Question

prove that sin theta-2sin3theta = (2cos3theta - cos theta)tan theta .

Answer 1

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Answer 2

$\sf{\underline{\boxed{\large{\blue{\mathsf{Solution}}}}}}$

$\bold{ sin\theta - 2sin^3\theta = ( 2 cos^3\theta - cos\theta ) tan \theta }$

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$\sf{\bold{\green{\underline{\underline{LHS}}}}}$

$\sf\implies sin\theta - 2sin^3 \theta$

• taking common

$\sf\implies sin\theta ( 1 - 2sin^2\theta )$

$\sf{\red{\boxed{\bold{ sin^2\theta = 1 - cos^2\theta}}}}$

$\sf\implies sin\theta [ 1 - 2 ( sin^2\theta )]$

$\sf\implies sin\theta [ 1 - 2 ( 1 - cos^2\theta )]$

$\sf\implies sin\theta ( 1 - 2 + 2cos^2\theta )$

$\sf\implies sin\theta ( 2cos^2\theta - 1 )$

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$\sf{\bold{\green{\underline{\underline{RHS}}}}}$

$\sf\implies ( 2cos^3\theta - cos\theta )tan\theta$

$\sf\implies cos\theta( 2cos^2\theta - 1 )tan\theta$

$\sf{\red{\boxed{\bold{tan\theta = \dfrac{sin\theta}{cos\theta}}}}}$

$\sf\implies ( 2cos^2\theta - 1 )cos\theta \times \dfrac{sin\theta}{cos\theta}$

$\sf\implies ( 2cos^2\theta - 1 )sin\theta$

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$\sf{\bold{\green{\underline{\underline{Compare}}}}}$

$\sf( 2cos^2\theta - 1 )sin\theta$ = $\sf ( 2cos^2\theta - 1 )sin\theta$

$\sf{\bold { LHS = RHS }}$

......Hence proved