Solve- sin 5-sin-sin 3-

As sin (+sin 1)=2sin(+1)2cos( 1)2 ⇒sin 5θ +sinθ =2sin 3θcos 2θ =sin 3θ ⇒sin 3θ (2cos 2θ 1)=0 ⇒sin 3θ =0 ⇒ 3θ =nπ ⇒θ =nπ3 or ...

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Byju's Answer

Standard XII

Mathematics

Sum of Trigonometric Ratios in Terms of Their Product

Solve: sin ...

Question

Solve: $sin 5 θ + sin θ = sin 3 θ$

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Solution

As $sin (+ sin 1) = 2 sin \frac{(+ 1)}{2} cos \frac{(- 1)}{2}$
$\Rightarrow sin 5 θ + sin θ = 2 sin 3 θ cos 2 θ = sin 3 θ$
$\Rightarrow sin 3 θ (2 cos 2 θ - 1) = 0$
$\Rightarrow sin 3 θ = 0$
$\Rightarrow 3 θ = n π$
$\Rightarrow θ = \frac{n π}{3}$
or $2 cos 2 θ - 1 = 0$
$\Rightarrow cos 2 θ = \frac{1}{2}$
$\Rightarrow 2 θ = 2 n π \pm \frac{π}{3}$
$\Rightarrow θ = n π \pm \frac{π}{6}$ .

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Similar questions

Q. Solve

sin θ + sin 5 θ = sin 3 θ; 0 \leq θ \leq π

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

\frac{sin 3 θ + 2 sin 5 θ + sin 7 θ}{sin θ + 2 sin 3 θ + sin 5 θ} = cos 2 θ + sin 2 θ . cot 3 θ

Q. If the value of

sin θ + sin 3 θ + sin 5 θ + . . . + sin (2 n - 1) θ = \frac{{sin}^{a} n θ}{sin θ}

Find

a .

Q. Find the number of roots of the equation

sin θ + sin 5 θ = sin 3 θ

[0, π]

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Solve: sin5θ+sinθ=sin3θ

As sin(+sin1)=2sin(+1)2cos(−1)2⇒sin5θ+sinθ=2sin3θcos2θ=sin3θ⇒sin3θ(2cos2θ−1)=0⇒sin3θ=0⇒3θ=nπ⇒θ=nπ3or 2cos2θ−1=0⇒cos2θ=12⇒2θ=2nπ±π3⇒θ=nπ±π6.

Solve: $sin 5 θ + sin θ = sin 3 θ$