 cos x. They are very useful in calculus for integrating trigonometric functions. But in this course memorizing them is not required.) Review Exercise 4.3. Use the half angle formulae above to find cos 45 o (= cosπ/4) and sin 45 o (= sinπ/4) by letting θ = 90 o (= π/). Recall: sin π = 1 and cos π = 0. Review Exercise 4.4. Use the half angle formulae (4.9) to find cos 15 o (= cosπ/1) and sin 15 o (= sin π/1) by letting θ = 30 o (= π/6). Recall that cos π 6 = 3/. (Simplifying your answer in the exercise is rather tricky.) Review Exercise 4.5. Find cos π/1 and sinπ/1 in two other ways: first, use π/1 = π/3 π/4; second, use π/1 = π/4 π/6. Compare answers with the last exercise. 3

4 Review Exercise 4.6. Recall that the tangent function and the secant function are defined via tanα = sin α cos α, sec α = 1 cos α. (The cotangent and cosecant functions are given by cotα = cos α sin α 1 tanα, csc α = 1 sin α. But they are rarely used.) Use sin α + cos α = 1 to show sec α = tan α + 1. Use the addition formulae for sine and cosine to obtain the addition formulae for tangent: tan(α + β) = tanα + tanβ tanα tanβ, tan(α β) = 1 tanαtanβ 1 + tanαtanβ, as well as the following double angle formula for tangent: tan α = tanα 1 tan α. (4.10) (A very good student can remember these identities. But you are not asked to do so.) Problem 4.7. Use the following figure to give another proof of (4.10) above. Figure 3. Application : the cosine law. Let ABC be an arbitrary triangle and let a = BC, b = AC, c = AB : 4

5 Figure 4. The cosine law in trigonometry says that c = a + b ab cos C. (4.11) To see this, let us put u = CB and v = CA. Then BA = v u. Hence c = AB = AB = v u = (v u) (v u) = u u + v v u v = u + v u v cos C = a + b ab cos C. The cosine law (4.11) tells you how to find the angles of a triangle if the lengths of its sides are given, and, given two sides and the angle between them, how to find the third side. Example. Given that three sides of a triangle are a = BC =, b = CA = and c = AB = 6, we want to find its angles. Applying the cosine law (4.11), we get ( 6) = +(1+ 3) (1+ 3) cos C, or 6 = (1+ 3) cos C, that is, + 3 = 4(1 + 3) cos C, which gives cos C = 1/ and hence C = 60 o, or π/3. Similarly, a = b +c bc cos A gives = (1+ 3) + 6 (1+ 3) 6 cos A, or 4 = (1 + 3) 6 cos A, giving us = (1 + 3) 6 cos A. So cos A = (1 + 3) 6 = ( 3 + 1) 3 = (3 + 3) = 1. Hence A = 45 o, or π/4. Since A + B + C = π (or 180 o ), we have B = π A + C = π π 3 π 4 = 5π 1. Contrasting the cosine law for a triangle, there is something called the sine law. Again, let ABC be a triangle with a = BC, b = AC, c = AB. The sine law says: sin A a = sin B b 5 = sin C c (4.1)

6 This law tells you in principle how two find the rest of a triangle if two angles and one side of it is given. Unlike the cosine law, you don t need any sophisticated machinery (such as dot products) to derive it. But you do need some cleverness. Through point A draw a line perpendicular to BC, and let D be the intersection of this line with BC. Let d = AD : The above figure tells us sin B = b d and sin C = Now (4.1) is more or less clear. and hence sin B b = 1 d Figure 5. sin C =. c Now we describe the way to define sin α and cos α for an arbitrary real number α by using the Cartesian coordinate system XOY ; (here O stands for the origin and X, Y stands for two coordinate axes). The trajectory of a point P = (x,y) satisfying x + y = 1 is the circle with radius 1 centered at the origin O, called the unit circle. We denote this circle by U; (nowaday, the standard notation for the unit circle is T, but the interpretation of letter T as one dimensional torus is too mysterious for us here). By using the set theoretical notation, we write U = {(x,y) R x + y = 1}. The above identity is read as follows: the symbol U stands for the set of all (ordered) pairs (x,y) of real numbers satisfying the identity x + y = 1. Draw the line segment OP and imagine that it turns like a the needle of a stop watch. Every minute the needle turns 360 o. After one minute and 15 seconds after its start, the needle has turned 450 o and it locates in the same position as the one after 15 seconds. However, there are two differences between the line segment OP and the needle of 6

7 a stop watch. The needle starts at the vertical direction, pointing upward, while OP starts at the horizontal direction, pointing towards right. Also, the needle turns in the clockwise direction, but the reverse direction, namely the anti clockwise direction, is considered to be positive. Now every real number α can be considered as the radian an angle, and, when converted to degree, we know how much OP should turn and where it should end up with. If we write (x, y) the terminal position of P after turning an angle of α, we set x = cos α, y = sinα. For example, take α = 7π/ Converting to degrees, we get 40 o, which is 360 o + 60 o. So the (terminal) location of OP is obtained by one full turn, followed by the 60 o (or π/3 in radian) turn. Consequently cos 7π 3 = cos π 3 = 1, sin 7π 3 = sin π 3 = 3. The graphs of the trigonometric functions sin x, cos x (called the sinusoidal curves) and tanx are given below 7

