InterviewSolution
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Prove that sin2x = {(2sinxcosx), (2tanx/(1 + tan2x))sin 2x = \(\begin{cases}2\sin\mathrm x\cos\mathrm x\\\frac{2\tan\mathrm x}{1+\tan^2\mathrm x}\end{cases}\) |
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Answer» We have sin(x + y) = sin xcosy + cosxsiny Replacing y by x we get sin 2x = 2sinx cosx Again sin2x = \(\frac{2\sin \mathrm \cos \mathrm x}{\sin^2\mathrm x+\cos^2\mathrm x}\) Dividing numerator and denominator by cos2x, We get sin2x = \(\cfrac{\frac{2\sin \mathrm x\cos \mathrm x}{\cos^2\mathrm x}}{1+\frac{\sin^2\mathrm x}{\cos^2\mathrm x}}\) = \(\frac{2\tan\mathrm x}{1+\tan^2\mathrm x}\) (i) \(\because\) sin(x + y) = sin x cos y + cos x sin y put y = x, we obtain sin 2x = sin x cos x + cos x sin x = 2sin x cos x (ii) \(\because\) sin 2x = 2sin x cos x = 2 \(\frac{\sin \mathrm x}{\cos \mathrm x}\) cos2x = \(\frac{2\tan\mathrm x}{\sec^2\mathrm x}\) (\(\because\) sec x = \(\frac{1}{\cos\mathrm x}\) & tan x = \(\frac{\sin\mathrm x}{\cos\mathrm x}\)) \(=\frac{2\tan\mathrm x}{1+\tan^2\mathrm x}\) (\(\because\) sec2x = 1 + tan2x) |
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