How do you simplify f(theta)=tan4theta+sin2theta+3cos8theta to trigonometric functions of a unit theta?

1 Answer
Jun 16, 2018

Use double angle formulae to obtain (4tantheta)/((1-tan^2theta)^3)(1-6tan^2theta+4tan^4theta)+2sinthetacostheta+3-96sin^2thetacos^2theta+384sin^4thetacos^4theta

Explanation:

Notice that the three multiples of theta are all powers of 2, so we can use double angle formulae to expand them. Start with

sin2theta=2sinthetacostheta and
cos2theta=cos^2theta-sin^2theta=1-2sin^2theta

From these deduce that

tan2theta=(sin2theta)/(cos2theta)=(2sinthetacostheta)/(cos^2theta-sin^2theta)=(2tantheta)/(1-tan^2theta)

We can now use these iteratively to deduce formulae for 4theta and 8theta.

tan4theta=(2tan2theta)/(1-tan^2 2theta)=(2((2tantheta)/(1-tan^2theta)))/(1-((2tantheta)/(1-tan^2theta))^2)
=(4tantheta)/(1-tan^2theta)*((1-tan^2theta)^2-4tan^2theta)/((1-tan^2theta)^2)

=(4tantheta)/((1-tan^2theta)^3)(1-6tan^2theta+4tan^4theta)

cos8theta=1-2sin^2 4theta=1-2(2sin2thetacos2theta)^2
=1-8sin^2 2thetacos^2 2theta=1-8sin^2 2theta(1-sin^2 2theta)
=1-8(2sinthetacostheta)^2(1-(2sinthetacostheta)^2)
=1-32sin^2thetacos^2theta(1-4sin^2thetacos^2theta)
=1-32sin^2thetacos^2theta+128sin^4thetacos^4theta

Combining these results:
f(theta)=tan4theta+sin2theta+3cos8theta=
(4tantheta)/((1-tan^2theta)^3)(1-6tan^2theta+4tan^4theta)+2sinthetacostheta+3-96sin^2thetacos^2theta+384sin^4thetacos^4theta