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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?

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    $\begingroup$ No, just draw a triangle with angles $\pi/3$, $\pi/6$, $\pi/2$ and label the edges appropriately as $1$, $2$ and $\sqrt{3}$. $\endgroup$ – Gregory Grant Dec 9 '15 at 16:52
  • $\begingroup$ How does one show that the angles of a triangle with those sides are $\pi/3, \pi/6, \pi/2$ (or vice versa)? That depends on what you're permitted to use in your demonstration. More context is needed from the OP, perhaps. $\endgroup$ – Brian Tung Dec 9 '15 at 16:56
  • $\begingroup$ Yes, use exponential function. Let $z:=e^{\frac{i\pi}3}$, $\operatorname{Re}z=\operatorname{cos}\frac\pi 3$ and $z^3=-1$. $\endgroup$ – Gyro Gearloose Dec 9 '15 at 17:02
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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 $\cos( \pi/3)=OH$.

enter image description here

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Not to confuse you. But.. $$x=\dfrac{\pi}{3}$$ $$3x=\pi$$ $$\sin 3x=\sin \pi$$ $$3\sin x-4\sin^3x=0$$ $$\sin x=0 \text{ or } \sin x = \dfrac{\sqrt{3}}{2} \text{ or } \sin x = \dfrac{-\sqrt{3}}{2}$$

$$\text{as } 0<\dfrac{\pi}{3}<\dfrac{\pi}{2} \text{, } \sin \dfrac{\pi}{3} = \dfrac{\sqrt{3}}{2}$$

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Without using triangles:
We use the following core formulas to prove this identity:

  1. Pythagorean identity: $cos^2(x) + sin^2(x) = 1$
  2. Angle transformation formula: $sin(x - y) = sin(x)cos(y) - cos(x)sin(y)$
  3. 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:
$cos^2(\frac{\pi}{3}) + sin^2(\frac{\pi}{3}) = 1 \Rightarrow sin^2(\frac{\pi}{3}) = 1 - cos^2(\frac{\pi}{3}) \stackrel{\frac{1}{2} = cos(\frac{\pi}{3})}{\Rightarrow} sin^2(\frac{\pi}{3}) = \frac{3}{4} \stackrel{(\sqrt)}{\Rightarrow} sin(\frac{\pi}{3}) = \frac{\sqrt3}{2}$ $\square$

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Hint. From the picture below and Pythagoras we get

$$h=\frac{\sqrt 3}{2}a\rightarrow \sin\frac\pi 3=\frac ha=\frac{\frac{\sqrt 3}{2}a}{a}=\frac{\sqrt 3}{2}.$$

enter image description here

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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

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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 $\sin(\pi/3) = \sqrt{3}/2$

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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 $sin(\frac{\pi}{3})=\frac{\sqrt3}{2}$

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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?

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