Question

How do you simplify `f(theta)=-tan4theta+sin2theta+cos4theta` to trigonometric functions of a unit `theta`?

Answer 1

`-(4 sin theta cos theta (1-2 sin^2 theta))/(1-8sin^2theta cos^2theta)`
`qquadqquad -2sin theta cos theta+1-8sin^2theta cos^2theta`

Explanation:

We start with the double angle identities :

`sin(2 theta) = 2 sin theta cos theta`

and

`cos(2 theta) = cos^2 theta - sin^2 theta = 2cos^2 theta-1 = 1-2sin^2 theta`

Now

`cos (4 theta) = cos(2times 2theta)`
`qquadqquadqquad = 1-2 sin^2(2theta)`
`qquadqquadqquad = 1-2(2 sin theta cos theta)^2`
`qquadqquadqquad = 1-8sin^2theta cos^2theta`

and

`sin(4 theta) = sin(2 times 2theta)`
`qquadqquadqquad = 2 sin(2theta)cos(2theta)`
`qquadqquadqquad = 2(2 sin theta cos theta)(1-2sin^2theta)`
`qquadqquadqquad = 4 sin theta cos theta (1-2 sin^2 theta)`

leading to

`tan(4theta) = sin(4 theta)/cos(4 theta)`
`qquadqquadqquad = (4 sin theta cos theta (1-2 sin^2 theta))/(1-8sin^2theta cos^2theta)`

Thus

`f(theta) = -tan(4 theta)+sin(2 theta)+cos(4 theta)`
`qquad\ quad=-(4 sin theta cos theta (1-2 sin^2 theta))/(1-8sin^2theta cos^2theta)`
`qquadqquad -2sin theta cos theta+1-8sin^2theta cos^2theta`