Question

How do you express `cot^3theta-cos^2theta-tan^3theta` in terms of non-exponential trigonometric functions?

Answer 1

`cos^3theta/sin^3theta - cos^2theta - sin^3theta/cos^3theta`

`(cos^6theta - sin^6theta)/(cos^3thetasin^3theta) - cos^2theta`

`(cos^6theta - sin^6theta - cos^5thetasin^3theta)/cos^3theta`

`((cos^2theta)^3 - (sin^2theta)^3 - costhetasintheta(cos^2theta)^2(sin^2theta))/(cos^2theta(costheta))`

Now use the power reduction formulae `cos^2x = (1 + cos2x)/2` and `sin^2x = (1 - cos2x)/2`

`(((1 + cos2theta)/2)^3- ((1 - cos2theta)/2)^3 - costhetasintheta((1 + cos2theta)/2)^2((1 - cos2theta)/2)^2)/((1 + cos2theta)/2costheta)`

This can be simplified further, but I'll leave the algebra up to you.

Hopefully this helps!

Answer 2

`cot^3theta-cos^2theta-tan^3theta=(16cot2theta)/(1-cos4theta)-2cot2theta-(1+cos2theta)/2`

Explanation:

`cot^3theta-cos^2theta-tan^3theta`

= `cot^3theta-tan^3theta-cos^2theta`

= `(cottheta-tantheta)(cot^2theta+tan^2theta+cotthetatantheta)-(1+cos2theta)/2`

= `(cottheta-tantheta)((cottheta+tantheta)^2-cotthetatantheta)-(1+cos2theta)/2`

= `((cos^2theta-sin^2theta)/(sinthetacostheta))((1/(sinthetacostheta))^2-1)-(1+cos2theta)/2`

= `(2cos2theta)/(sin2theta)((2/(sin2theta))^2-1)-(1+cos2theta)/2`

= `2cot2theta(4/(sin^2 2theta)-1)-(1+cos2theta)/2`

= `2cot2theta(8/(1-cos4theta)-1)-(1+cos2theta)/2`

= `(16cot2theta)/(1-cos4theta)-2cot2theta-(1+cos2theta)/2`