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JEE MainMathematicsToughTrigonometry

The values
of θ\theta satisfying sin⁡7θ=sin4θ−sinθ\sin{7\theta} = sin4\theta - sin\theta and 0<θ<π2 are πa,πb.0 < \theta < \frac{\pi}{2}\ are\ \frac{\pi}{a},\frac{\pi}{b}. Then
the value of (a+b-10) is

Show answer and solution
sin⁡7θ+sin⁡θ=sin⁡4θ\sin{7\theta} + \sin\theta = \sin{4\theta}
2sin⁡4θcos⁡3θ=sin⁡4θ2\sin{4\theta\cos{3\theta}} = \sin{4\theta}
2sin⁡4θcos⁡3θ−sin⁡4θ=02\sin{4\theta\cos{3\theta}} - \sin{4\theta} = 0
sin⁡4θ[2cos⁡3θ−1]=0\sin{4\theta\left\lbrack 2\cos{3\theta} - 1 \right\rbrack} = 0
sin⁡4θ=0 or cos3θ=12\sin{4\theta} = 0{\ or\ cos}{3\theta} = \frac{1}{2}
⇒θ=π4\Rightarrow \theta = \frac{\pi}{4}  and
 ⇒θ=π9\Rightarrow \theta = \frac{\pi}{9}
θ=π9,π4\theta = \frac{\pi}{9},\frac{\pi}{4}

Source: Grass-G / M-18 Mathematics Paper 27 Nov Third.docx

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