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7.3 Proving Trigonometric Identities

Prove identities using other identities.
Use the product-to-sum identities and the sum-to-product identities to derive other identities.

The Logic of Proving Identities

We outline two algebraic methods for proving identities.

Method 1. Start with either the left side or the right side of the equation and obtain the other side. For example, suppose you are trying to prove that the equation P=Q $P = Q$ is an identity. You might try to produce a string of statements (R1, R2,… or T1, T2,…) $(R_{1}, R_{2}, \dots or T_{1}, T_{2}, \dots)$ like the following, which start with P and end with Q or start with Q and end with P:

P = = ⋮ = R 1 R 2 Q or Q = = ⋮ = T 1 T 2 P .

$\begin{matrix} P & = & R_{1} & or & Q & = & T_{1} \\ = & R_{2} & = & T_{2} \\ ⋮ & ⋮ \\ = & Q & = & P . \end{matrix}$

Method 2. Work with each side separately until you obtain the same expression. For example, suppose you are trying to prove that P=Q $P = Q$ is an identity. You might be able to produce two strings of statements like the following, each ending with the same statement S.

P = = ⋮ = R 1 R 2 S or Q = = ⋮ = T 1 T 2 S .

$\begin{matrix} P & = & R_{1} & or & Q & = & T_{1} \\ = & R_{2} & = & T_{2} \\ ⋮ & ⋮ \\ = & S & = & S . \end{matrix}$

The number of steps in each string might be different, but in each case the result is S.

A first step in learning to prove identities is to have at hand a list of the identities that you have already learned. Such a list is on the inside back cover of this text. Ask your instructor which ones you are expected to memorize. The more identities you prove, the easier it will be to prove new ones. A list of helpful hints is shown at left.

Proving Identities

In what follows, method 1 is used in Examples 1, 3, and 4, and method 2 is used in Examples 2 and 5.

Example 1

Prove the identity 1+sin 2θ=(sin θ+cos θ)2 $1 + sin 2 θ = {(sin θ + c o s θ)}^{2}$ .

Solution

Let’s use method 1. We begin with the right side and obtain the left side:

(sin θ + cos θ) 2 = = = sin 2 θ + 2 sin θ cos θ + cos 2 θ 1 + 2 sin θ cos θ 1 + sin 2 θ . S q u a r i n g R e c a l l i n g t h e i d e n t i t y s i n 2 x + c o s 2 x = 1 a n d s u b t i t u t i n g U s i n g s i n 2 x = 2 s i n x c o s x

$\begin{matrix} {(sin θ + cos θ)}^{2} & = & {sin}^{2} θ + 2 sin θ cos θ + {cos}^{2} θ & Squaring \\ = & 1 + 2 sin θ cos θ & Recalling the identity {sin}^{2} x + {cos}^{2} x = 1 and subtituting \\ = & 1 + sin 2 θ . & Using sin 2 x = 2 sin x cos x \end{matrix}$

We could also begin with the left side and obtain the right side:

1 + sin 2 θ = = = 1+ 2 sin cos θ sin 2 θ + 2 sin θ cos θ + cos 2 θ (sin θ + cos θ) 2 . U s i n g s i n 2 x = 2 s i n x c o s x R e p l a c i n g 1 w i t h s i n 2 θ + c o s 2 θ F a c t o r i n g

$\begin{matrix} 1 + sin 2 θ & = & 1+ 2 sin cos θ & Using sin 2 x = 2 sin x cos x \\ = & {sin}^{2} θ + 2 \sin θ \cos θ + {cos}^{2} θ & Replacing 1 with {sin}^{2} θ + {cos}^{2} θ \\ = & {(sin θ + \cos θ)}^{2} . & Factoring \end{matrix}$

Now Try Exercises 13 and 19.

Example 2

Prove the identity

sin 2 x tan 2 x = tan 2 x - sin 2 x .

