JEE Main · 2022 · 25 Jun 2022 · Shift 2 · Trigonometric Ratios and Identities

The value of

2 sin (1 2^{\circ}) - sin (7 2^{\circ})

is :

Answer: (D)

Solution

Rewrite

2 sin 1 2^{\circ} - sin 7 2^{\circ} = sin 1 2^{\circ} + sin 1 2^{\circ} - sin 7 2^{\circ}

. Using the sum-to-product formula:

sin 1 2^{\circ} + sin 1 2^{\circ} = 2 sin 1 2^{\circ}

. Alternatively,

sin 1 2^{\circ} - sin 7 2^{\circ} = - 2 cos (\frac{1 2 ^{\circ} + 7 2 ^{\circ}}{2}) sin (\frac{7 2 ^{\circ} - 1 2 ^{\circ}}{2}) = - 2 cos 4 2^{\circ} sin 3 0^{\circ} = - 2 cos 4 2^{\circ} \cdot \frac{1}{2} = - cos 4 2^{\circ} = - sin 4 8^{\circ}

. So

sin 1 2^{\circ} - sin 4 8^{\circ} = - 2 cos 3 0^{\circ} sin 1 8^{\circ} = - 2 \cdot \frac{3}{2} \cdot \frac{5 - 1}{4} = - \frac{3 ( 5 - 1 )}{4} = \frac{3 ( 1 - 5 )}{4}

JEE Main · 2022 · 25 Jun 2022 · Shift 2 · Single correct option