Question

Prove Sin^2 72° - Sin^2 36° =√5/4
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Answer 1

Here, we will use,

sin 18^@ = (sqrt5-1)/4

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sin^2(72^@) -sin^2(60^@)

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=sin^2(90^@ -18^@) - sin^2(60^@)

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=cos^2 (18^@)- sin^2(60^@)

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=1-sin^2(18^@)- sin^2(60^@)

<br> Putting values of

sin 18^@ and sin 60^@

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=1-( (sqrt5-1)/4)^2 -(sqrt3/2)^2

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=1-((6-2sqrt5)/16)-3/4

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=(16-6+2sqrt5-12)/16

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=(2sqrt5-2)/16

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:. sin^2(72^@) -sin^2(60^@) =(sqrt5-1)/8

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