Prove that (sin² 72⁰-sin² 60⁰)=(√5-1)/8.

Question

Prove that [latex] \left(\sin ^{2} 72^{\circ}-\sin ^{2} 60^{\circ}\right)=\frac{(\sqrt{5}-1)}{8} [/latex].

Accepted Answer

We have

[latex]\begin{aligned} ext { LHS } & =\left(\sin ^{2} 72^{\circ}-\sin ^{2} 60^{\circ} ight) \& =\frac{1}{2}\left(2 \sin ^{2} 72^{\circ}-2 \sin ^{2} 60^{\circ} ight)\ \& =\frac{1}{2}\left\{\left(1-\cos 144^{\circ} ight)-\left(1-\cos 120^{\circ} ight) ight\} \quad\left[\because 2 \sin ^{2} heta=(1-\cos 2 heta) ight] \ \& =\frac{1}{2}\left(\cos 120^{\circ}-\cos 144^{\circ} ight) \ \& =-\frac{1}{4}-\frac{1}{2} \cos 144^{\circ}=-\frac{1}{4}-\frac{1}{2} \cos \left(180^{\circ}-36^{\circ} ight)\left[\because \cos 120^{\circ}=-\frac{1}{2} ight] \ \& =-\frac{1}{4}+\frac{1}{2} \cos 36^{\circ}=-\frac{1}{4}+\frac{(\sqrt{5}+1)}{8} \quad\left[\because \cos 36^{\circ}=\frac{(\sqrt{5}+1)}{4} ight] \ \& =\frac{(\sqrt{5}-1)}{8}= ext { RHS. }\end{aligned}[/latex]

Question 15.4.23: Prove that (sin² 72⁰-sin² 60⁰)=(√5-1)/8....

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