Prove that-sin 2 72-sin 2 60-5-1-8

We have, sin^2 72^∘ sin^2 60^∘ = sin^2 (90^∘ 18^∘) (√(3)/2 )^2 = cos^2 18^∘ 3/4 = (√(10+2 √(5))/4 )^2 3/4 [∵ cos 18^∘ = √(10+2 √(5))/4 ] =10+2√(5)/ ...

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Standard IX

Mathematics

Values of Trigonometric Ratios

Prove that:si...

Question

Prove that:

$s i n^{2} 72^{\circ} - s i n^{2} 60^{\circ} = \frac{\sqrt{5} - 1}{8}$

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Solution

We have,

$s i n^{2} 72^{\circ} - s i n^{2} 60^{\circ} = s i n^{2} (90^{\circ} - 18^{\circ}) - {(\frac{\sqrt{3}}{2})}^{2} = c o s^{2} 18^{\circ} - \frac{3}{4}$

$= {(\frac{\sqrt{10 + 2 \sqrt{5}}}{4})}^{2} - \frac{3}{4} [∵ c o s 18^{\circ} = \frac{\sqrt{10 + 2 \sqrt{5}}}{4}] = \frac{10 + 2 \sqrt{5}}{4} - \frac{3}{4} = \frac{10 + 2 \sqrt{5} - 12}{4} = \frac{2 \sqrt{5} - 2}{16} = \frac{\sqrt{5} - 1}{8}$

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{sin}^{2} 72^{\circ} - {sin}^{2} 60^{\circ} = \frac{\sqrt{5} - 1}{8}

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Prove that: sin2 72∘−sin2 60∘=√5−18

We have, sin2 72∘−sin2 60∘=sin2 (90∘−18∘)−(√32)2=cos2 18∘−34 =(√10+2√54)2−34[∵ cos 18∘=√10+2√54]=10+2√54−34=10+2√5−124=2√5−216=√5−18

Prove that:

$s i n^{2} 72^{\circ} - s i n^{2} 60^{\circ} = \frac{\sqrt{5} - 1}{8}$