${sin}^{2} x {tan}^{2} x = {tan}^{2} x - {sin}^{2} x .$

Solution

For this proof, we are going to work with each side separately using method 2. We try to obtain the same expression on each side. In actual practice, you might work on one side for a while, then work on the other side, and then go back to the first side. In other words, you work back and forth until you arrive at the same expression. Let’s start with the right side.

tan 2 x - sin 2 x = = = = = = sin 2 x cos 2 x - sin 2 x sin 2 x cos 2 x - sin 2 x \cdot cos 2 x cos 2 x sin 2 x - sin 2 x cos 2 x cos 2 x sin 2 x ( 1 - cos 2 x ) cos 2 x sin 2 x sin 2 x cos 2 x sin 4 x cos 2 x R e c a l l i n g t h e i d e n t i t y t a n x = s i n x c o s x a n d s u b s t i t u t i n g M u l t i p l y i n g b y 1 i n o r d e r t o s u b t r a c t C a r r y i n g o u t t h e s u b t r a c t i o n F a c t o r i n g R e c a l l i n g t h e i d e n t i t y 1 - c o s 2 x 5 s i n 2 x a n d s u b s t i t u t i n g

$\begin{matrix} {tan}^{2} x - {sin}^{2} x & = & \frac{{sin}^{2} x}{{cos}^{2} x} - {sin}^{2} x & Recalling the identity tan x = \frac{sin x}{cos x} and substituting \\ = & \frac{{sin}^{2} x}{{cos}^{2} x} - {sin}^{2} x \cdot \frac{{cos}^{2} x}{{cos}^{2} x} & Multiplying by 1 in order to subtract \\ = & \frac{{sin}^{2} x - {sin}^{2} x {cos}^{2} x}{{cos}^{2} x} & Carrying out the subtraction \\ = & \frac{{sin}^{2} x (1 - {cos}^{2} x)}{{cos}^{2} x} & Factoring \\ = & \frac{{sin}^{2} x {sin}^{2} x}{{cos}^{2} x} & Recalling the identity 1 - {cos}^{2} x 5 {sin}^{2} x and substituting \\ = & \frac{{sin}^{4} x}{{cos}^{2} x} \end{matrix}$

At this point, we stop and work with the left side, sin2 x tan2 x ${sin}^{2} x {tan}^{2} x$ , of the original identity and try to end with the same expression that we ended with on the right side:

sin 2 x tan 2 x = = sin 2 x sin 2 x cos 2 x sin 4 x cos 2 x . R e c a l l i n g t h e i d e n t i t y t a n x = s i n x c o s x a n d s u b s t i t u t i n g

$\begin{matrix} {sin}^{2} x {tan}^{2} x & = & {sin}^{2} x \frac{{sin}^{2} x}{{cos}^{2} x} & Recalling the identity tan x = \frac{sin x}{cos x} and substituting \\ = & \frac{{sin}^{4} x}{{cos}^{2} x} . \end{matrix}$

We have obtained the same expression from each side, so the proof is complete.

Now Try Exercise 25.

Example 3

Prove the identity

sin 2 x sin x - cos 2 x cos x = sec x .

$\frac{sin 2 x}{sin x} - \frac{cos 2 x}{cos x} = sec x .$

Solution

sin 2 x sin x - cos 2 x cos x = = = = = = = 2 sin x cos x sin x - cos 2 x - sin 2 x cos x 2 cos x - cos 2 x - sin 2 x cos x 2 cos 2 x cos x - cos 2 x - sin 2 x cos x 2 cos 2 x - cos 2 x + sin 2 x cos x cos 2 x + sin 2 x cos x 1 cos x sec x U s i n g d o u b l e - a n g l e i d e n t i t i e s S i m p l i f y i n g M u l t i p l y i n g 2 c o s x b y 1, o r c o s x / c o s x S u b t r a c t i n g U s i n g a P y t h a g o r e a n i d e n t i t y R e c a l l i n g a b a s i c i d e n t i t y

$\begin{matrix} \frac{sin 2 x}{sin x} - \frac{cos 2 x}{cos x} & = & \frac{2 sin x cos x}{sin x} - \frac{{cos}^{2} x - {sin}^{2} x}{cos x} & Using double-angle identities \\ = & 2 cos x - \frac{{cos}^{2} x - {sin}^{2} x}{cos x} & Simplifying \\ = & \frac{2 {cos}^{2} x}{cos x} - \frac{{cos}^{2} x - {sin}^{2} x}{cos x} & Multiplying 2 cos x by 1, or cos x / cos x \\ = & \frac{2 {cos}^{2} x - {cos}^{2} x + {sin}^{2} x}{cos x} & Subtracting \\ = & \frac{{cos}^{2} x + {sin}^{2} x}{cos x} \\ = & \frac{1}{cos x} & Using a Pythagorean identity \\ = & sec x & Recalling a basic identity \end{matrix}$

Now Try Exercise 15.

