How do you use the half angle identity to find sin 105?

2 Answers
Aviv S.
Mar 20, 2018

#sin105^@=sqrt(2+sqrt3)/2#

Explanation:

We need to use the half angle formula:

#sin(theta/2)=+-sqrt((1-costheta)/2)#

In this case, we want to find #sin(105^@)#, so that's what we want #sin(theta/2)# to equal. To find out what our #theta# is, set these to equal to each other:

#sin(105^@)=sin(theta/2)#

#105^@=theta/2#

#210^@=theta#

This is our #theta#. Now, we can use the half angle formula:

#color(white)=sin(105^@)#

#=sin(210^@/2)#

#=+-sqrt((1-cos(210^@))/2)#

#=+-sqrt((1-(-sqrt3/2))/2)#

#=+-sqrt((1+sqrt3/2)/2)#

#=+-sqrt((1+sqrt3/2)/2)#

#=+-sqrt((2+sqrt3)/4)#

#=+-sqrt(2+sqrt3)/sqrt4#

#=+-sqrt(2+sqrt3)/2#

Since #105^@# is in quadrant II, we know that our answer will be positive that angle is above the #x#-axis (and we are taking the sine). Therefore:

#sin105^@=sqrt(2+sqrt3)/2#

We can check our answer using a calculator (be sure it is in degrees mode):

https://www.desmos.com/calculator

sankarankalyanam
Mar 20, 2018

#color(blue)(sin (105) = = +-(1/2) sqrt(2 + sqrt3)#

Explanation:

http://www2.clarku.edu/~djoyce/trig/identities.html

Given #theta / 2 = 105^@ = (7pi)/12#

#theta = (7pi)/6#

#sin (theta/2) = +- sqrt((1 - cos theta)/2)#

#sin ((7pi)/12) = +- sqrt((1-cos (((7pi)/12)*2))/2)#

#sin ((7pi)/12) = +- sqrt((1 - cos ((7pi)/6))/2)#

But # cos ((7pi)/6) = cos (pi + pi/6) = - cos (pi/6)#

#:. sin ((7pi)/12) = +- sqrt((1 + cos(pi/6))/2)#

#=> +- sqrt((1 + sqrt(3)/2)/2)#

#=> +- sqrt((2 + sqrt3)/4) = +-(1/2) sqrt(2 + sqrt3)#

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