Trigonometry Values Pi/6

Given a Right Triangle with angles defined to be $\alpha$ and $2\alpha$, it must be true that

$$\alpha+2\alpha+{\textstyle{1\over 2}}\pi = \pi,$$ (1)

so $\alpha = {\pi/6}$. Define the hypotenuse to have length $1$ and the side opposite $\alpha$ to have length $x$, then the side opposite $2\alpha$ has length $\sqrt{1-x^2}$. This gives $\sin\alpha\equiv x$ and

$$\sin(2\alpha) = \sqrt{1-x^2}.$$ (2)

But

$$\sin(2\alpha)=2\sin\alpha\cos\alpha=2x\sqrt{1-x^2},$$ (3)

so we have

$$\sqrt{1-x^2}=2x\sqrt{1-x^2}\,.$$ (4)

This gives $2x=1$, or

$$\sin\left({\pi\over 6}\right)= {\textstyle{1\over 2}}.$$ (5)

$\cos(\pi/6)$ is then computed from

$$\cos\left({\pi\over 6}\right)= \sqrt{1 - \sin^2\left({\pi \o... ...{\textstyle{1\over 2}})^2} = {\textstyle{1\over 2}}\sqrt{3}\,.$$ (6)

Summarizing,

$$\sin\left({\pi\over 6}\right) = {\textstyle{1\over 2}}$$ (7)

$$\cos\left({\pi\over 6}\right) = {\textstyle{1\over 2}}\sqrt{3}$$ (8)

$$\tan\left({\pi\over 6}\right) = {\textstyle{1\over 3}}\sqrt{3}\,.$$ (9)

See also Hexagon, Hexagram