Example 4

Prove the identity

sec t - 1 t sec t = 1 - cos t t .

$\frac{sec t - 1}{t sec t} = \frac{1 - cos t}{t} .$

Solution

We use method 1, starting with the left side. Note that the left side involves sec t, whereas the right side involves cos t, so it might be wise to make use of a basic identity that involves these two expressions: sec t=1/cos t $sec t = 1 / cos t$ .

sec t - 1 t sec t = = = = 1 cos t - 1 t 1 cos t (1 cos t - 1) \cdot cos t t 1 t - cos t t 1 - cos t t S u b s t i t u t i n g 1 / c o s t f o r s e c t M u l t i p l y i n g

$\begin{matrix} \frac{sec t - 1}{t sec t} & = & \frac{\frac{1}{cos t} - 1}{t \frac{1}{cos t}} & Substituting 1/cos t for sec t \\ = & (\frac{1}{cos t} - 1) \cdot \frac{cos t}{t} \\ = & \frac{1}{t} - \frac{cos t}{t} & Multiplying \\ = & \frac{1 - cos t}{t} \end{matrix}$

We started with the left side and obtained the right side, so the proof is complete.

Now Try Exercise 5.

Example 5

Prove the identity

cot ϕ + c s c ϕ = sin ϕ 1 - cos ϕ .

$cot ϕ + c s c ϕ = \frac{sin ϕ}{1 - cos ϕ} .$

Solution

We are again using method 2, beginning with the left side:

cot ϕ + c s c ϕ = = cos ϕ sin ϕ + 1 sin ϕ 1 + cos ϕ sin ϕ . U s i n g b a s i c i d e n t i t i e s A d d i n g

$\begin{matrix} cot ϕ + c s c ϕ & = & \frac{cos ϕ}{sin ϕ} + \frac{1}{sin ϕ} & Using basic identities \\ = & \frac{1 + cos ϕ}{sin ϕ} . & Adding \end{matrix}$

At this point, we stop and work with the right side of the original identity:

sin ϕ 1 - cos ϕ = = sin ϕ 1 - cos ϕ \cdot 1 + cos ϕ 1 + cos ϕ sin ϕ ( 1 + cos ϕ ) 1 - cos 2 ϕ . M u l t i p l y i n g b y 1

$\begin{matrix} \frac{sin ϕ}{1 - cos ϕ} & = & \frac{sin ϕ}{1 - cos ϕ} \cdot \frac{1 + cos ϕ}{1 + cos ϕ} & Multiplying by 1 \\ = & \frac{sin ϕ (1 + cos ϕ)}{1 - {cos}^{2} ϕ} . \end{matrix}$

Continuing, we have

sin ϕ 1 - cos ϕ = = sin ϕ ( 1 + cos ϕ ) sin 2 ϕ 1 + cos ϕ sin ϕ . U s i n g s i n 2 x = 1 - c o s 2 x S i m p l i f y i n g

$\begin{matrix} \frac{sin ϕ}{1 - cos ϕ} & = & \frac{sin ϕ (1 + cos ϕ)}{{sin}^{2} ϕ} & Using {sin}^{2} x = 1 - {cos}^{2} x \\ = & \frac{1 + cos ϕ}{sin ϕ} . & Simplifying \end{matrix}$

The proof is complete since we obtained the same expression from each side.

Now Try Exercise 29.

Product-to-Sum and Sum-to-Product Identities

On occasion, it is convenient to convert a product of trigonometric expressions to a sum, or the reverse. The following identities are useful in this connection.

We can derive product-to-sum identities (1) and (2) using the sum and difference identities for the cosine function:

cos (x + y) cos (x - y) = = cos x cos y - sin x sin y, cos x cos y + sin x sin y . S u m i d e n t i t y D i f f e r e n c e i d e n t i t y

$\begin{matrix} cos (x + y) & = & cos x cos y - sin x sin y, & Sum identity \\ cos (x - y) & = & cos x cos y + sin x sin y . & Difference identity \end{matrix}$

Subtracting the sum identity from the difference identity, we have

cos (x - y) - cos (x + y) 1 2 [cos (x - y) - cos (x + y)] = = 2 sin x sin y sin x sin y . S u b t r a c t i n g M u l t i p l y i n g b y 1 2

$\begin{matrix} cos (x - y) - cos (x + y) & = & 2 sin x sin y & Subtracting \\ \frac{1}{2} [cos (x - y) - cos (x + y)] & = & sin x sin y . & Multiplying by \frac{1}{2} \end{matrix}$

Thus, sin x sin y=12 [cos (x−y)−cos (x+y)] $sin x sin y = \frac{1}{2} [cos (x - y) - cos (x + y)]$ .

