<s>[INST] Calculate the exact value of $\sin\frac{11\pi}{8}$ Calculate the exact value of $\sin\frac{11\pi}{8}$. 
The formula $\sin^2x=\frac12(1–\cos2x)$ may be helpful.
I was thinking of using the Angle-Sum and -Difference Identity:
$\sin(\alpha-\beta)=\sin\alpha\cos\beta-\cos\alpha\sin\beta$
For instance: $\alpha=\frac{20\pi}{8}$ and $\beta=\frac{9\pi}{8}$ 
Am I on the right track?
 [/INST] Another idea: Let be $x = 3\pi/8$
$$0 = \sin(3\pi) = \sin(8x) = 2\sin(4x)\cos(4x) = \cdots $$
$$= (-128\sin^7 x + 192\sin^5 x - 80\sin^3 x + 8\sin x)\cos x.$$
Then, $s = \sin x$ is a root of
$$0 = 128s^7 - 192s^5 + 80s^3 - 8s = 8(16s^6 - 24s^4 + 10s^2 - 1)s = 8(8s^4 - 8s^2 + 1)(2s^2 - 1)s.$$
The roots of the biquadratic are
$$\pm\sqrt{\frac12\pm\frac{\sqrt2}4}.$$
Finally, you must discard the wrong solutions using that $s>1/2$ (why this is true?).
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