Prove the following:sin 36° = \(\frac{\sqrt{10-2\sqrt5}}{4}\)sin 36�

1.	Prove the following:sin 36° = \(\frac{\sqrt{10-2\sqrt5}}{4}\)sin 36° = (√10-2√5)/4
Answer» We know that, sin² θ = 1 – cos θ sin² 36° = 1 – cos² 36° = 1-(\(\frac{\sqrt{5+1}}{4}\))² ^{=\(\frac{16-(5+1+2\sqrt5)}{16}\)} = ^{\(\frac{10-2\sqrt5}{16}\)} ∴ sin 36° = \(\frac{\sqrt{10-2\sqrt5}}{4}\)……[∵ sin 36° is positive]

Prove the following:sin 36° = \(\frac{\sqrt{10-2\sqrt5}}{4}\)sin 36° = (√10-2√5)/4

Answer»

We know that, sin² θ = 1 – cos θ sin² 36° = 1 – cos² 36°

= 1-(\(\frac{\sqrt{5+1}}{4}\))²

^{=\(\frac{16-(5+1+2\sqrt5)}{16}\)}

= ^{\(\frac{10-2\sqrt5}{16}\)}

∴ sin 36° = \(\frac{\sqrt{10-2\sqrt5}}{4}\)……[∵ sin 36° is positive]