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### Double- and Half-Angle Identities

```
Date: 11/24/98 at 23:34:03
From: Jeff Hefner
Subject: Double- and Half-Angle Identities

I am having trouble understanding the purpose of double and half
angles. For example, use double-angle or half-angle identities to find
the exact value of each:

1) sin 5pi/12
2) cos 7pi/8
```

```
Date: 11/25/98 at 11:58:17
From: Doctor Anthony
Subject: Re: Double- and Half-Angle Identities

In general, we use the double- and half-angle formulas when we want to
calculate the sine, cosine, etc, of an angle whose value isn't readily
known. For example, we know that cos(30) = sqrt(3)/2. But what is
cos(15)?

Recall the half-angle formula for cosine:

cos(2 theta) = cos^2(theta) - sin^2(theta) = 2 cos^2(theta) - 1

For sine it is:

sin(2 theta) = 2 sin(theta) cos(theta)

Then we know that sqrt(3)/2 = cos(30) = 2 cos^2(15) - 1. So:

cos(15) = sqrt((1 + sqrt(3))/4)

Turning to your first problem, for the first one, note in degrees
sin(5 pi/12) = sin(75) = cos(15).

Then we have:

cos(30) = 2 cos^2(15) - 1
sqrt(3)/2 = 2 cos^2(15) - 1
2 cos^2(15) = 1 + sqrt(3)/2

2+sqrt(3)
cos^2(15) = ----------
4

sqrt(2+sqrt(3))
cos(15)  =  ---------------
2

So we must find sqrt(2+sqrt(3)). Consider:

[sqrt(3)+1]^2 = 3 + 1 + 2 sqrt(3) = 4 + 2 sqrt(3)
= 2[2 + sqrt(3)]

and so:

2 + sqrt(3) = [sqrt(3) + 1]^2/2
sqrt([2+sqrt(3)]) = [sqrt(3)+1]/sqrt(2)

Therefore:

sqrt(3) + 1
sin (5 pi/12) = cos(15) = -----------
2 sqrt(2)

For the second problem, cos(7pi/8):

In degrees cos(7.pi/8) = -cos(22.5).
Thus:

2 tan(22.5)
tan(45) =  ---------------- = 1
1 - tan^2(22.5)

2 tan(22.5) = 1 - tan^2(22.5)

tan^2(22.5) + 2 tan(22.5) - 1 = 0

tan(22.5) = [-2 + sqrt(4+4)]/2  = [-2+ 2sqrt(2)]/2
= sqrt(2)-1

tan^2(22.5) = 2 + 1 - 2 sqrt(2) = 3 - 2 sqrt(2)

sec^2(22.5) = 1 + tan^2(22.5) = 4 - 2 sqrt(2)

cos^2(22.5) = 1/[4 - 2 sqrt(2)]

So cos(7 pi/8) = -1/[sqrt(4 - 2 sqrt(2)]

- Doctor Anthony, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Trigonometry

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