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

2 Answers
Nghi N.
Jun 1, 2016

#- 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#

mason m
Jun 3, 2016

#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#

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)#

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