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?
 A: Hint:
It's  a simple geometric result. See the figure and note that the triangle $OPM$ is equilateral.  
If $OP =1$ than $\sin (\pi/3)=PH$ and $\cos( \pi/3)=OH$. 

A: Not to confuse you. But..
$$x=\dfrac{\pi}{3}$$
$$3x=\pi$$
$$\sin 3x=\sin \pi$$
$$3\sin x-4\sin^3x=0$$
$$\sin x=0 \text{ or } \sin x = \dfrac{\sqrt{3}}{2} \text{ or } \sin x = \dfrac{-\sqrt{3}}{2}$$
$$\text{as } 0<\dfrac{\pi}{3}<\dfrac{\pi}{2} \text{, } \sin \dfrac{\pi}{3} = \dfrac{\sqrt{3}}{2}$$
A: 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:
$cos^2(\frac{\pi}{3}) + sin^2(\frac{\pi}{3}) = 1  \Rightarrow sin^2(\frac{\pi}{3}) = 1 - cos^2(\frac{\pi}{3}) \stackrel{\frac{1}{2} = cos(\frac{\pi}{3})}{\Rightarrow} sin^2(\frac{\pi}{3}) = \frac{3}{4} \stackrel{(\sqrt)}{\Rightarrow} sin(\frac{\pi}{3}) = \frac{\sqrt3}{2}$
A: Hint. From the picture below and Pythagoras we get
$$h=\frac{\sqrt 3}{2}a\rightarrow \sin\frac\pi 3=\frac ha=\frac{\frac{\sqrt 3}{2}a}{a}=\frac{\sqrt 3}{2}.$$

A: consider an equilateral triangle and and construct one hight of this triangle then we have a right triangle and we get $\sin(\pi/3)=\frac{h}{a}$ with $h=\frac{\sqrt{3}}{2}a$ we get  the searched term, where a is the side length of the triangle 
A: Consider an equilateral triangle of side length $a$. We will compute the area by two different ways to compute the value of $\sin(\pi/3)$.


*

*Method $1$: Since the semi-perimeter is $s = \dfrac{3a}2$, using Heron's formula we have the area of the triangle to be $\dfrac{\sqrt3}4a^2$

*Method $2$: The height of the altitude is $a\sin(\pi/3)$ and hence the area is $\dfrac12 \cdot a \cdot a\sin(\pi/3)$.


Comparing the two, we obtain that $\sin(\pi/3) = \sqrt{3}/2$
A: Note that:
$sin(3\theta)=3sin(\theta)-4sin^{3}(\theta)$
Let $\theta =\frac{\pi}{3}$ and let $sin(\frac{\pi}{3})=x$
Then:
$0=x(3-4x^{2})$.
Hence either $x=0$ or $x^{2}=\frac{3}{4}$
But since $sin(0)=0$ and the $sine$ function has a period of $2\pi$ then we must conclude that $sin(\frac{\pi}{3})=\frac{\sqrt3}{2}$
A: Hint: a the sum of the angles of a triangle is $180^\circ$ (or $\pi$ radians). If all of the angles were equal, what would the angle measure of each angle be? Can you find its altitude?
