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Use the appropriate addition formula to find the exact value of the expression.


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It's the pi sign but I wasn't sure how to make it on this computer – user62229 Feb 14 '13 at 1:13
$\pi$ = $\pi$ $\;\sin\left(\frac{11}{12}\pi\right)$ = $\sin\left(\frac{11}{12}\pi\right)$ – amWhy Feb 14 '13 at 1:15
I was trying to find the derivative but that doesn't seem to work – user62229 Feb 14 '13 at 1:20

$(1)$ $$\sin\left(\frac{11}{12}\pi\right) = \sin\left(\frac 5{12}\pi + \frac 6{12} \pi \right) = \sin\left(\frac 5{12}\pi + \frac 12 \pi\right)$$ You can use the sum-of-angles formula for $\sin(a + b)$:

$$\sin(a + b) = \sin(a)\cos(b) + \cos(a)\sin(b)$$

$$\sin\left(\frac 5{12}\pi + \frac 12 \pi\right) = \sin\left(\frac 5{12} \pi\right)\cos\left(\frac 12 \pi\right) + \cos\left(\frac 5{12} \pi\right)\sin\left(\frac 12 \pi\right)$$

$(2)$ Notice that the first term in the sum is multiplied by $\cos\left(\frac{\pi}{2}\right) = 0.$ And since $\sin\left(\frac{\pi}{2}\right) = 1$ you need only compute $\;\cos\left(\frac{5 \pi}{12}\right)\;$. Indeed, seeing that $$\cos\left(\frac{5\pi}{12}\right) = \cos\left(\frac{2\pi}{12} + \frac{3\pi}{12}\right) = \cos\left(\frac{\pi}6 + \frac{\pi}4\right)\tag{1}$$

we can use the "angle-sum formula" for cosine: $$\cos(a + b) = \cos a \cos b - \sin a \sin b$$ $$\cos\left(\frac{\pi}6 + \frac{\pi}4\right) = \cos\left(\frac{\pi}6\right)\cos\left(\frac{\pi}4\right) - \sin\left(\frac{\pi}6\right)\sin\left(\frac{\pi}4\right)$$

Now the computation is one with which you should be familiar: $$\sin\left(\frac{\pi}{6}\right) = \frac12,\;\;\cos\left(\frac{\pi}6\right) = \sqrt 3/2,\;\; \cos\left(\frac{\pi}4 \right) = \sin\left(\frac{\pi}4\right) = 1/\sqrt 2$$

You can scroll over the greyed out line below to check your answer.

$$\;\sin\left(\frac{11 \pi}{12}\right)\;\;=\;\;\frac{\sqrt 3 -1}{2\sqrt 2}$$

You can also use the fact that $$\sin\left(x +\frac {\pi}{2}\right) = \cos(x)$$

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Hint: Try $$\sin\left(\frac{2\pi}{12}+\frac{3\pi}{12}+\frac{6\pi}{12}\right).$$From here, use the addition formula, $\sin(\alpha+\beta)=\sin(\alpha)\cos(\beta)+\sin(\beta)\cos(\alpha)$, to get $$\sin\left(\frac{2\pi}{12}+\frac{3\pi}{12}\right)\cos\left(\frac{6\pi}{12}\right)+\sin\left(\frac{6\pi}{12}\right)\cos\left(\frac{2\pi}{12}+\frac{3\pi}{12}\right).$$ As you can see, you will need to use the addition formula one more time, and then note that $$\frac{2\pi}{12}=\frac{\pi}{6},\quad\frac{3\pi}{12}=\frac{\pi}{4},\quad\text{and}\quad \frac{6\pi}{12}=\frac{\pi}{2}.$$

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Do I add them together – user62229 Feb 14 '13 at 1:21
Use the addition formula, $\sin(x+y)=\sin(x)\cos(y)+\sin(y)\cos(x)$. Here, let $x=2\pi/12+3\pi/12$ and $y=6\pi/12$. You will be able to use the sum-angle formula one more time after this application. – Clayton Feb 14 '13 at 1:23

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