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This is a sub-question of my homework. I'll take care of the other steps involved! Here's my work so far from being given $\sin(11\pi/8)$

I first converted the radian value of the sine parameter which is 247.5 degrees. If i put that on the unit circle it gives me a reference triangle with the angles: 22.5, 67.5 and 90. I then noticed that $22.5 = 1/2 \cdot 45$. I then concluded that if the $\sin(45^\circ) = \frac{\sqrt{2}}{2}$ then the opposite side would shrink by 1/2. So, the triangle's sides would be 1/2(opp 22.5) 1(opp 67.5) and by Pythagoras theorem $\sqrt{5}/2$ (opp 90). The sin of this triangle would then be $\frac{1}{2} /\frac{\sqrt{5}}{2}$ which is about $0.4472135955$.

I also thought of using a half-angle identity and came to the conclusion that $\sin(11\pi/8)$ was $\frac{\sqrt{2 - \sqrt{2}}}{2}$ which is about $0.38268344324$

Neither of these answers is correct! my calculator gives the decimal answer of $-0.9238795325$ to $\sin(11\pi/8)$.

I simply need an exact answer to this simple trig function.

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  • $\begingroup$ First argument does not work. Show us how you arrived at the answer for the half angle formula. If you can fix the error there, I feel like you'll have it. $\endgroup$ – John Molokach Nov 26 '15 at 0:47
  • $\begingroup$ Since $\pi/2 < 11\pi/8 < 3\pi/2$, we must have the sine negative. $\endgroup$ – GEdgar Nov 26 '15 at 1:37
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Remember the values for which sine is positive or negative. You would approach this from using half angle identities. Try checking your work, it should evaluate to $$-\frac{\sqrt{2 + \sqrt2}}2$$

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$$ \sin \frac{11\pi}8 = - \sin \frac{3\pi}8 \\ =-\sqrt{1-\cos^2\frac{3\pi}8} \\ =-\sqrt{\frac{1-\cos\frac{3\pi}4}2} \\ =-\sqrt{\frac{1+\frac{\sqrt{2}}2}2} \\ =-\frac{\sqrt{2+\sqrt{2}}}2 $$

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