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

InterviewSolution

1.	Prove that : `sin^2(72^@) - sin^2 (60^@) = (sqrt5 - 1)/8`
Answer» 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` `=1-((6-2sqrt5)/16)-3/4` `=(16-6+2sqrt5-12)/16` `=(2sqrt5-2)/16` `:. sin^2(72^@) -sin^2(60^@) =(sqrt5-1)/8`

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