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

L=sin2(π16)−sin2(π8) (∵sin2θ=1−cos2θ2)⇒L=(1−cos(π8)2)−(1−cos(π4)2)⇒L=12[cos(π4)−cos(π8)]⇒L=122−12cos(π8)M=cos2(π16)−sin2(π8)⇒M=(1+cos(π8)2)−(1−cos(π4)2)⇒M=12cos(π8)+122

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

L=sin2(π16)−sin2(π8) (∵sin2θ=1−cos2θ2)⇒L=(1−cos(π8)2)−(1−cos(π4)2)⇒L=12[cos(π4)−cos(π8)]⇒L=122−12cos(π8)M=cos2(π16)−sin2(π8)⇒M=(1+cos(π8)2)−(1−cos(π4)2)⇒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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$M = \frac{1}{2 \sqrt{2}} + \frac{1}{2} \cos \frac{π}{8}$

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

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

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

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Detailed Solution

$L = \sin^{2} (\frac{π}{16}) - \sin^{2} (\frac{π}{8}) \begin{array}{l} (∵ \sin^{2} θ = \frac{1 - \cos 2 θ}{2}) \\ \Rightarrow L = (\frac{1 - \cos (\frac{π}{8})}{2}) - (\frac{1 - \cos (\frac{π}{4})}{2}) \\ \Rightarrow L = \frac{1}{2} [\cos (\frac{π}{4}) - \cos (\frac{π}{8})] \end{array}$

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

$\begin{array}{l} \Rightarrow M = (\frac{1 + \cos (\frac{π}{8})}{2}) - (\frac{1 - \cos (\frac{π}{4})}{2}) \\ \Rightarrow M = \frac{1}{2} \cos (\frac{π}{8}) + \frac{1}{2 \sqrt{2}} \end{array}$