Question

If x = 30°, verify that

(i) `\tan 2x=\frac{2\tan x}{1-\tan ^{2}x`

(ii) `\sin x=\sqrt{\frac{1-\cos 2x}{2}}`

Answer 1

(i) When x = 30°, we have 2x = 60° .

∴ tan 2x = tan 60° = √3

`\text{And, }\frac{2\tan x}{1-\tan ^{2}x}=\frac{2\tan 30^\circ}{1-\tan ^{2}30^\circ }`

`=\frac{2\times \frac{1}{\sqrt{3}}}{1-( \frac{1}{\sqrt{3}})^{2}}`

`=\frac{2/\sqrt{3}}{1-\frac{1}{3}}=\frac{2/\sqrt{3}}{2/3}=\frac{2}{\sqrt{3}}\times\frac{3}{2}=\sqrt{3}`

`\therefore \tan 2x=\frac{2\tan x}{1-\tan ^{2}x`

(ii) When x = 30°, we have 2x = 60°.

`\therefore \sqrt{\frac{1-\cos 2x}{2}}=\sqrt{\frac{1-\cos 60^\circ }{2}}`

`\sqrt{\frac{1-\frac{1}{2}}{2}}=\sqrt{\frac{1}{4}}=\frac{1}{2}`

And, sinx = sin30° = 1/2

`\therefore \sin x=\frac{\sqrt{1-\cos 2x}}{2}`