Sin- -sin 2- 1-cos- -cos 2- -tan-

sinθ+sin2θ1+cosθ+cos2θ =sinθ+2sinθcosθ1+cosθ+2cos^2θ 1 =sinθ(1+2cosθ)cosθ(1+2cosθ)=tanθ ...

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Trigonometric Ratios of Allied Angles

sinθ +sin 2θ ...

Question

$\frac{sin θ + sin 2 θ}{1 + cos θ + cos 2 θ} = tan θ$

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Solution

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Prove that:

$\frac{c o s θ}{(1 - t a n θ)} - \frac{s i n^{2} θ}{(c o s θ - s i n θ)} = (c o s θ + s i n θ)$

\cos θ + \cos 2 θ + \cos 3 θ = 0

\cos θ + \cos 3 θ - \cos 2 θ = 0

\sin θ + \sin 5 θ = \sin 3 θ

\cos θ \cos 2 θ \cos 3 θ = \frac{1}{4}

\cos θ + \sin θ = \cos 2 θ + \sin 2 θ

\sin θ + \sin 2 θ + \sin 3 = 0

\sin θ + \sin 2 θ + \sin 3 θ + \sin 4 θ = 0

\sin 3 θ - \sin θ = 4 \cos^{2} θ - 2

\sin 2 θ - \sin 4 θ + \sin 6 θ = 0

\frac{sin θ}{sin 2 θ} = \frac{cos θ}{cos 2 θ}

\frac{sin θ + sin 2 θ}{cos θ + cos 2 θ}

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