How do you simplify #sin4theta-cot2theta# to trigonometric functions of a unit #theta#?

1 Answer
Shwetank Mauria
Jun 23, 2016

#sin4theta-cot2theta#

= #4sinthetacos^3theta-4sin^3thetacostheta-1/2(cottheta-tantheta)#

Explanation:

we use #sin2A=2sinAcosA# and #cos2A=cos^2A-sin^2A#

Hence #sin4theta-cot2theta=2sin2thetacos2theta-(cos2theta)/(sin2theta)#

= #4sinthetacostheta(cos^2theta-sin^2theta)-(cos^2theta-sin^2theta)/(2sinthetacostheta)#

= #4sinthetacos^3theta-4sin^3thetacostheta-(costheta)/(2sintheta)+(sintheta)/(2costheta)#

= #4sinthetacos^3theta-4sin^3thetacostheta-1/2(cottheta-tantheta)#

Related questions
See all questions in Double Angle Identities
Impact of this question
1064 views around the world
You can reuse this answer
Creative Commons License