Question

How do you express `f(theta)=-sin(theta/4)-cos(theta/2)+6sin(theta/2)` in terms of trigonometric functions of a whole theta?

Answer 1

With `theta in [0.pi], f(theta)= -sqrt((1-sqrt((1+cos theta)/2))/2)-sqrt((1+cos theta)/2)`

`+6sqrt((1-cos theta)/2)`

Explanation:

I assume that `theta in [0, pi]` so that

#theta/2 and theta/4 in [0, pi/2]

Use `sin A =+-sqrt((1-cos 2A)/2) and cos A = +-sqrt((1+cos 2A)/2)`

Under the assumption, prefixing signs are both positive, for our

application

`f(theta)=-sqrt((1-cos (theta/2))/2)-sqrt((1+cos theta)/2)`

`+6sqrt((1-cos theta)/2)`

`=-sqrt((1-sqrt((1+cos theta)/2))/2)-sqrt((1+cos theta)/2)`

`+6sqrt((1-cos theta)/2)`