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Miscellaneous Exercise 3 · Q117

Q.Show that 2sin⁡−135=tan⁡−12472\sin^{-1}\dfrac{3}{5} = \tan^{-1}\dfrac{24}{7}.

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Let sin⁡−135=θ\sin^{-1}\dfrac35=\theta, so sin⁡θ=35,cos⁡θ=45,tan⁡θ=34\sin\theta=\dfrac35,\cos\theta=\dfrac45,\tan\theta=\dfrac34. tan⁡2θ=2tan⁡θ1−tan⁡2θ=2⋅341−916=3/27/16=32×167=247\tan2\theta=\dfrac{2\tan\theta}{1-\tan^2\theta}=\dfrac{2\cdot\frac34}{1-\frac9{16}}=\dfrac{3/2}{7/16}=\dfrac{3}{2}\times\dfrac{16}{7}=\dfrac{24}{7}. Since θ∈(0,π2)\theta\in\left(0,\dfrac{\pi}{2}\right) (as sin⁡θ=3/5>0\sin\theta=3/5>0 and θ\theta is a principal sin⁡−1\sin^{-1} value), 2θ∈(0,π)2\theta\in(0,\pi); and since tan⁡2θ=24/7>0\tan2\theta=24/7>0, in fact $2\theta\in\left(0,\dfrac{\pi}{2} …

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