Q44 / 90 · JEE Main 2022 (25 Jun Shift 2) · Mathematics

The value of $2 \sin 12 ° - \sin 72 °$ is

A
$\frac{\sqrt{5} (1 - \sqrt{3})}{4}$
B
$\frac{1 - \sqrt{5}}{8}$
C
$\frac{\sqrt{3} (1 - \sqrt{5})}{2}$
Correct answer
D
$\frac{\sqrt{3} (1 - \sqrt{5})}{4}$

Solution

To find $\Rightarrow 2 \sin 12 ° - \sin 72$

$\Rightarrow \sin 12 ° + \sin 12 ° - \sin 72$

$\Rightarrow \sin 12 ° + 2 \cos 42 ° \sin (- 30 °)$ $\Rightarrow \sin 12 ° - \cos 42 °$

$\Rightarrow \sin 12 ° - \sin 48 °$ $\Rightarrow 2 \cos (30 °) \sin (- 18 °)$

$\Rightarrow - \sqrt{3} \sin 18 ° = - \sqrt{3} (\frac{\sqrt{5} - 1}{4})$ $= \sqrt{3} (\frac{1 - \sqrt{5}}{4})$

Hence, the correct answer is option: 'D'.

The value of 2sin12°-sin72° is

Solution

Relevant Questions

The value of $2 \sin 12 ° - \sin 72 °$ is