How do you simplify $sin 4 θ - cot 2 θ$ $sin4theta-cot2theta$ to trigonometric functions of a unit $θ$ $theta$ ?

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How do you simplify $sin 4 θ - cot 2 θ$ $sin4theta-cot2theta$ to trigonometric functions of a unit $θ$ $theta$ ?

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Answer 1 · 2016-06-23T14:11:34

$sin 4 θ - cot 2 θ$ $sin4theta-cot2theta$

= $4 sin θ {cos}^{3} θ - 4 {sin}^{3} θ cos θ - \frac{1}{2} (cot θ - tan θ)$ $4sinthetacos^3theta-4sin^3thetacostheta-1/2(cottheta-tantheta)$

Explanation:

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

Hence $sin 4 θ - cot 2 θ = 2 sin 2 θ cos 2 θ - \frac{cos 2 θ}{sin 2 θ}$ $sin4theta-cot2theta=2sin2thetacos2theta-(cos2theta)/(sin2theta)$

= $4 sin θ cos θ ({cos}^{2} θ - {sin}^{2} θ) - \frac{{cos}^{2} θ - {sin}^{2} θ}{2 sin θ cos θ}$ $4sinthetacostheta(cos^2theta-sin^2theta)-(cos^2theta-sin^2theta)/(2sinthetacostheta)$

= $4 sin θ {cos}^{3} θ - 4 {sin}^{3} θ cos θ - \frac{cos θ}{2 sin θ} + \frac{sin θ}{2 cos θ}$ $4sinthetacos^3theta-4sin^3thetacostheta-(costheta)/(2sintheta)+(sintheta)/(2costheta)$

= $4 sin θ {cos}^{3} θ - 4 {sin}^{3} θ cos θ - \frac{1}{2} (cot θ - tan θ)$ $4sinthetacos^3theta-4sin^3thetacostheta-1/2(cottheta-tantheta)$

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How do you simplify $sin 4 θ - cot 2 θ$ $sin4theta-cot2theta$ to trigonometric functions of a unit $θ$ $theta$ ?

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How do you simplify $sin 4 θ - cot 2 θ$ $sin4theta-cot2theta$ to trigonometric functions of a unit $θ$ $theta$ ?