Find the sum of the possible values of -0- - such that sin -sin 4 -sin 7 -0 are

The correct option is D 55π/18 We have sin(θ)+sin(4 θ)+sin(7 θ)=0 ⇒ 2sin(4 θ)cos(3 θ)+sin(4 θ) =0 [Using sin C+sin D=2sinC+D/2cosC D/2] ⇒sin(4 θ)(2cos(3 θ)+1) ...

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Standard XII

Mathematics

Basic Trigonometric Identities

Find the sum ...

Question

Find the sum of the possible values of $θ ϵ (0, π)$ such that $sin (θ) + sin (4 θ) + sin (7 θ) = 0$ are

$\frac{55 π}{18}$

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$\frac{55 π}{16}$

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$\frac{65 π}{18}$

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$\frac{65 π}{14}$

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Solution

The correct option is A
$\frac{55 π}{18}$

We have

$sin (θ) + sin (4 θ) + sin (7 θ) = 0$

$\Rightarrow 2 sin (4 θ) cos (3 θ) + sin (4 θ) = 0$ [Using $sin C + sin D = 2 sin \frac{C + D}{2} cos \frac{C - D}{2}$ ]

$\Rightarrow sin (4 θ) (2 cos (3 θ) + 1) = 0$

$\Rightarrow sin (4 θ) = 0$ or $cos (3 θ) = - \frac{1}{2} = cos (\frac{2 π}{3})$

$\Rightarrow θ = \frac{π}{4}, \frac{π}{2}, \frac{3 π}{4}$ or $3 θ = \frac{2 π}{3}, 2 π \pm \frac{2 π}{3}$

$\Rightarrow θ = \frac{2 π}{9}, \frac{π}{4}, \frac{4 π}{9}, \frac{π}{2}, \frac{3 π}{4}, \frac{8 π}{9}$

$\Rightarrow$ Sum $= \frac{2 π}{9} + \frac{π}{4} + \frac{4 π}{9} + \frac{π}{2} + \frac{3 π}{4} + \frac{8 π}{9}$

$= \frac{8 π + 9 π + 16 π + 18 π + 27 π + 32 π}{36} = \frac{110 π}{36} = \frac{55 π}{18}$

Suggest Corrections

Find the sum of the possible values of θϵ(0,π) such that sin(θ)+sin(4θ)+sin(7θ)=0 are

Find the sum of the possible values of $θ ϵ (0, π)$ such that $sin (θ) + sin (4 θ) + sin (7 θ) = 0$ are