Trigonometric identities

Transcription

1 Trigonometric identities An identity is an equation that is satisfied by all the values of the variable(s) in the equation. For example, the equation (1 + x) = 1 + x + x is an identity. If you replace x by any specific number, the left hand side, which is (1 + x), will give the same value as the right hand side, which is 1 + x + x. In particular, (1 + ( )) is ( ) = 4 and we get the same number = 4 when we replace x in 1 + x + x by ( ). We have already encountered the following trigonometric identities: (a) tan x = sin x (b) sec x = 1 (c) csc x = 1 sin x (d) cot x = 1 tan x The trigonometric identity sin x + cos x = 1 The trig identity sin x + cos x = 1 follows from the Pythagorean theorem. In the figure below, an angle x is drawn. The side opposite the angle has length a, the adjacent side has length b, and the hypotenuse has length h. h a x b Therefore, sin x = a h and = b. It follows that h ( (sin x) + () a ) ( b = + = h h) a h + b h = a + b h The Pythagorean theorem asserts that a + b = h. It follows that (sin x) + () = a + b h = h h = 1 For convenience, (sin x) and () are written more briefly as sin x and cos x respectively. Therefore we have derived the identity sin x + cos x = 1 By simply re-arranging the terms, it may be written as sin x = 1 cos x OR cos x = 1 sin x NOTE that the identity DOES NOT simplify to sin x + = 1. Indeed there are many numbers x such that sin x + 1. For example, sin ( ) ( π + cos π ) = Two useful identities are derived from sin x + cos x = 1 by dividing both sides of the identity by sin x or cos x. 1

2 If we divide both sides of sin x + cos x = 1 by cos x, the result is sin x cos x + cos x = 1 ( ) ( ) sin x 1 cos x OR + 1 = Since sin x = tan x and 1 = sec x, it follows that (tan x) + 1 = (sec x). For convenience, we write (tan x) and (sec x) more briefly as tan x and sec x. Therefore we have verified that tan x + 1 = sec x If we divide both sides of sin x + cos x = 1 by sin x, the result is sin x sin x + sin x = 1 ( ) ( 1 sin x OR 1 + = sin x sin x Since sin x = cot x and 1 sin x = csc x, it follows that 1 + (cot x) = (csc x). For convenience, we write (cot x) and (csc x) more briefly as cot x and csc x. Therefore we have verified that cot x + 1 = csc x You will be required to verify identities. One way to do so is to start with one side of the identity and through a series of algebraic operations, derive the other side of the identity. Here is an example: Example 1 To verify the identity sin x + = sec x. We start with the left hand side because there is more we can do with it than the right hand side. Denote the left hand side by LHS and the right hand side by RHS. Then LHS = sin x + = sin x + = sin x + cos x = sec x = RHS (since we need a common denominator in order to add the two fractions) = 1 (since sin x + cos x = 1) We have verified that both sides of the identity are equal. This verifies the identity. Exercise 1. Verify each identity: sec x (a) = tan x cot x (b) tan x + cot x = sec x csc x (c)sec x ( 1 sin x ) = 1 (d) sec x tan x = sec x csc x (e) + sin x tan x = sec x (f) sec x = tan x sin x (g) csc x cot x = sec x csc x (h) sin x ( 1 + cot x ) = 1 (i) sin x 1 tan = cos x x (j) 1. Verify that = sin x tan x (k) sec x = tan x sin x (l) sin x tan x + = sin x + cot x sin x + sin x sin x = sec x csc x. Write cos 4 x sin 4 x as a difference of two squares, factor it and use the result to show that cos 4 x sin 4 x = cos x sin x. )

3 Identities for differences or sums of angles In this section we walk you through a derivation of the identity for cos(y x). This will be used to derive others involving sums or differences of angles. Contrary to what one would expect, it is NOT TRUE that cos(y x) = cos y for all angles x and y. Almost any pair of angles x and y gives cos(y x) cos y. For example, the choice y = 10 and x = 0 gives cos (y x) = cos (90 ) = 0 which is not equal to cos 10 cos 0, (you can easily check that). To derive the identity, start with a circle of radius 1 and center (0, 0), labelled P in Figure 1 below. Figure 1 Consider a line P Q that makes an angle y, in the second quadrant, with the positive horizontal axis. Figure Can you see why the coordinates of Q must be (cos y, sin y)? The figure below includes a line P R making an angle x in the first quadrant, with the positive horizontal