8 8

9 Example. (This example is harder, but it is important for students in natural sciences or engineering.) The superposition of several sinusoidal curves (representing several monochromatic waves) with the same period T in general can be written as f(x) = A sin πx T + B cos πx T, where A and B are certain constants. We are asked to rewrite it in the form f(x) = C sin π(x + φ) T so that the amplitude C and phase shift φ can be read off immediately. More precisely, we are asked how to calculate C and a in the last identity in terms of A and B. Let us begin with the observation (this is slick) that f(x) can be rewritten as f(x) = u v with u = (A, B) and v = (cos θ, sin θ), where we put θ = πx/t for notational simplicity. Now u = A + B and v = cos θ + sin θ = 1. Hence Y = u v u v = A + B. This shows that the amplitude of f(x) cannot exceed A + B. It suggests to us to write down C = A + B and f(x) = C(a sin θ +b cos θ) with a = A/C and b = B/C. Now a +b = (A/C) +(B/C) = (A +B )/C = (A +B )/ A + B = 1. This shows that the point P = (a,b) on the unit circle and hence if we write πφ/t for the angle between OX and OP, we would have a = cos(πφ/t) and b = sin(πφ/t). So f(x) = C(cos πφ T which is the required expression. sin πx T + sin πφ T cos πx T = C cos π(x + φ) T Application 3: the area formula for a parallelogram. Let u = (a, b) and v = (c,d) be given vectors in the Cartesian plane. Let P be the parallelogram spanned by u and v; (u and v lie on two neighboring edges of P, as shown in the following figure.) 9

Figure 7. We want to find an explicit expression for the area of P in terms of the components a, b, c, d of the vectors u and v. Let θ be the angle between u and v.

10 Figure 7. We want to find an explicit expression for the area of P in terms of the components a, b, c, d of the vectors u and v. Let θ be the angle between u and v. Considering u as the base of the parallelogram P in the above figure, the height should be h = v sin θ. So the area A of the parallelogram P is given by A = u h = u v sin θ (4.13) On the other hand, the dot product u v of u and v is given by u v = u v cos θ (4.14) Compare the right hand sides of (4.13) and (4.14). The discrepancy is sin θ vs cos θ. As we know, the basic relation between sinθ and cos θ is sin θ+cos θ = 1. This suggests to take squares of both sides of (4.13) and (4.14) and then add the resulting identities A + (u v) = ( u v ) sin θ + ( u v ) cos θ = ( u v ) = u v. Now, from u = (a,b), v = (c,d) we get u v = ac + bd, u = a + b and v = c + d. So the identity A + (u v) = u v gives A + (ac + bd) = (a + b )(c + d ) (The next line is some algebraic manipulation. I presume that you are familiar with things like (a + b) = a + ab + b.) Thus A = (a + b )(c + d ) (ac + bd) = a c + a d + b c + b d (a c + acbd + b d ) 10

11 = a d + b c acbd = (ad) (ad)(bc) + (bc) = (ad bc). (4.15) Taking the square root of both sides, we have A = ac bd (4.16) Recall that, for any real number a, the square root of the square of a is a, not a: a = a. Certainly, if we know ahead of time that a 0, then a = a and hence we should write a = a. For example, A above stands for the area of P, which is surely 0 and consequently A = A. However, ad bc could be negative and hence the square root of its square should be written as ad bc. The expression ad bc has a name. It is called the determinant of the matrix It is denoted by deta or a c [ ] a b A =. [ b a d deta = det c ] b = ad bc. d Now (4.16) above tells us that the area of the parallelogram spanned by vectors u = (a,b) and v = (c,d) is equal to the absolute value of the determinant a b ad bc. Exercise 4.8. In each of the following parts, calculate the area of the parallelogram spanned by the given vectors u and v. Draw the parallelogram on graphic paper and check your answer but counting boxes. (a) u = (4, 1), v = (, 3); (b) u = (1, ), v = (3, 1); (c) u = (1, 4), v = (4, 1); (d) u = (, 6), v = (6, 6). Exercise 4.9. In each of the following parts, calculate the area enclosed by the triangle ABC with the given vertices. (Hint: Take one half of the area of the parallelogram spanned by AB and AC.) Draw this triangular on graphic paper and see winch way is easier, by calculating a determinant or by counting boxes. (a) A = (, 1), B = (5, 3), C = (1, 4); (b) A = (3, 1), B = (, 3), C = ( 1, ). Now we study some basic properties of determinants: 11

