How do you use half angle formula to find #sin 67.5#?

1 Answer
Shantelle
May 4, 2018

#sqrt(2+sqrt2)/2#

Explanation:

#sin67.5^@#

The half angle formula for #sin(theta/2) = +-sqrt((1-costheta)/2)#.

The #+-# sign depends on which quadrant your angle is in. Since #67.5^@# is in the 1st quadrant and sine is positive in that quadrant, we know that it is positive.

#67.5^@ * 2 = 135^@#, so:
#sin(theta/2) = sqrt((1- cos135^@)/2)#

# # #quadquadquadquadquadquadquad=sqrt((1-(-sqrt2/2))/2)#

# # #quadquadquadquadquadquadquad=sqrt((2/2+sqrt2/2)/2)#

# # #quadquadquadquadquadquadquad=sqrt(((2+sqrt2)/2)/2)#

# # #quadquadquadquadquadquadquad=sqrt((2+sqrt2)/4)#

# # #quadquadquadquadquadquadquad=sqrt(2+sqrt2)/2#

Hope this helps!

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