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

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Answer 1 · 2018-07-11T00:36:22

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Explanation:

Let $\frac{π}{8} = x$ $pi/8=x$ then $8 x = π and 4 x = \frac{π}{2}$ $8x=pi and 4x=pi/2$

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

$= {cos}^{2} x + {cos}^{2} (3 x) + {cos}^{2} (5 x) + {cos}^{2} (7 x)$ $=cos^2x+cos^2(3x)+cos^2(5x)+cos^2(7x)$

$= {cos}^{2} x + {cos}^{2} (3 x) + {cos}^{2} (4 x + x) + {cos}^{2} (4 x + 3 x)$ $=cos^2x+cos^2(3x)+cos^2(4x+x)+cos^2(4x+3x)$

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

$= {cos}^{2} x + {cos}^{2} (3 x) + {sin}^{2} (x) + {sin}^{2} (3 x) = 1 + 1 = 2 = R H S$ $=cos^2x+cos^2(3x)+sin^2(x)+sin^2(3x)=1+1=2=RHS$

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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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Explanation:

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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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Impact of this question

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