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:


For instance: $\alpha=\frac{20\pi}{8}$ and $\beta=\frac{9\pi}{8}$

Am I on the right track?

  • $\begingroup$ Why not just use the formula suggested in the problem? $\endgroup$ – Ian Nov 24 '16 at 16:53
  • $\begingroup$ Well I was curious if my suggestion was right and also isn't the formula wrong? I copied it down word by word and I think it is a typo from my teacher. Since wouldn't $sin^2(x)= 1- cos^2(x)$ $\endgroup$ – user392659 Nov 24 '16 at 16:56
  • $\begingroup$ No, it's right. Note the $2x$ in the argument of the cosine. $\endgroup$ – Ian Nov 24 '16 at 17:51
  • $\begingroup$ The actual intent of the question: $\sin(11 \pi/8)=\pm \sqrt{ \frac{1-\cos(2(11\pi/8))}{2}}$. That leaves you to compute $\cos(11\pi/4)$ and then resolve which sign you should have with the $\pm$. $\endgroup$ – Ian Nov 24 '16 at 19:21


Now $0<3\pi/8<\pi/2$

Using $\cos2A=1-2\sin^2A,2\sin3\pi/8=+\sqrt{2(1-\cos3\pi/4)}$

Finally, $\cos3\pi/4=\cos(\pi-\pi/4)=-\cos\pi/4=?$

  • $\begingroup$ wait is the ? at the end mean that you have left it incomplete for me to work it out for myself ? $\endgroup$ – user392659 Nov 24 '16 at 17:03
  • $\begingroup$ @user392659. I bet he did it on purpose ! $\endgroup$ – Claude Leibovici Nov 24 '16 at 17:21

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 $\tan A=\dfrac{1-\cos2A}{\sin2A},\tan\dfrac{3\pi}8=\dfrac{1-\cos\dfrac{3\pi}4}{\sin\dfrac{3\pi}4}$





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?).

  • $\begingroup$ ...That's way more complicated than this question had to be... $\endgroup$ – Ian Nov 24 '16 at 19:20
  • $\begingroup$ @Ian, true, but the process is straightforward and applicable to similar problems. $\endgroup$ – Martín-Blas Pérez Pinilla Nov 24 '16 at 20:42

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