# Reducing simple trigonometric expression

I have an expression: $\sin(\pi/3) - \sin(\pi/6)$. I have it reducing to $3 / \sqrt{2} - 1/2$, but my text has $3 \sqrt{3} - 3$. How does my eq become their eq?

Edited the eq

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There must be a typo in your statement. What you have written looks like 0 to me. – Geoff Robinson Aug 26 '12 at 15:05
Your answer suggests that the original expression was actually $\sin\left(\frac{\pi}3\right)-\cos\left(\frac{\pi}3\right)$, not $\sin\left(\frac{\pi}3\right)-\sin\left(\frac{\pi}3\right)$, which is obviously $0$. However, $\sin\left(\frac{\pi}3\right)=\frac{\sqrt3}2$, not $\frac3{\sqrt2}$, so $\sin\left(\frac{\pi}3\right)-\cos\left(\frac{\pi}3\right)=\frac{\sqrt3}2-\frac{‌​1}{2}$. – Brian M. Scott Aug 26 '12 at 15:11
I have formatted your question using LaTeX and have left the (probably unintended) original question. Please take a moment to see what question you intended to ask and try to use LaTeX in the future. – Austin Mohr Aug 26 '12 at 15:34

Starting with their work $$\dfrac{\Delta y}{\Delta t} = \dfrac{\sin(\dfrac{\pi}{3}) - \sin(\dfrac{\pi}{6})}{\dfrac{\pi}{3} - \dfrac{\pi}{6}}$$
This then simplifies as $$\dfrac{\dfrac{\sqrt{3}}{2} - \dfrac{1}{2}}{\dfrac{\pi}{6}} = \dfrac{\sqrt{3}-1}{2} \cdot \dfrac{6}{\pi} = \dfrac{6(\sqrt{3}-1)}{2} \cdot \dfrac{1}{\pi} = \dfrac{3 \sqrt{3} - 3}{\pi}.$$
$$\frac{\sin\left(\frac\pi3\right)-\sin\left(\frac\pi6\right)}{\frac\pi3-\frac\pi6}=\frac{\frac{\sqrt3}2-\frac12}{\frac\pi6}=\frac{\sqrt3-1}2\cdot\frac{6}{\pi}=\frac{3(\sqrt3-1)}{\pi}$$