How do you express #f(theta)=cos(theta/4)+csc(theta/2)+sin(theta/2)# in terms of trigonometric functions of a whole theta?

1 Answer
Abdullah M.
Oct 21, 2017

# f(x) = (sqrt((sqrt(cosx+1)+sqrt2)/(2sqrt2))) + sqrt((2+cosx)/(2cosx)) +sqrt((cosx-1)/2) #

Explanation:

Let #theta# = x

# cos(x/4)# ---> #cos(x/2) = 2cos^2(x/4) - 1 # --> #sqrt((sqrt(cosx+1)+sqrt2)/(2sqrt2)) #

# sinx(x/2)# --> #cos(x)=1-sin^2(x/2)# --> #sqrt((cosx-1)/2)#

#csc(x/2)# --> #sec(x)=1-2csc^2(x/2)# --> #sqrt((2+cosx)/(2cosx))#

# f(x) = (sqrt((sqrt(cosx+1)+sqrt2)/(2sqrt2))) + sqrt((2+cosx)/(2cosx)) +sqrt((cosx-1)/2) #

I could only come up to this, i'm pretty sure it can be simplified further, however, i've done the bulk of it.

Related questions
See all questions in Half-Angle Identities
Impact of this question
1781 views around the world
You can reuse this answer
Creative Commons License