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

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Answer 1 · 2016-06-23T14:11:34

$sin4theta-cot2theta$

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

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

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

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

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

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

How do you simplify sin4theta-cot2thetasin4theta-cot2theta to trigonometric functions of a unit thetatheta?