Adding the cosine sum and difference identities, we have

cos (x - y) + cos (x + y) 1 2 [cos (x - y) + cos (x + y)] = = 2 cos x cos y cos x cos y . A d d i n g M u l t i p l y i n g b y 1 2

$\begin{matrix} cos (x - y) + cos (x + y) & = & 2 cos x cos y & Adding \\ \frac{1}{2} [cos (x - y) + cos (x + y)] & = & cos x cos y . & Multiplying by \frac{1}{2} \end{matrix}$

Thus, cos x cos y=12 [cos (x−y)+cos (x+y)] $cos x cos y = \frac{1}{2} [cos (x - y) + cos (x + y)]$ .

Identities (3) and (4) can be derived in a similar manner using the sum and difference identities for the sine function.

Example 6

Find an identity for 2 sin 3θ cos 7θ $2 sin 3 θ cos 7 θ$ .

Solution

We will use the identity

sin x \cdot cos y = 1 2 [sin (x + y) + sin (x - y)] .

$sin x \cdot cos y = \frac{1}{2} [sin (x + y) + sin (x - y)] .$

Here x=3θ $x = 3 θ$ and y=7θ $y = 7 θ$ . Thus,

2 sin 3 θ cos 7 θ = = = 2 \cdot 1 2 [sin (3 θ + 7 θ) + sin (3 θ - 7 θ)] sin 10 θ + sin (- 4 θ) sin 10 θ - sin 4 θ . U s i n g s i n (- θ) = - s i n θ

$\begin{matrix} 2 sin 3 θ cos 7 θ & = & 2 \cdot \frac{1}{2} [sin (3 θ + 7 θ) + sin (3 θ - 7 θ)] \\ = & sin 10 θ + sin (- 4 θ) \\ = & sin 10 θ - sin 4 θ . & Using sin (- θ) = - sin θ \end{matrix}$

Now Try Exercise 37.

The sum-to-product identities (5)–(8) can be derived using the product-to-sum identities. Proofs are left to the exercises.

Example 7

Find an identity for cos θ+cos 5θ $cos θ + cos 5 θ$ .

Solution

We will use the identity

cos y + cos x = 2 cos x + y 2 cos x - y 2 .

$cos y + cos x = 2 cos \frac{x + y}{2} cos \frac{x - y}{2} .$

Here x=5θ $x = 5 θ$ and y=θ $y = θ$ . Thus,

cos θ + cos 5 θ = = 2 cos 5 θ + θ 2 cos 5 θ - θ 2 2 cos 3 θ cos 2 θ .

$\begin{matrix} cos θ + cos 5 θ & = & 2 cos \frac{5 θ + θ}{2} cos \frac{5 θ - θ}{2} \\ = & 2 cos 3 θ cos 2 θ . \end{matrix}$

Now Try Exercise 35.

7.3 Exercise Set

Prove the identity.

