Question

How do you evaluate `sin ((7pi)/8)` using the half angle formula?

Answer 1

`- sqrt(2 - sqrt2)/2`

Explanation:

Trig table, unit circle -->
`sin ((7pi)/8) = sin (-pi/8 + 2pi) = - sin (pi/8)`
Find `sin (pi/8)` by using trig identity:
`cos 2a = 1 - 2sin^2 a.`
`cos (pi/4) = sqrt2/2 = 1 - 2sin^2 (pi/8)`
`2sin^2 (pi/8) = 1 - sqrt2/2 = (2 - sqrt2)/2`
`sin^2 (pi/8) = (2 - sqrt2)/4`
`sin (pi/8) = sqrt(2 -sqrt2)/2`
The negative answer is rejected because sin (pi/8) is positive.
Finally,
`sin ((7pi)/8) = - sqrt(2 - sqrt2)/2`

Answer 2

`sqrt(2-sqrt2)/2`

Explanation:

This can also be shown through the sine half-angle formula:

`sin(x/2)=+-sqrt((1-cos(x))/2)`

Here, since we want to find `sin((7pi)/8)`, we know that `x/2=(7pi)/8` and `x=(7pi)/4`.

`sin((7pi)/8)=sqrt((1-cos((7pi)/4))/2)`

Note that the `+-` sign has just turned into a positive sign: the sine of `(7pi)/8` will be positive since `(7pi)/8` is in the second quadrant.

`sin((7pi)/8)=sqrt((1-(sqrt2/2))/2)=sqrt((2-sqrt2)/4)=sqrt(2-sqrt2)/2`

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We can show that `sin(x/2)=+-sqrt((1-cos(x))/2)` using the cosine double-angle formula.

`cos(2x)=1-2sin^2(x)`

This is the same as saying

`cos(x)=1-2sin^2(x/2)`

The argument of the cosine function is double that of the sine function--just expressed differently.

Solving for `sin(x/2)`, we see that

`2sin^2(x/2)=1-cos(x)`

`sin^2(x/2)=(1-cos(x))/2`

`sin(x/2)=+-sqrt((1-cos(x))/2)`