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?
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?
On
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
On
Consider an equilateral triangle of side length $a$. We will compute the area by two different ways to compute the value of $\sin(\pi/3)$.
Comparing the two, we obtain that $\sin(\pi/3) = \sqrt{3}/2$
On
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 $sin(\frac{\pi}{3})=\frac{\sqrt3}{2}$
On
Not to confuse you. But.. $$x=\dfrac{\pi}{3}$$ $$3x=\pi$$ $$\sin 3x=\sin \pi$$ $$3\sin x-4\sin^3x=0$$ $$\sin x=0 \text{ or } \sin x = \dfrac{\sqrt{3}}{2} \text{ or } \sin x = \dfrac{-\sqrt{3}}{2}$$
$$\text{as } 0<\dfrac{\pi}{3}<\dfrac{\pi}{2} \text{, } \sin \dfrac{\pi}{3} = \dfrac{\sqrt{3}}{2}$$
Hint. From the picture below and Pythagoras we get
$$h=\frac{\sqrt 3}{2}a\rightarrow \sin\frac\pi 3=\frac ha=\frac{\frac{\sqrt 3}{2}a}{a}=\frac{\sqrt 3}{2}.$$