1. sec x−sin x tan x=cos x $sec x - sin x tan x = cos x$
2. 1+cos θsin θ+sin θcos θ=cos θ+1sin θ cos θ $\frac{1 + cos θ}{sin θ} + \frac{sin θ}{cos θ} = \frac{cos θ + 1}{sin θ cos θ}$
3. 1−cos xsin x=sin x1+cos x $\frac{1 - cos x}{sin x} = \frac{sin x}{1 + cos x}$
4. 1+tan y1+cot y=sec ycsc y $\frac{1 + tan y}{1 + cot y} = \frac{sec y}{csc y}$
5. 1+tan θ1−tan θ+1+cot θ1−cot θ=0 $\frac{1 + tan θ}{1 - tan θ} + \frac{1 + cot θ}{1 - cot θ} = 0$
6. sin x+cos xsec x+csc x=sin xsec x $\frac{sin x + cos x}{sec x + csc x} = \frac{sin x}{sec x}$
7. cos2 α+cot αcos2 α−cot α=cos2 α tan α+1cos2 α tan α−1 $\frac{{cos}^{2} α + cot α}{{cos}^{2} α - cot α} = \frac{{cos}^{2} α tan α + 1}{{cos}^{2} α tan α - 1}$
8. sec 2θ=sec2 θ2−sec2 θ $sec 2 θ = \frac{{sec}^{2} θ}{2 - {sec}^{2} θ}$
9. 2 tan θ1+tan2 θ=sin 2θ $\frac{2 tan θ}{1 + {tan}^{2} θ} = sin 2 θ$
10. cos (u−v)cos u sin v=tan u+cot v $\frac{cos (u - v)}{cos u sin v} = tan u + cot v$
11. 1−cos 5θ cos 3θ−sin 5θ sin 3θ=2 sin2 θ $1 - cos 5 θ cos 3 θ - sin 5 θ sin 3 θ = 2 {sin}^{2} θ$
12. cos4 x−sin4 x=cos 2x ${cos}^{4} x - {sin}^{4} x = cos 2 x$
13. 2 sin θ cos3 θ+2 sin3 θ cos θ=sin 2θ $2 sin θ {cos}^{3} θ + 2 {sin}^{3} θ cos θ = sin 2 θ$
14. tan 3t−tan t1+tan 3t tan t=2 tan t1−tan2 t $\frac{tan 3 t - tan t}{1 + tan 3 t tan t} = \frac{2 tan t}{1 - {tan}^{2} t}$
15. tan x−sin x2 tan x=sin2 x2 $\frac{tan x - sin x}{2 tan x} = {sin}^{2} \frac{x}{2}$
16. cos3 β−sin3 βcos β−sin β=2+sin 2β2 $\frac{{cos}^{3} β - {sin}^{3} β}{cos β - s i n β} = \frac{2 + sin 2 β}{2}$
17. sin (α+β) sin (α−β)=sin2 α−sin2 β $sin (α + β) sin (α - β) = {sin}^{2} α - {sin}^{2} β$
18. cos2 x(1−sec2 x)=−sin2 x ${cos}^{2} x (1 - {sec}^{2} x) = - {sin}^{2} x$
19. tan θ(tan θ+cot θ)=sec2 θ $tan θ (tan θ + cot θ) = {sec}^{2} θ$
20. cos θ+sin θcos θ=1+tan θ $\frac{cos θ + s i n θ}{cos θ} = 1 + tan θ$
21. 1+cos2 xsin2 x=2 csc2 x−1 $\frac{1 + {cos}^{2} x}{{sin}^{2} x} = 2 {csc}^{2} x - 1$
22. tan y+cot ycsc y=sec y $\frac{tan y + cot y}{csc y} = sec y$
23. 1+sin x1−sin x+sin x−11+sin x=4 sec x tan x $\frac{1 + sin x}{1 - sin x} + \frac{sin x - 1}{1 + sin x} = 4 sec x tan x$
24. tan θ−cot θ=(sec θ−csc θ) (sin θ+cos θ) $tan θ - cot θ = (sec θ - csc θ) (sin θ + cos θ)$
25. cos2 α cot2 α=cot2 α−cos2 α ${cos}^{2} α {cot}^{2} α = {cot}^{2} α - {cos}^{2} α$
26. tan x+cot xsec x+csc x=1cos x+sin x $\frac{tan x + cot x}{sec x + csc x} = \frac{1}{cos x + sin x}$
27. 2 sin2 θ cos2 θ+cos4 θ=1−sin4 θ $2 {sin}^{2} θ {cos}^{2} θ + {cos}^{4} θ = 1 - {sin}^{4} θ$
28. cot θcsc θ−1=csc θ+1cot θ $\frac{cot θ}{csc θ - 1} = \frac{csc θ + 1}{cot θ}$
29. 1+sin x1−sin x=(sec x+tan x)2 $\frac{1 + sin x}{1 - sin x} = {(sec x + tan x)}^{2}$
30. sec4 s−tan2 s=tan4 s+sec2 s ${sec}^{4} s - {tan}^{2} s = {tan}^{4} s + {sec}^{2} s$
31. Verify the product-to-sum identities (3) and (4) using the sine sum and difference identities.
32. Verify the sum-to-product identities (5)–(8) using the product-to-sum identities (1)–(4).

