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Prove that: sin^2(pi/18)+sin^2(pi/9)+sin...

Prove that: `sin^2(pi/18)+sin^2(pi/9)+sin^2([7pi]/18)+sin^2([4pi]/9)=2`

Text Solution

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`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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