4 axis. Figure The coordinates of R are (, sin x). A standard notation for the length of the line segment QR is QR. Use the distance formula to show that QR simplifies to QR = (cos y + sin y sin x) (1) If you rotate the circle in Figure counter-clockwise until the ray P R merges into the positive horizontal axis the result is Figure 4 below. Figure 4 Angle QP R is y x, therefore Q has coordinates (cos (y c), sin (y x)). Clearly R has coordinates (1, 0). It follows that the length of QR is also given by QR = [cos(y x) 1] + [sin (y 1)] Show that this simplifies to QR = cos(y x) () 4

5 Comparing (1) and () reveals that cos(y x) = (cos y + sin y sin x) What do you conclude about cos(y x)? You should obtain cos (y x) = cos y + sin y sin x This is our first identity for the cosine of the difference of two angles. Example We know the exact values of cos 0, sin 0, cos 45 and sin 45. We may use them to calculate cos 15. The result: cos 15 = cos (45 0 ) = cos 45 cos 0 + sin 45 sin 0 = + 1 = = 4 Example 4 We are given that x is an angle in the first quadrant with = 1, y is an angle in the third 4 quadrant with cos y = 1 and we have to determine cos (y x). Since cos (y x) = cos y + sin y sin x, we have to find sin x and sin y. Diagrams will help. In Figure 5 we have a right triangle with a hypotenuse of length 4 and a horizontal side of length 1 because = horizontal coordinate length of hypotenuse = 1 4 The vertical side of the triangle is labelled b because it is not known. Pythagorean theorem: 1 + b = 4 We may determine b using the Therefore b = 15, hence b = ± 15. We take the positive sign because the vertical coordinate is positive in the first quadrant. Therefore b = 15 and so sin x = In Figure 6 we have drawn a rectangle with a hypotenuse with length and a horizontal coordinate 1 because cos y = horizontal coordinate length of hypotenuse = 1 = 1 The vertical coordinate is unknown, therefore we labelled it a. By the Pythagorean theorem = ( 1) + a = 1 + a Therefore a = 8, hence a = ± 8. In this case we must take the negative sign because the vertical coordinate is negative in the third quadrant. Therefore sin y = 8 Figure 5 Figure 6 5

6 It follows that cos (y x) = cos y + sin y sin x = ( ) ( 1 1 ) + 4 Other sum/difference identities ( ) ( ) = cos (y + x) = cos y sin y sin x. To derive this, we use the fact that cos( A) = cos A and sin( B) = sin B for all angles A and B. Therefore cos (y + x) = cos (y ( x)) = cos y cos( x) + sin y sin( x) = cos y sin y sin x Example 5 cos 75 = cos ( ) = cos 0 cos 45 sin 0 sin 45 = 1 6 = 4 sin (y x) = sin y cos y sin x. To derive it, we use the fact that cos(90 A) = sin A and sin(90 B) = cos B for any angles A and B. Therefore sin(y x) = cos (90 (y x)) = cos ((90 y) + x) = cos(90 y) sin(90 y) sin x = sin y cos y sin x Example 6 sin 15 = sin (45 0 ) = sin 45 cos 0 cos 45 sin 0 = 1 6 = 4 sin (y + x) = sin y sin x+cos y sin x. We derive this using the fact that cos( A) = cos A and sin( B) = sin B for all angles A and B. Therefore sin (y + x) = sin (y ( x)) = sin y cos( x) cos y sin( x) = sin y cos y( sin x) = sin y + cos y sin x Example 7 sin 75 = sin ( ) = sin 0 cos 45 + cos 0 sin 45 = = 4 tan y + tan x tan (y + x) = 1 tan y tan x sin(x ± y) tan (y ± x) = cos(x ± y). tan y tan x and tan (y x) =. These follows from the fact that 1 + tan y tan x Example 8 tan 75 = tan ( ) = tan 0 + tan 45 1 tan 0 tan 45 = Exercise = 1 1. You are given that x and y are angles in the first and fourth quadrants respectively, with sin x = 1 1 and cos y = 4 5. Draw the two angles then determine the exact values of the following expressions: a) b) sin y c) tan x d) tan y e) sin (x + y) f) sin (x y) g) cos (x + y) h) cos (x y) i) tan (x y) j) tan (x y) k) csc (x y) l) cot (x + y). If x is an angle in the first quadrant with sin x = 4 5, and y is an angle in the fourth quadrant with cos y =, determine the exact value of each expression: a. sin (x y) b. cos (x + y) c. sin (x + y) d. tan (x y) e. sec (x y) f. cot (x + y) 6