Use the product-to-sum identities and the sum-to-product identities to find identities for each of the following.

33. sin 3θ−sin 5θ $sin 3 θ - sin 5 θ$
34. sin 7x−sin 4x $sin 7 x - sin 4 x$
35. sin 8θ+sin 5θ $sin 8 θ + sin 5 θ$
36. cos θ−cos 7θ $cos θ - cos 7 θ$
37. sin 7u sin 5u $sin 7 u sin 5 u$
38. 2 sin 7θ cos 3θ $2 sin 7 θ cos 3 θ$
39. 7 cos θ sin 7θ $7 cos θ sin 7 θ$
40. cos 2t sin t $cos 2 t sin t$
41. cos 55° sin 25° $cos 55 ° sin 25 °$
42. 7 cos 5θ cos 7θ $7 cos 5 θ cos 7 θ$

Use the product-to-sum identities and the sum-to-product identities to prove each of the following.

43. $sin 4 θ + sin 6 θ = cot θ (cos 4 θ - cos 6 θ)$
44. $tan 2 x (cos x + cos 3 x) = sin x + sin 3 x$
45. $cot 4 x (sin x + sin 4 x + sin 7 x) = \cos x + cos 4 x + cos 7 x$
46. $tan \frac{x + y}{2} = \frac{sin x + sin y}{cos x + cos y}$
47. $cot \frac{x + y}{2} = \frac{sin y - sin x}{cos x - cos y}$
48. $tan \frac{θ + ϕ}{2} tan \frac{ϕ - θ}{2} = \frac{cos θ - cos ϕ}{cos θ + \cos ϕ}$
49. $tan \frac{θ + ϕ}{2} (sin θ - sin ϕ)$

$= \tan \frac{θ - ϕ}{2} (sin θ + sin ϕ)$
50. $sin 2 θ + sin 4 θ + sin 6 θ = 4 cos θ cos 2 θ sin 3 θ$

Skill Maintenance

For each function:

a) Graph the function.
b) Determine whether the function is one-to-one.
c) If the function is one-to-one, find an equation for its inverse.
d) Graph the inverse of the function. [5.1]

51. $f (x) = 3 x - 2$
52. $f (x) = x^{3} + 1$
53. $f (x) = x^{2} - 4, x \geq 0$
54. $f (x) = \sqrt{x + 2}$

Solve.

55. $2 x^{2} = 5 x$ [3.2]
56. $3 x^{2} + 5 x - 10 = 18$ [3.2]
57. $x^{4} + 5 x^{2} - 36 = 0$ [3.2]
58. $x^{2} - 10 x + 1 = 0$ [3.2]
59. $\sqrt{x - 2} = 5$ [3.4]
60. $x = \sqrt{x + 7} + 5$ [3.4]

Synthesis

Prove the identity.

61. $ln | tan x | = - ln | cot x |$
62. $ln | sec θ + tan θ | = - ln | sec θ - tan θ |$
63. $log (cos x - sin x) + log (cos x + sin x) = log cos 2 x$
64. Mechanics. The following equation occurs in the study of mechanics:

$sin θ = \frac{I_{1} cos ϕ}{\sqrt{{(I_{1} cos ϕ)}^{2} + {(I_{2} sin ϕ)}^{2}}} .$

It can happen that $I_{1} = I_{2}$ . Assuming that this happens, simplify the equation.
65. Alternating Current. In the theory of alternating current, the following equation occurs:

$R = \frac{1}{ω C (tan θ + t a n ϕ)} .$

Show that this equation is equivalent to

$R = \frac{cos θ cos ϕ}{ω C sin (θ + ϕ)} .$
66. Electrical Theory. In electrical theory, the following equations occur:

$E_{1} = \sqrt{2} E_{t} cos (θ + \frac{π}{P})$

and

$E_{2} = \sqrt{2} E_{t} cos (θ - \frac{π}{P}) .$

Assuming that these equations hold, show that

$\frac{E_{1} + E_{2}}{2} = \sqrt{2} E_{t} cos θ cos \frac{π}{P}$

and

$\frac{E_{1} - E_{2}}{2} = - \sqrt{2} E_{t} sin θ sin \frac{π}{P} .$

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Table of Contents for 7.3 Proving Trigonometric Identities

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Table of Contents for
7.3 Proving Trigonometric Identities