Question

show that cos²π/8+cos²3π/8+cos²5π/8+cos²7π/8=2​

Answer 1

cos2A≡2cos²A-1 so cos²A≡½(1+cos2A).

Therefore, we can substitute for the squares of the cosines:

cos²(π/8)=½(1+cos(π/4))=½(1+√2/2);

cos²(3π/8)=½(1+cos(3π/4))=½(1-√2/2);

cos²(5π/8)=½(1+cos(5π/4))=½(1-√2/2);

cos²(7π/8)=½(1+cos(7π/4))=½(1+√2/2);

Add these together and we get the result 4/2=2 QED

hope it helps !!

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Answer 2

Solution-

LHS =

$\bf \: {cos}^{2} \frac{\pi}{8} + {cos}^{2} \frac{3\pi}{8} + {cos}^{2} \frac{5\pi}{8} + {cos}^{2} \frac{7\pi}{8}$

$= \sf \: {cos}^{2} \frac{\pi}{8} + {cos}^{2} \frac{3\pi}{8} + {cos}^{2} \left[\pi - \frac{3\pi}{8} \right] + {cos}^{2} \left[ \pi - \frac{\pi}{8} \right] \\ \\ = \sf \: {cos}^{2} \frac{\pi}{8} + {cos}^{2} \frac{3\pi}{8} + \left[ - cos \frac{3\pi}{8} \right]^{2} + {\left[ - cos \frac{\pi}{8} \right]}^{2} \\ \\ = \sf \: {cos}^{2} \frac{\pi}{8} + {cos}^{2} \frac{3\pi}{8} + {cos}^{2} \frac{3\pi}{8} + {cos}^{2} \frac{\pi}{8} \\ \\ = \sf \: 2 {cos}^{2} \frac{\pi}{8} + 2 {cos}^{2} \frac{3\pi}{8} \\ \\ = \sf \: \left[1 + cos \frac{\pi}{8} \right] + \left[ 1 + cos \frac{3\pi}{8} \right] \\ \\ \sf \because \: 2 {cos}^{2} x = 1 + cos2x \\ \\ = \sf \: 2 + \frac{1}{ \sqrt{2} } + \left[ - \frac{1}{ \sqrt{2} } \right] \\ \\ = \sf \: 2$