Mathematics › Trigonometric Equations › Solving Trigonometric Equation
We have,sin5θ=sin3θ+2θ⇒sin5θ=sin3θcos2θ+cos3θsin2θ⇒sin5θ=3sinθ-4sin3θ2cos2θ-1+4cos3θ-3cosθ2sinθcosθ⇒sin5θ=6sinθcos2θ-3sinθ-8sin3θcos2θ+4sin3θ+8sinθcos4θ-6sinθcos2θ⇒sin5θ=-3sinθ-8sin2θsinθcos2θ+4sin2θsinθ+8sinθcos4θ⇒sin5θ=-3sinθ-81-cos2θsinθcos2θ+41-cos2θsinθ+8sinθcos4θ⇒sin5θ=-3sinθ-8sinθcos2θ+8sinθcos4θ+4sinθ-4sinθcos2θ+8sinθcos4θ⇒sin5θ=16sinθcos4θ-12sinθcos2θ+sinθ
We have,
sin5θ=sin3θ+2θ
⇒sin5θ=sin3θcos2θ+cos3θsin2θ
⇒sin5θ=3sinθ-4sin3θ2cos2θ-1+4cos3θ-3cosθ2sinθcosθ
⇒sin5θ=6sinθcos2θ-3sinθ-8sin3θcos2θ+4sin3θ+8sinθcos4θ-6sinθcos2θ
⇒sin5θ=-3sinθ-8sin2θsinθcos2θ+4sin2θsinθ+8sinθcos4θ
⇒sin5θ=-3sinθ-81-cos2θsinθcos2θ+41-cos2θsinθ+8sinθcos4θ
⇒sin5θ=-3sinθ-8sinθcos2θ+8sinθcos4θ+4sinθ-4sinθcos2θ+8sinθcos4θ
⇒sin5θ=16sinθcos4θ-12sinθcos2θ+sinθ
Asked in: AP EAMCET 2022 (04 Jul Shift 2)
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