Question

Prove that sin² 72° - sin² 60° = (√5 - 1)/8.

Answer 1

To Prove :-

→ sin² 72° - sin² 60° = (√5 - 1)/8

Proof :-

$\bigstar$ LHS :

$\sf{sin^2 72 ^\circ - sin^2 60^\circ}$

$\sf{= sin^2(90^\circ - 18^\circ)-sin^2 60^\circ }\\\\\sf{= cos^2 18^\circ-sin^260^\circ}\\\\\sf{= (1-sin^218 ^\circ) - sin^2 60^\circ}$

We know,

◘ sin 18° = (√5 - 1)/4 and sin 60° = √3/2

$\sf{= \bigg[1- \bigg(\dfrac{\sqrt{5}-1}{4} \bigg)^2\bigg] - \bigg(\dfrac{\sqrt{3}}{2} \bigg)^2 }\\\\\sf{= \bigg[1-\bigg(\dfrac{5+1-2\sqrt{5}}{16} \bigg) \bigg] -\dfrac{3}{4} }\\\\\sf{= 1- \bigg(\dfrac{6-2\sqrt{5}}{16} \bigg) -\dfrac{3}{4} }\\\\\sf{= \dfrac{16-6+2\sqrt{5}-12 }{16} }\\\\\sf{= \dfrac{2\sqrt{5} -2}{16} }\\\\\sf{= 2 \bigg(\dfrac{\sqrt{5}-1}{16} \bigg)}\\\\\sf{= \dfrac{\sqrt{5}-1}{8} }$

$\bigstar$ LHS = RHS

Proved! ✔

Answer 2

To Prove :-

→ sin² 72° - sin² 60° = (√5 - 1)/8

Proof :-

LHS :

We know,

◘ sin 18° = (√5 - 1)/4 and sin 60° = √3/2

LHS = RHS

Proved! ✔