How do you express f(theta)=sin^2(theta)+6cot^2(theta)-8cos^4thetaf(θ)=sin2(θ)+6cot2(θ)8cos4θ in terms of non-exponential trigonometric functions?

1 Answer
Jul 10, 2016

We know that, cos 2x=2cos^2x-1=1-2sin^2xcos2x=2cos2x1=12sin2x

From these, we get, sin^2x=(1-cos2x)/2,...........(1) and, cos^2x=(1+cos2x)/2.................(2)

Now, cos^4x=(cos^2x)^2={(1+cos2x)/2}^2=1/4{1+2cos2x+cos^2(2x)}

Here, to convert cos^2(2x) into multiple angle, we will reuse (2), and, put 2x in place of x, to get,

cos^2(2x)=(1+cos2(2x))/2=(1+cos4x)/2.

Therefore, cos^4x=1/4{1+2cos2x+(1+cos4x)/2}=1/8(3+4cos2x+cos4x).........................(3)

Finally, we have, cot^2x=cos^2x/sin^2x=(1+cos2x)/(1-cos2x).....(4)

Using, (1),(3) and (4), we have,

f(theta)=1/2(1-cos2theta)+6{(1+cos2theta)/(1-cos2theta)}-(3+4cos2theta+cos4theta)

We can still go further and express f(theta) as a rational function, as follows :

f(theta)={(1-cos2theta)^2+12(1+cos2theta)-2(1-cos2theta)(3+4cos2theta+cos4theta)}/{2(1-cos2theta)

Where, Nr.=1-2cos2theta+cos^2(2theta)+12+12cos2theta-2(3+4cos2theta+cos4theta-3cos2theta-4cos^2(2theta)-cos4thetacos2theta)

=13+10cos2theta+cos^2(2theta)-6-2cos2theta-2cos4theta+8cos^2(2theta)+2cos4thetacos2theta

=7+8cos2theta-2cos4theta+9cos^2(2theta)+cos(4theta+2theta)+cos(4theta-2theta)

=7+8cos2theta-2cos4theta+9{(1+cos4theta)/2}+cos6theta+cos2theta
=1/2(23+18cos2theta+5cos4theta+2cos6theta)

:. f(theta)=(23+18cos2theta+5cos4theta+2cos6theta)/(4(1-cos2theta))