How do you simplify cot8thetacot8θ to trigonometric functions of a unit thetaθ?

1 Answer
Dec 7, 2017

cot8theta=(cot^8theta-28cot^6theta+58cot^4theta-28cot^2theta+1)/(8cot^7theta-56cot^5theta+56cot^3theta-8cottheta)cot8θ=cot8θ28cot6θ+58cot4θ28cot2θ+18cot7θ56cot5θ+56cot3θ8cotθ

Explanation:

We can use double angle identity cot2A=(cot^2A-1)/(2cotA)cot2A=cot2A12cotA

As such cot8theta=(cot^2 4theta-1)/(2cot 4theta)cot8θ=cot24θ12cot4θ

= (((cot^2 2theta-1)/(2cot2theta))^2-1)/(2(cot^2 2theta-1)/(2cot2theta))(cot22θ12cot2θ)212cot22θ12cot2θ

= ((cot^2 2theta-1)^2-(2cot2theta)^2)/(4cot2theta(cot^2 2theta-1)(cot22θ1)2(2cot2θ)24cot2θ(cot22θ1)

= (cot^4 2theta+1-2cot^2 2theta-4cot^2 2theta)/(4cot^3 2theta-4cot2theta)cot42θ+12cot22θ4cot22θ4cot32θ4cot2θ

= (cot^4 2theta-6cot^2 2theta+1)/(4cot^3 2theta-4cot2theta)cot42θ6cot22θ+14cot32θ4cot2θ

= (((cot^2theta-1)/(2cottheta))^4-6((cot^2theta-1)/(2cottheta))^2+1)/(4((cot^2theta-1)/(2cottheta))^3-4((cot^2theta-1)/(2cottheta)))(cot2θ12cotθ)46(cot2θ12cotθ)2+14(cot2θ12cotθ)34(cot2θ12cotθ)

= ((cot^2theta-1)^4-24cot^2theta(cot^2theta-1)^2+(2cottheta)^4)/(8cottheta(cot^2theta-1)^3-32cot^3theta(cot^2theta-1))(cot2θ1)424cot2θ(cot2θ1)2+(2cotθ)48cotθ(cot2θ1)332cot3θ(cot2θ1)

= ((cot^8theta-4cot^6theta+6cot^4theta-4cot^2theta+1)-24cot^2theta(cot^4theta-2cot^2theta+1)+4cot^4theta)/(8cottheta(cot^6theta-3cot^4theta+3cot^2theta-1)-32cot^5theta+32cot^3theta)(cot8θ4cot6θ+6cot4θ4cot2θ+1)24cot2θ(cot4θ2cot2θ+1)+4cot4θ8cotθ(cot6θ3cot4θ+3cot2θ1)32cot5θ+32cot3θ

= (cot^8theta-28cot^6theta+58cot^4theta-28cot^2theta+1)/(8cot^7theta-56cot^5theta+56cot^3theta-8cottheta)cot8θ28cot6θ+58cot4θ28cot2θ+18cot7θ56cot5θ+56cot3θ8cotθ