Question

How do you simplify `f(theta)=cot(theta/2)+3sin(theta/4+pi/2)` to trigonometric functions of a unit `theta`?

Answer 1

`f(theta)=1/2(cot(theta/4)-tan(theta/4))+3cos(theta/4)`

Explanation:

Given:
`f(theta)=cot(theta/2)+3sin(theta/4+pi/2)`
`cot(theta/2)=cos(theta/2)/sin(theta/2)`
`sin(theta/4+pi/2)=sin(pi/2+theta/4)`
`=cos(theta/4)`
`3sin(theta/4+pi/2)=3cos(theta/4)`
Thus,
`cot(theta/2)+3sin(theta/4+pi/2)=cos(theta/2)/sin(theta/2)+3cos(theta/4)`
`cos(theta/2)=cos2(theta/4)`

`cos(theta/2)=cos^2(theta/4)-sin^2(theta/4)`
`sin(theta/2)=sin2(theta/4)`
`sin(theta/2)=2sin(theta/4)cos(theta/4)`
Thus,
`cot(theta/2)=(cos^2(theta/4)-sin^2(theta/4))/(2sin(theta/4)cos(theta/4)`
`cot(theta/2)=1/2(cot(theta/4)-tan(theta/4))`

Now,

`cot(theta/2)+3sin(theta/4+pi/2)=1/2(cot(theta/4)-tan(theta/4))+3cos(theta/4)`
`f(theta)=1/2(cot(theta/4)-tan(theta/4))+3cos(theta/4)`