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
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.