InterviewSolution
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InterviewSolution
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Prove that:`sin^2(pi/18)+sin^2(pi/9)+sin^2([7pi]/18)+sin^2([4pi]/9)=2` |
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Answer» `L.H.S. = sin^2(pi/18)+sin^2(pi/9)+sin^2((7pi)/18)+sin^2((4pi)/9)` Now, As `sin x = cos((pi/2)-x)` `:. sin^2((7pi)/18) = cos^2(pi/2-(7pi)/18) = cos^2(pi/9)` `:.sin^2((4pi)/9) = cos^2(pi/2-(4pi)/9) = cos^2(pi/18)` So, putting these values, `L.H.S. = sin^2(pi/18)+sin^2(pi/9)+cos^2(pi/9)+cos^2(pi/18)` `=(sin^2(pi/18)+cos^2(pi/18))+(sin^2(pi/9)+cos^2(pi/9))` `1+1 = 2` |
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