How to prove ?

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How to prove ?

${cos}^{2} (\frac{π}{8}) + {cos}^{2} (\frac{3 π}{8}) + {cos}^{2} (\frac{5 π}{8}) + {cos}^{2} (\frac{7 π}{8}) = 2$ $cos^2(pi/8)+cos^2((3pi)/8)+cos^2((5pi)/8)+cos^2((7pi)/8)=2$

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Answer 1 · 2018-07-05T16:46:31

${cos}^{2} (\frac{π}{8}) + {cos}^{2} (\frac{3 π}{8}) + {cos}^{2} (\frac{5 π}{8}) + {cos}^{2} (\frac{7 π}{8}) = 2$ $cos^2(pi/8)+cos^2((3pi)/8)+cos^2((5pi)/8)+cos^2((7pi)/8)=2$

Using cofunction identities:
$cos x = sin (\frac{π}{2} - x)$ $cosx=sin(pi/2-x)$

${cos}^{2} (\frac{π}{8}) + {sin}^{2} (\frac{π}{2} - \frac{3 π}{8}) + {sin}^{2} (\frac{π}{2} - \frac{5 π}{8}) + {cos}^{2} (\frac{7 π}{8}) = 2$ $cos^2(pi/8)+sin^2(pi/2-(3pi)/8)+sin^2(pi/2-(5pi)/8)+cos^2((7pi)/8)=2$

${cos}^{2} (\frac{π}{8}) + {sin}^{2} (\frac{π}{8}) + {sin}^{2} (\frac{- π}{8}) + {cos}^{2} (\frac{7 π}{8}) = 2$ $cos^2(pi/8)+sin^2(pi/8)+sin^2((-pi)/8)+cos^2((7pi)/8)=2$

${sin}^{2} (- \frac{π}{8})$ $sin^2(-pi/8)$ and ${sin}^{2} (\frac{7 π}{8})$ $sin^2((7pi)/8)$ will yield the same values, so we can just substitute:

${cos}^{2} (\frac{π}{8}) + {sin}^{2} (\frac{π}{8}) + {sin}^{2} (\frac{7 π}{8}) + {cos}^{2} (\frac{7 π}{8}) = 2$ $cos^2(pi/8)+sin^2(pi/8)+sin^2((7pi)/8)+cos^2((7pi)/8)=2$

Recall ${sin}^{2} x + {cos}^{2} x = 1$ $sin^2x+cos^2x=1$

$1 + 1 = 2$ $1+1=2$

$2 = 2$ $2=2$

Answer 2 · 2018-07-05T16:55:01

Please see the proof below

Explanation:

Apply the half angle formula

${cos}^{2} (\frac{a}{2}) = \frac{1 + cos a}{2}$ $cos^2(a/2)=(1+cosa)/2$

Therefore,

${cos}^{2} (\frac{π}{8}) = \frac{1}{2} (1 + cos (\frac{π}{4})) = \frac{1}{2} (1 + \frac{\sqrt{2}}{2})$ $cos^2(pi/8)=1/2(1+cos(pi/4))=1/2(1+sqrt2/2)$

${cos}^{2} (\frac{3}{8} π) = \frac{1}{2} (1 + cos (\frac{3}{4} π)) = \frac{1}{2} (1 - \frac{\sqrt{2}}{2})$ $cos^2(3/8pi)=1/2(1+cos(3/4pi))=1/2(1-sqrt2/2)$

${cos}^{2} (\frac{5}{8} π) = \frac{1}{2} (1 + cos (\frac{5}{4} π)) = \frac{1}{2} (1 - \frac{\sqrt{2}}{2})$ $cos^2(5/8pi)=1/2(1+cos(5/4pi))=1/2(1-sqrt2/2)$

${cos}^{2} (\frac{7}{8} π) = \frac{1}{2} (1 + cos (\frac{7}{4} π)) = \frac{1}{2} (1 + \frac{\sqrt{2}}{2})$ $cos^2(7/8pi)=1/2(1+cos(7/4pi))=1/2(1+sqrt2/2)$

So,

${cos}^{2} (\frac{π}{8}) + {cos}^{2} (\frac{3}{8} π) + {cos}^{2} (\frac{5}{8} π) + {cos}^{2} (\frac{7}{8} π)$ $cos^2(pi/8)+cos^2(3/8pi)+cos^2(5/8pi)+cos^2(7/8pi)$

$= \frac{1}{2} (1 + \frac{\sqrt{2}}{2}) + \frac{1}{2} (1 - \frac{\sqrt{2}}{2}) + \frac{1}{2} (1 - \frac{\sqrt{2}}{2}) + \frac{1}{2} (1 + \frac{\sqrt{2}}{2})$ $=1/2(1+sqrt2/2)+1/2(1-sqrt2/2)+1/2(1-sqrt2/2)+1/2(1+sqrt2/2)$

$= \frac{1}{2} + \frac{1}{2} + \frac{1}{2} + \frac{1}{2}$ $=1/2+1/2+1/2+1/2$

$= 2$ $=2$

$Q E D$ $QED$

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How to prove ?

${cos}^{2} (\frac{π}{8}) + {cos}^{2} (\frac{3 π}{8}) + {cos}^{2} (\frac{5 π}{8}) + {cos}^{2} (\frac{7 π}{8}) = 2$ $cos^2(pi/8)+cos^2((3pi)/8)+cos^2((5pi)/8)+cos^2((7pi)/8)=2$

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How to prove ?

cos2(π8)+cos2(3π8)+cos2(5π8)+cos2(7π8)=2cos^2(pi/8)+cos^2((3pi)/8)+cos^2((5pi)/8)+cos^2((7pi)/8)=2

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Explanation:

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Impact of this question

${cos}^{2} (\frac{π}{8}) + {cos}^{2} (\frac{3 π}{8}) + {cos}^{2} (\frac{5 π}{8}) + {cos}^{2} (\frac{7 π}{8}) = 2$ $cos^2(pi/8)+cos^2((3pi)/8)+cos^2((5pi)/8)+cos^2((7pi)/8)=2$