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?
 A: $\sin(\pi+3\pi/8)=-\sin3\pi/8$
Now $0<3\pi/8<\pi/2$
Using $\cos2A=1-2\sin^2A,2\sin3\pi/8=+\sqrt{2(1-\cos3\pi/4)}$
Finally, $\cos3\pi/4=\cos(\pi-\pi/4)=-\cos\pi/4=?$
A: As $\dfrac{11\pi}8=\pi+\dfrac{3\pi}8$ which lies in the third Quadrant,
$\sin<0,\cos <0,\tan>0$
Now $\tan\dfrac{11\pi}8=\cdots=\tan\dfrac{3\pi}8$
Using $\tan A=\dfrac{1-\cos2A}{\sin2A},\tan\dfrac{3\pi}8=\dfrac{1-\cos\dfrac{3\pi}4}{\sin\dfrac{3\pi}4}$
$\sin\dfrac{3\pi}4=\sin\left(\pi-\dfrac\pi4\right)=\sin\dfrac\pi4=?$
$\cos\dfrac{3\pi}4=\cos\left(\pi-\dfrac\pi4\right)=-\cos\dfrac\pi4=?$
$$\dfrac{\sin\dfrac{11\pi}8}{\sqrt2+1}=\dfrac{\cos\dfrac{11\pi}8}1=-\sqrt{\dfrac{\sin^2\dfrac{11\pi}8+\cos^2\dfrac{11\pi}8}{(\sqrt2+1)^2+1^2}}=-\dfrac1{\sqrt{2\sqrt2(\sqrt2+1)}}$$
A: 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?).
