Prove sin 3-sin-sin 2-cos 3-cos-cos 3-tan 2-

[ sin #160; 3θ +sin #160; θ ...

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Standard XII

Mathematics

Sum of n Terms

Prove sin 3...

Question

Prove $\frac{sin 3 θ + sin θ + sin 2 θ}{cos 3 θ + cos θ + cos 3 θ} = tan 2 θ$

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Solution

$\begin{matrix} \frac{sin 3 θ + sin θ + sin 2 θ}{cos 3 θ + cos θ + cos 2 θ} = tan 2 θ ⎡ ⎢ ⎢ ⎣ \begin{matrix} sin x + sin y = 2 sin \frac{x + y}{2} cos \frac{z - y}{2} cos x + cos y = 2 cos \frac{x + y}{2} cos \frac{x + y}{2} \end{matrix} ⎤ ⎥ ⎥ ⎦ L H S = \frac{2 sin \frac{3 θ +}{2} cos \frac{3 θ}{2} + sin 2 θ}{2 cos \frac{3 θ + θ}{2} cos \frac{3 - θ}{2} + cos 2 θ} L H S = \frac{2 sin 2 θ cos θ + sin 2 θ}{2 cos 2 θ cos θ + cos 2 θ} L H S = \frac{sin 2 θ (2 cos θ + 1)}{cos 2 θ (2 cos θ + 1)} = tan 2 θ = R H S \end{matrix}$

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$Solve the following equations : (i) c o s θ + c p s 2 θ + c o s 3 θ = 0 (i i) c o s θ + c o s 3 θ - c o s 2 θ = 0 (i i i) s i n θ + s i n 5 θ = s i n 3 θ (i v) c o s θ c o s 2 θ c o s 3 θ = \frac{1}{4} (v) c o s θ + s i n θ = c o s 2 θ + s i n 2 θ (v i) s i n θ + s i n 2 θ + s i n 3 θ = 0 (v i i) s i n θ + s i n 2 θ + s i n 3 θ + s i n 4 θ = 0 (v i i i) s i n 3 θ - s i n θ = 4 c o s^{2} θ - 2 (i x) s i n 2 θ - s i n 4 θ + s i n 6 θ = 0$

$\frac{c o s^{3} θ + s i n^{3} θ}{c o s θ + s i n θ} + \frac{c o s^{3} θ - s i n^{3} θ}{c o s θ - s i n θ} = 2$

Q. Solve:

\frac{sin 3 θ + cos 3 θ}{cos θ - sin θ}

Q. Simplify :

\frac{{sin}^{3} θ + {cos}^{3} θ}{sin θ + cos θ} + sin θ cos θ

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Prove sin3θ+sinθ+sin2θcos3θ+cosθ+cos3θ=tan2θ

sin3θ+sinθ+sin2θcos3θ+cosθ+cos2θ=tan2θ⎡⎢ ⎢⎣sinx+siny=2sinx+y2cosz−y2cosx+cosy=2cosx+y2cosx+y2⎤⎥ ⎥⎦LHS=2sin3θ+2cos3θ2+sin2θ2cos3θ+θ2cos3−θ2+cos2θLHS=2sin2θcosθ+sin2θ2cos2θcosθ+cos2θLHS=sin2θ(2cosθ+1)cos2θ(2cosθ+1)=tan2θ=RHS

Prove $\frac{sin 3 θ + sin θ + sin 2 θ}{cos 3 θ + cos θ + cos 3 θ} = tan 2 θ$