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Question: If \(\sin\theta = \frac{12}{13},(0 < \theta < \frac{\pi}{2})\) and \(\cos\varphi = - \frac{3}{5},\le...

If sin⁡θ=1213,(0<θ<π2)\sin\theta = \frac{12}{13},(0 < \theta < \frac{\pi}{2}) and cos⁡φ=−35,(π<φ<3π2)\cos\varphi = - \frac{3}{5},\left( \pi < \varphi < \frac{3\pi}{2} \right). Then sin⁡(θ+φ)\sin(\theta + \varphi) will be

A

−5661\frac{- 56}{61}

B

−5665\frac{- 56}{65}

C

165\frac{1}{65}

D

– 56

Answer

−5665\frac{- 56}{65}

Explanation

Solution

We have sin⁡θ=1213\sin\theta = \frac{12}{13} cos⁡θ=1−sin⁡2θ=1−(1213)2=513\cos\theta = \sqrt{1 - \sin^{2}\theta} = \sqrt{1 - \left( \frac{12}{13} \right)^{2}} = \frac{5}{13}

and cos⁡φ=−35,sin⁡φ=1−925=−45\cos\varphi = \frac{- 3}{5},\sin\varphi = \sqrt{1 - \frac{9}{25}} = \frac{- 4}{5}, [∵π<φ<3π2]\left\lbrack \because\pi < \varphi < \frac{3\pi}{2} \right\rbrack

Now, sin⁡(θ+φ)=sin⁡θ.cos⁡φ+cos⁡θ.sin⁡φ\sin(\theta + \varphi) = \sin\theta.\cos\varphi + \cos\theta.\sin\varphi

=(1213)(−35)+(513)(−45)=−3665−2065=−5665= \left( \frac{12}{13} \right)\left( \frac{- 3}{5} \right) + \left( \frac{5}{13} \right)\left( \frac{- 4}{5} \right) = \frac{- 36}{65} - \frac{20}{65} = \frac{- 56}{65}.