12 1. If we add a multiple of a row (or a column) to the other, the determinant is unchanged. For example, if we add to the first row a, b by the multiple of the second row c, d, say mc, md, then the determinant remains the same: a + mc b + md = a b. (4.17) To see this, notice that the left hand side is (a + mc)d (b + md)c, which is equal to ad + mcd (bc + mdc) = ad + mcd bc mdc = ad bc. In practice we often use a special case of (3.17). For example, one special case is obtained by letting m = 1: a c b d = a b Example. Find the determinant D = Solution: If we follow the definition, we start with D = , which needs some heavy computation to get the answer. So we prefer to use the basic property of determinants explained here to proceed as follows: D = = = = =. Here I spell out all small steps, many of which can be skipped.. The effect of switching two rows (or columns) of a determinant is a change in sign: a b = a b, b a d c = a b. Indeed, the first identity is nothing but cb ad = (ad bc), which is clearly valid. The second identity is also easy to check. 3. A common factor from any row or any column in a determinant can be taken out; for example, a c mb md = m a c b d. (4.18) 1

13 Example. We would like to simplify the determinant 6a 3b D = 8c 4d. There is a common factor of 3 in the first row and a common factor of 4 in the second row. Hence we can proceed as follows a b D = 3 4 = 3 4 a b = 4(ad bc). Exercise Evaluate or simplify each of the following determinants: cosα sin α cos(α/) sin(α/) (a) (b) sin α cos α (c) sin(α/) cos(α/) a/b c/d ab bc a + b b + c (d) d/c b/a (e) ad cd (f) a + d c + d. Sketch the vectors u = (cosα, sin α), v = ( sin α, cos α) with α = 30 o. Check that u v. Use your figure to explain your answer in part (b). Exercise (a) Verify (4.18) above. (b) Verify the following identity a b A C B D = aa + bb ac + bd ca + db cc + dd. Now let us go back to (4.15). The computation there tells us that (a + b )(c + d ) = (ac + bd) + (ad bc). (4.19) This identity is quite fascinating. Let us call a +b a sum of two squares. Then c +d is another sum of two squares. Their product is still a sum of two squares u + v with u = ac+bd and v = ad bc. For example, (3 +5 )(7 + ) = This is of course fabricated from (4.19). Without (4.18), it looks very mysterious! Exercise 4.1. How is the identity (3 + 5 )(7 + ) = fabricated? Answers to Exercises (outline) 13

14 = cos α + sin α, cos(α) = cos α sin α, sin α = sin α cos α, and 0 = sinαcos α cos α sin α. 4.. The first identity is obtained from cos(α) = cos α sin α by the substitution sin α = 1 cos α; the second one from the same identity by using the substitution cos α = 1 sin α. The rest is done by letting α = θ/ in the last two identities and then making the necessary rearrangement Letting θ = π/ in the last exercise, we have cos π 4 = 1 + cosπ/ = 1 and sin π 4 = 1 cos π/ = 1. Hence cos π 4 = sin π 4 = Letting θ = π/6 in Exercise 4., we have cos π 1 = 1 + cosπ/6 = 1 + 3/ = and Thus we have sin π 1 = 1 cos π 1 = cos π 1 = + 3 = 3 4 and sin π 3 1 = Now we use the second approach: cos π ( π 1 = cos 3 π ) = cos π 4 3 cos π 4 + sin π 3 sin π 4 = =. 4 sin π ( π 1 = sin 3 π ) = sin π 4 3 cos π 4 cos π 3 sin π 3 4 = 1 6 =. 4 The answers appear to be different from those of the previous exercise. Actually they are the same. This can be seen by squaring, e.g. ( ) 6 = = ( ) =

15 Next, 4.6. We have 1 + tan α = 1 + ( ) sin α = 1 + sin α cosα cos α = cos α + sin α cos α tan(α ± β) = sin(α ± β) cos(α ± β) = sin α cos β ± cos α sin β cos α cos β sin α sin β = 1 cos α = sec α. Dividing both numerator and denominator of the last fraction by cosαcosβ, we obtain the required identity Take a point P(a,b) on the unit circle (thus a + b = 1). Let O be the origin as usual and let A be the point ( 1, 0). Let α be the angle between the line AP and the x axis. Then the angle between OP and the x axis is α. Thus we have So, using b = 1 a, we have tanα = b 1 + a and tanα = b a. tanα 1 tan α = b/(1 + a) b(1 + a) = 1 (b/(1 + a)) (1 + a) b = b(1 + a) 1 + a + a (1 a ) = b(1 + a) b(1 + a) = a + a a(1 + a) = b a = tan α (a) 14 (b) 5 (c) 15 (d) (a) 11/ (b) 3/ (a) 0 (b) 1 (c) cosα (d) 0 (e) 0 (f) (b d)(c a) The right hand side is (aa + bb)(cc + dd) (ac + bd)(ca + db) = aacc + aadd + bbcc + bbdd (acca + acdb + bdca + bddb) = aadd + bbcc acdb bdca = (aadd acdb) + (bbcc bdca) = ad(ad BC) + bc(bc AD) = (ad bc)(ad BC) which is the left hand side. 15

16 4.1. (This question is just for fun. It will not be asked in the exam) Let u = (3, 5) and v = (7, ). Then u = 3 + 5, v = 7 +, 3 5 u v = = 31 and A = 7 = 9. The identity is obtained from u v = (u v) + A. 16

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