If L = ^216-^28 and M = ^216-^28, then:

Question

If $L = \sin^2\dfrac\pi{16}-\sin^2\dfrac\pi8$ and $M = \cos^2\dfrac\pi{16}-\sin^2\dfrac\pi8$, then:

Accepted Answer

$$L+M = \left(\sin^2\tfrac\pi{16}+\cos^2\tfrac\pi{16}\right) - 2\sin^2\tfrac\pi8 = 1-2\sin^2\tfrac\pi8 = \cos\tfrac\pi4 = \frac1{\sqrt2}$$

$$M-L = \cos^2\tfrac\pi{16}-\sin^2\tfrac\pi{16} = \cos\tfrac\pi8$$

Solving simultaneously:

$$M = \frac12\left(\frac1{\sqrt2}+\cos\tfrac\pi8\right) = \frac1{2\sqrt2}+\frac12\cos\tfrac\pi8, \qquad L = \frac1{2\sqrt2}-\frac12\cos\tfrac\pi8$$

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