<s>[INST] Show $\sin(\frac{\pi}{3})=\frac{1}{2}\sqrt{3}$ I have to show that 
$$\sin\left(\frac{\pi}{3}\right)=\frac{1}{2}\sqrt{3}$$
and 
$$\cos\left(\frac{\pi}{3}\right)=\frac{1}{2}$$
Should I use the exponential function?
 [/INST] Without using triangles:
We use the following core formulas to prove this identity:


*

*Pythagorean identity: $cos^2(x) + sin^2(x) = 1$

*Angle transformation formula: $sin(x - y) = sin(x)cos(y) - cos(x)sin(y)$

*Double-angle formula: $sin(2x) = 2sin(x)cos(x)$
To show: $sin(\frac{\pi}{3}) = \frac{\sqrt3}{2}$
We start by showing the following identity: $sin(\pi - x) = sin(x)$
$sin(\pi - x) \stackrel{2.}{=} sin(\pi)cos(x) - cos(\pi)sin(x)$
$\stackrel{sin(\pi) = 0}{\Rightarrow}$
$\stackrel{cos(\pi) = -1}{\Rightarrow}$
$= sin(x)$
We use $x = \frac{\pi}{3}$ and get: $sin(\frac{2\pi}{3}) = sin(\frac{\pi}{3})$
Now: $sin(2\frac{\pi}{3}) \stackrel{3.}{=} 2sin(\frac{\pi}{3})cos(\frac{\pi}{3})$
$\stackrel{/sin(\frac{\pi}{3})}{\Rightarrow}$
$\stackrel{/2)}{\Rightarrow}$
$\frac{1}{2} = cos(\frac{\pi}{3})$
Finally we use 1. to reach the desired result:
$cos^2(\frac{\pi}{3}) + sin^2(\frac{\pi}{3}) = 1  \Rightarrow sin^2(\frac{\pi}{3}) = 1 - cos^2(\frac{\pi}{3}) \stackrel{\frac{1}{2} = cos(\frac{\pi}{3})}{\Rightarrow} sin^2(\frac{\pi}{3}) = \frac{3}{4} \stackrel{(\sqrt)}{\Rightarrow} sin(\frac{\pi}{3}) = \frac{\sqrt3}{2}$
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