How do you simplify #cos4theta-cot2theta# to trigonometric functions of a unit #theta#?

Question

1455 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-05-30T15:07:28

Please see the explanation below

First, compute #cos(4theta)#

#cos(4theta)=cos(2theta+2theta)#

#=cos(2theta)cos(2theta)-sin(2theta)sin(2theta)#

#=(cos^2theta-sin^2theta)^2-(2sinthetacostheta)^2#

#=cos^4theta-2cos^2thetasin^2theta+sin^4theta-4sin^2thetacos^2theta#

#=cos^4theta-6sin^2thetacos^2theta+sin^4theta#

#=cos^4theta-6(1-cos^2theta)cos^2theta+(1-cos^2theta)^2#

#=cos^4theta-6cos^2theta+6cos^4theta+1-2cos^2theta+cos^4theta#

#=1-8cos^2theta+8cos^4theta#

Then, compute #cot(2theta)#

#cot(2theta)=1/tan(2theta)=1/((2tantheta)/(1-tan^2theta))#

#=(1-tan^2theta)/(2tantheta)#

#=1/(2tantheta)-1/2tantheta#

#=1/2cottheta-1/(2cottheta)#

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

Finally,

#cos(4theta)-cot(2theta)=1-8cos^2theta+8cos^4theta-1/2((cot^2theta-1)/(cottheta))#