If L=sin2(π16)−sin2(π8) and M=cos2(π16)−sin2(π8), then

L=sin2(π16)−sin2(π8) =1−cos(π/8)2−(1−cos(π/4)2) =12[cos(π4)−cos(π8)]L=122−12cos(π8)M=cos2(π16)−sin2(π8) =1+cos(π/8)2−(1−cos(π/4)2) =12(cos(π/8)+cos(π4))M=12cos(π8)+122

If L=sin2(π16)−sin2(π8) and M=cos2(π16)−sin2(π8), then

L=sin2(π16)−sin2(π8) =1−cos(π/8)2−(1−cos(π/4)2) =12[cos(π4)−cos(π8)]L=122−12cos(π8)M=cos2(π16)−sin2(π8) =1+cos(π/8)2−(1−cos(π/4)2) =12(cos(π/8)+cos(π4))M=12cos(π8)+122

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If $L = \sin^{2} (\frac{π}{16}) - \sin^{2} (\frac{π}{8})$ and $M = \cos^{2} (\frac{π}{16}) - \sin^{2} (\frac{π}{8}),$ then

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An Intiative by Sri Chaitanya

$M = \frac{1}{2 \sqrt{2}} + \frac{1}{2} \cos \frac{π}{8}$

$L = \frac{1}{4 \sqrt{2}} - \frac{1}{4} \cos \frac{π}{8}$

$M = \frac{1}{4 \sqrt{2}} + \frac{1}{4} \cos \frac{π}{8}$

$L = - \frac{1}{2 \sqrt{2}} + \frac{1}{2} \cos \frac{π}{8}$

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Detailed Solution

$\begin{array}{l} L = \sin^{2} (\frac{π}{16}) - \sin^{2} (\frac{π}{8}) \\ = \frac{1 - \cos (π / 8)}{2} - (\frac{1 - \cos (π / 4)}{2}) \\ = \frac{1}{2} [\cos (\frac{π}{4}) - \cos (\frac{π}{8})] \\ L = \frac{1}{2 \sqrt{2}} - \frac{1}{2} \cos (\frac{π}{8}) \\ M = \cos^{2} (\frac{π}{16}) - \sin^{2} (\frac{π}{8}) \\ = \frac{1 + \cos (π / 8)}{2} - (\frac{1 - \cos (π / 4)}{2}) \end{array}$