# Exact Value of sin 7½°

How to find the exact value of sin 7½° using the value of cos 15°?

Solution:

7½° lies in the first quadrant.

Therefore, sin 7½° is positive.

For all values of the angle A we know that, cos (α - β) = cos α cos β + sin α sin β.

Therefore, cos 15° = cos (45° - 30°)

cos 15° = cos 45° cos 30° + sin 45° sin 30°

= $$\frac{1}{√2}$$∙$$\frac{√3}{2}$$ + $$\frac{1}{√2}$$∙$$\frac{1}{2}$$

= $$\frac{√3}{2√2}$$ + $$\frac{1}{2√2}$$

= $$\frac{√3 + 1}{2√2}$$

Again for all values of the angle A we know that, cos A = 1 - 2 sin$$^{2}$$$$\frac{A}{2}$$

⇒ 1 - cos A = 2 sin$$^{2}$$ $$\frac{A}{2}$$

⇒ 2 sin$$^{2}$$ $$\frac{A}{2}$$ = 1 - cos A

⇒ 2 sin$$^{2}$$ 7½˚ = 1 - cos 15°

⇒ sin$$^{2}$$ 7½˚ = $$\frac{1 - cos 15°}{2}$$

⇒ sin$$^{2}$$ 7½˚ = $$\frac{1 - \frac{√3 + 1}{2√2}}{2}$$

⇒ sin$$^{2}$$ 7½˚ = $$\frac{2√2 - √3 - 1}{4√2}$$

⇒ sin 7½˚ = $$\sqrt{\frac{4 - √6 - √2}{8}}$$, [Since sin 7½° is positive]

⇒ sin 7½˚ = $$\frac{\sqrt{4 - √6 - √2}}{2√2}$$

Therefore, sin 7½˚ = $$\frac{\sqrt{4 - √6 - √2}}{2√2}$$