Question

Prove that sin2x = {(2sinxcosx), (2tanx/(1 + tan²x))

sin 2x = \(\begin{cases}2\sin\mathrm x\cos\mathrm x\\\frac{2\tan\mathrm x}{1+\tan^2\mathrm x}\end{cases}\)

Answer 1

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 = \(\cfrac{\frac{2\sin \mathrm x\cos \mathrm x}{\cos^2\mathrm x}}{1+\frac{\sin^2\mathrm x}{\cos^2\mathrm x}}\)

= \(\frac{2\tan\mathrm x}{1+\tan^2\mathrm x}\)

Answer 2

(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)