How do you simplify #cot8theta# to trigonometric functions of a unit #theta#?

1 Answer
Dec 7, 2017

#cot8theta=(cot^8theta-28cot^6theta+58cot^4theta-28cot^2theta+1)/(8cot^7theta-56cot^5theta+56cot^3theta-8cottheta)#

Explanation:

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

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

= #(((cot^2 2theta-1)/(2cot2theta))^2-1)/(2(cot^2 2theta-1)/(2cot2theta))#

= #((cot^2 2theta-1)^2-(2cot2theta)^2)/(4cot2theta(cot^2 2theta-1)#

= #(cot^4 2theta+1-2cot^2 2theta-4cot^2 2theta)/(4cot^3 2theta-4cot2theta)#

= #(cot^4 2theta-6cot^2 2theta+1)/(4cot^3 2theta-4cot2theta)#

= #(((cot^2theta-1)/(2cottheta))^4-6((cot^2theta-1)/(2cottheta))^2+1)/(4((cot^2theta-1)/(2cottheta))^3-4((cot^2theta-1)/(2cottheta)))#

= #((cot^2theta-1)^4-24cot^2theta(cot^2theta-1)^2+(2cottheta)^4)/(8cottheta(cot^2theta-1)^3-32cot^3theta(cot^2theta-1))#

= #((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)#

= #(cot^8theta-28cot^6theta+58cot^4theta-28cot^2theta+1)/(8cot^7theta-56cot^5theta+56cot^3theta-8cottheta)#