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\begin{aligned} &= \sin\left(\frac{\color{red}{1}}{\color{red}{3}}\pi\right) - \frac{\sin\left(\frac{\color{red}{5}}{\color{red}{3}}\pi\right)}{\color{red}{5}}\\ &= \frac{\color{red}3\sqrt{\color{red}3}}{\color{red}5}. \end{aligned}

I understand sin and cos but I am having trouble getting from the top line to the bottom line. I tried to look up sign at $\dfrac{1}{3}\pi$ and $\dfrac{5}{3}\pi$ and plugging them in but can't seem to figure it out. Can anyone explain or show me the proper steps?

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    $\begingroup$ use $\sin(\pi+\frac{2 \pi}{3})=-\cos (\frac{2 \pi}{3})$ $\endgroup$ – Alex Jan 27 '14 at 1:32
  • $\begingroup$ I am trying to get the bottom line as my answer, so its a little different $\endgroup$ – Chris Jan 27 '14 at 1:40
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Use transformation identities: $$ \sin(\frac{5 \pi}{3})=\sin (2 \pi- \frac{\pi}{3})=-\sin \frac{\pi}{3}=-\frac{\sqrt{3}}{2} $$

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I suspect the problem you are having is that you are measuring the angles in degrees, but they are given in radians. Note that $1$ rad $= \frac{180}{\pi}^\circ$, so $\frac{\pi}{3} = 60^\circ$ and $\frac{5\pi}{3} = 300^\circ$.

So our question asks

$\sin 60^\circ - \frac{\sin 300^\circ}{5}$

You know that $\sin 60^\circ = \frac{\sqrt{3}}{2}$ and $\sin 300^\circ = -\frac{\sqrt{3}}{2}$ so plug them in:

$\frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{10} = \frac{3 \sqrt{3}}{5}$

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Sin(pi/3) triangle

solve sin(pi/3)

Source: Weisstein, Eric W. "Trigonometry Angles--Pi/3." From MathWorld--A Wolfram Web Resource. http://mathworld.wolfram.com/TrigonometryAnglesPi3.html

So
$sin(\frac{\pi}{3})=\frac{\sqrt{3}}{2}$
and
$\frac{sin(\frac{5\pi}{3})}{5}=\frac{-sin(\frac{\pi}{3})}{5}=\frac{-\frac{\sqrt{3}}{2}}{5}$
therefore
$\frac{\sqrt{3}}{2}-\frac{-\frac{\sqrt{3}}{2}}{5}=\frac{5\sqrt{3}}{10}+\frac{\sqrt{3}}{10}=\frac{5\sqrt{3}+\sqrt{3}}{10}=\frac{\sqrt{3}(5+1)}{10}=\frac{\sqrt{3}(6)}{10}=\frac{\sqrt{3}(3)}{5}$

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