How do you find #cos(1/2sin^-1(sqrt3/2))#?

1 Answer
Bdub
Oct 13, 2016

#cos(1/2 sin^-1(sqrt3/2))=sqrt3/2#

Explanation:

#cos(1/2 sin^-1(sqrt3/2))#

The restriction for the range of arcsin x is #[-pi/2,pi/2]#. Since the argument is positive it means that our triangle is in quadrant I with the opposite side of #sqrt 3# and hypotenuse 2 and therefore the adjacent is 1. Note that we do't need to know the angle even though we can tell what it is. Let's just call the angle #theta# then we need to find #cos (1/2 theta)#. So we use the formula #cos (1/2theta)=sqrt(1/2(1+costheta)# to evaluate #cos (1/2 theta)#. That is,

#cos (1/2 theta)=sqrt(1/2(1+costheta)#

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

#=sqrt(1/2(3/2)#

#=sqrt(3/4)=sqrt3/2#

Related questions
See all questions in Basic Inverse Trigonometric Functions
Impact of this question
2516 views around the world
You can reuse this answer
Creative Commons License