7 The Double Angle Formulas These are formulas that enable us to calculate sin x,, tan x and their reciprocals, once we know the values of sin x, and tan x, hence the term "double angle formulas". They are derived from the identities for sums of angles. If we replace y by x in the identity sin (x + y) = sin x cos y + sin y, we get sin (x + x) = sin x = sin x + sin x = sin x Thus This is our first "double angle formula" sin x = sin x () If we replace y by x in the identity Thus cos (x + y) = cos y sin x sin y, we get cos (x + x) = = sin x sin x = cos x sin x = cos x sin x (4) There are two other versions of this formula obtained by using the identity sin x + cos x = 1. If we solve for sin x to get sin x = 1 cos x then substitute into (4) we get I.e. = cos x sin x = = cos x (1 cos x) = cos x 1 = cos x 1 If, on the other hand, we solve for cos x to get cos x = 1 sin x then substitute into (4) we get I.e. = cos x sin x = 1 sin x sin x = 1 sin x = 1 sin x Which one of these three formulas for to use depends on the problem at hand. If we replace y by x In the identity tan(x + y) = Thus tan(x + x) = tan x = tan x = tan x + tan y 1 tan x tan y we get tan x + tan x 1 tan x tan x = tan x 1 tan x tan x 1 tan x Example 10 To determine sin x, and tan x given that x is an angle in the second quadrant with sin x =. The angle is shown below. The horizontal coordinate a is unknown but we easily calculate it using the Pythagorean theorem. Thus + a = which translates into a = 5. This implies that a = ± 5. Since the horizontal coordinate must be negative, we must take a = 5. 7

8 From the figure, = a = 5 and tan x = a = 5. Therefore sin x = sin x = ( ) ( ) 5 = ( = cos x 1 = tan x = tan x = 5 tan x 1 tan x = ) 1 = = 1 9. ) ( ) = 1 ( Rules for Changing Product to Sum/Differences = = 0 5 = 4 5 In some problems, it is easier to deal with sums of two trigonometric functions than their product. In such cases, we use the "Product to Sum" rules to convert products like sin x cos y, sin x sin y and cos y into sums or differences involving sine or cosine terms. We use the sum/difference rules we have already derived which are: sin A cos B + cos A sin B = sin (A + B) (5) sin A cos B cos A sin B = sin (A B) (6) cos A cos B + sin A sin B = cos (A B) (7) Adding (5) to (6) gives Adding (7) to (8) gives cos A cos B sin A sin B = cos (A + B) (8) sin A cos B = 1 [sin(a + B) + sin(a B)] cos A cos B = 1 [cos(a B) + cos(a + B)] Subtracting (8) from (7) gives sin A sin B = 1 [cos(a B) cos(a + B)] Example 11 To write sin x sin 5x, cos 4x cos 5x and sin x as sums/differences of trig functions. We use sin A sin B = 1 [cos(a B) cos(a + B)] to write sin x sin 5x as a sum of trig functions. Take A = x and B = 5x. Then sin x sin 5x = 1 [cos( x) cos 8x] = 1 [cos x cos 8x] For cos 4x cos 5x, we use cos A cos B = 1 [cos(a B) + cos(a + B)] with A = 4x and B = 5x. The result is cos 4x cos 5x = 1 [ + cos 9x] For sin x, we first write it as sin x and use sin A cos B = 1 [sin(a + B) + sin(a B)] with A = x and B = x. The result is sin x = 1 [sin x + sin x] = 1 [sin x + sin x] 8

