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Question 154 of 177

Q.Show that: sin⁡−1(817)+sin⁡−1(35)=sin⁡−1(7785)\sin^{-1}\left(\dfrac{8}{17}\right) + \sin^{-1}\left(\dfrac{3}{5}\right) = \sin^{-1}\left(\dfrac{77}{85}\right).

Maharashtra MsbshseMaharashtra HSC (MSBSHSE) Board 2020Subjective· 3mImportance★★★★★
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Let A=sin⁡−1817A=\sin^{-1}\frac{8}{17}, B=sin⁡−135B=\sin^{-1}\frac{3}{5}; compute sin⁡(A+B)\sin(A+B).

Let A=sin⁡−1817A=\sin^{-1}\dfrac{8}{17}, so sin⁡A=817\sin A = \dfrac{8}{17}, and since AA is acute, cos⁡A=1517\cos A = \dfrac{15}{17} (as 8,15,178,15,17 form a Pythagorean triple).

Let B=sin⁡−135B=\sin^{-1}\dfrac{3}{5}, so sin⁡B=35\sin B = \dfrac35, cos⁡B=45\cos B = \dfrac45 (as 3,4,53,4,5 form a Pythagorean triple).

sin⁡(A+B)=sin⁡Acos⁡B+cos⁡Asin⁡B=817⋅45+1517⋅35=3285+4585=7785\sin(A+B) = \sin A\cos B + \cos A\sin B = \dfrac8{17}\cdot\dfrac45 + \dfrac{15}{17}\cdot\dfrac35 = \dfrac{32}{85}+\dfrac{45}{85} = \dfrac{77}{85}

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