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Prove that : sin^2(72^@) - sin^2 (60^@) ...

Prove that : `sin^2(72^@) - sin^2 (60^@) = (sqrt5 - 1)/8`

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Here, we will use, ` sin 18^@ = (sqrt5-1)/4`
`sin^2(72^@) -sin^2(60^@)`
`=sin^2(90^@ -18^@) - sin^2(60^@)`
`=cos^2 (18^@)- sin^2(60^@)`
`=1-sin^2(18^@)- sin^2(60^@)`
Putting values of `sin 18^@ and sin 60^@`
`=1-( (sqrt5-1)/4)^2 -(sqrt3/2)^2`
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