9 Sum/Difference to Products There are problems that require us to change a sum or difference of sine or cosine terms into a product of sine and/or cosine terms. In other words, we may want to do the opposite of what we did in the above section. We use the same identities as above but now we write them as: sin (A + B) = sin A cos B + cos A sin B (9) sin (A B) = sin A cos B cos A sin B (10) cos (A + B) = cos A cos B sin A sin B (11) cos(a B) = cos A cos B + sin A sin B (1) We then introduce variables P = A + B and Q = A B. We have to write A and B in terms of these new variables. It turns out that A = 1 (P + Q) and B = 1 (P Q) Adding (??) to 10 gives a sum of sines which is sin (A + B) + sin(a B) = sin A cos B Using the new variables P and Q this result may be written as sin P + sin Q = [ sin 1 (P + Q) cos 1 (P Q)] Subtracting 10 from (??) to gives a difference of sines which is Adding (11) to 1 gives a sum of cosines which is sin P sin Q = [ cos 1 (P + Q) sin 1 (P Q)] cos P + cos Q = [ cos 1 (P + Q) cos 1 (P Q)] Subtracting 1 from (11) to gives a difference of cosines which is cos P cos Q = [ sin 1 (P + Q) sin 1 (P Q)] Example 1 To write cos x + cos 5x, sin x + sin x, sin 5x sin x and cos x as products of trig functions: Using the above identities: The Half Angle Formulas x + cos 5x = cos 1 (x + 5x) cos 1 (x 5x) = cos 4x sin x + sin x = sin 1 (x + x) cos 1 (x x) = sin x sin 5x sin x = cos 1 (5x + x) sin 1 (5x x) = cos 7 x sin x cos x = sin 1 (x + x) sin 1 (x x) = sin 5 x sin 1 x These are formulas that enable us to calculate sin 1 x, cos 1 x, tan 1 x and their reciprocals, once we know the values of sin x, and tan x, hence the term "half angle formulas". They are derived from the two identities. cos y = 1 sin y and cos y = cos y 1 Take cos y = 1 sin y. Rearrange it as sin y = 1 cos y. Now solve for sin y. The result is 1 cos y sin y = ± 9

10 Finally, replace y with 1 x to get sin 1 x = ± 1 The sign is dictated by the position of the angle 1 x. If it is in the first or second quadrant, (where the sine function is positive), take the positive sign. If it is in the third or fourth quadrant, take the negative sign. To derive the half angle formula for the cosine function, take the identity cos y = cos y 1 and solve for cos y. The result is 1 + cos y cos y = ± As you would expect, replace y by 1 x to get cos 1 x = ± 1 + Again the sign is dictated by the position of the angle 1 x. If it is in the first or fourth quadrant, (where the cosine function is positive), take the positive sign. If it is in the second or third quadrant, take the negative sign. We do not need to do any heavy lifting to determine the half angle formula for the tangent function. We simply use the identity tan 1 x = sin 1 x cos 1 x = ± = ± 1 +. Likewise, you should expect the sign to be dictated by the position of 1 x. If it is in the first or third quadrant, take the positive sign. If it is in the second or fourth quadrant, take the negative sign. Example 1 To determine sin x,, tan x, sin 1 x, cos 1 x and tan 1 x given that x is an angle in the fourth quadrant with = 1 1. The angle is shown below. The vertical coordinate a may be obtained from 1 + a = 1 which translates into a = 5. This implies that a = ±5. Since the coordinate must be negative, we must take a = 5. From the figure, sin x = a = 5 1 and tan x = 5 1. Therefore sin x = sin x = ( ( 1) 5 1 ) 1 = = cos x sin x = =

11 tan x = tan x = tan x 1 tan x = ( ) 1 1 ( ) = = = = 1 5 Before calculating sin 1 x, cos 1 x and tan 1 x, we note that 1 x is an angle between 15 and 180, (because x is between 70 and 60 ), therefore it is in the second quadrant. This implies that sin 1 x is positive but cos 1 x and tan 1 x are both negative. Therefore sin 1 x = 1 1 = = 6 cos 1 x = 1 + = tan 1 x = 1 + = Exercise 14 5 = 6 1 = 5 = You should know the exact value of cos 45. Use it and a half angle formula to determine the exact value of cos.5.. x is an angle in the third quadrant with = 1. Draw the angle then use a half angle formula to determine the exact values of sin ( 1 x) and cos ( 1 x).. If x is an angle in the first quadrant with sin x = 1, and y is an angle in the fourth quadrant with cos y = 5, determine the exact value of each expression: (a) sin(x y) (b) (c) tan(x + y) (d) tan 1 y 4. You are given that x is an angle in the second quadrant and = 5 1 (a) Draw the angle and calculate the exact values of sin x and tan x. (b) Now calculate the exact values of the following: (a) sin x (b) (c) tan x (d) tan 1 x (c) You have enough information to calculate sec x. Calculate it. 5. You are given that y is an angle in the third quadrant and tan y = 4. (a) Draw the angle and calculate the exact value of cos y and sin y. (b) Calculate the exact value, (no calculator), of the following: (a) sin 1 y (b) cos 1 y (c) tan 1 y (c) You have enough information to calculate the exact value of cot 1 y. Calculate it. 11