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Without the aid of a calculator, compute $\sin\frac{5\pi}{12}+\sin\frac{\pi}{12}$

The only method in doing this that I know was to just plug it in into the calculator, but I really had no clue on how to do this without the aid of a calculator,

I was thinking more along the lines of using the Taylor polynomial series but that really didn't work out for me though.

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    $\begingroup$ Hint: half-angle formulae. $\endgroup$ – Sean Roberson May 25 '17 at 2:46
  • $\begingroup$ Taylor series will only get you an approximation of the value; they will not allow you to compute it exactly $\endgroup$ – qbert May 25 '17 at 3:34
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An easy way to find the value is to use the identity $$\sin A + \sin B = 2 \sin\frac{(A+B)}{2} \cos\frac{(A-B)}{2}$$

which is equivalent to product the formula.

Plugging in the values for $A$ and $B$, we can easily see it to be $$ 2\sin\frac{\pi}{4} \cos\frac{\pi}{6} = \sqrt\frac{3}{2}$$

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Use the product formula:

$$\sin(a)\cos(b) = \frac{1}{2}(\sin(a+b) + \sin(a-b))$$

let $a = \frac{3\pi}{12}$ and $b = \frac{2\pi}{12}$

Then

$$\sin(\frac{5\pi}{12}) + \sin(\frac{\pi}{12})= \sin(\frac{3\pi}{12}+\frac{2\pi}{12}) + sin(\frac{3\pi}{12}-\frac{2\pi}{12}) $$ $$=2\sin(\frac{3\pi}{12})\cos(\frac{2\pi}{12})=2\sin(\frac{\pi}{4})\cos(\frac{\pi}{6})$$

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$$\sin\frac{5\pi}{12}+\sin\frac{\pi}{12}=\sqrt{\frac{1-\cos\frac{5\pi}{6}}{2}}+\sqrt{\frac{1-\cos\frac{\pi}{6}}{2}}=$$ $$=\sqrt{\frac{1+\frac{\sqrt3}{2}}{2}}+\sqrt{\frac{1-\frac{\sqrt3}{2}}{2}}=\frac{\sqrt{4+2\sqrt3}+\sqrt{4-2\sqrt3}}{2\sqrt2}=$$ $$=\frac{\sqrt3+1+\sqrt3-1}{2\sqrt2}=\sqrt{\frac{3}{2}}.$$

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One more method:

\begin{align*} \left(\sin\frac{5\pi}{12}+\sin\frac{\pi}{12}\right)^2&=\left(\cos\frac{\pi}{12}+\sin\frac{\pi}{12}\right)^2\\ &=\cos^2\frac{\pi}{12}+\sin^2\frac{\pi}{12}+2\sin\frac{\pi}{12}\cos\frac{\pi}{12}\\ &=1+\sin\frac{\pi}{6}\\ &=\frac{3}{2} \end{align*}

As $\displaystyle \sin\frac{5\pi}{12}>0$ and $\displaystyle \sin\frac{\pi}{12}>0$,

$$\sin\frac{5\pi}{12}+\sin\frac{\pi}{12}=\sqrt{\displaystyle \frac{3}{2}}=\frac{\sqrt{6}}{2}$$

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