# Useful Triangles for Trigonometry

## $$\sin45$$, $$\cos45$$ and $$\tan45$$

Sketch an isosceles right-angled triangle and mark the equal sides as length $$1$$.

Next, as it is isoseles and we know one of the angles is $$90^{\circ}$$, the other two must be $$45^{\circ}$$.

From Pythagoras' theorem $$a^2+b^2=c^2$$, it can easily be seen that the hypotenuse is of length $$\sqrt{2}$$.

Now we just need our SOH CAH TOA;

$$\sin45=\frac{\text{opposite}}{\text{hypotenuse}}=\frac{1}{\sqrt{2}}$$
$$\cos45=\frac{\text{adjacent}}{\text{hypotenuse}}=\frac{1}{\sqrt{2}}$$
$$\tan45=\frac{\text{opposite}}{\text{adjacent}}=\frac{1}{1}=1$$

## $$\sin30$$, $$\cos30$$, $$\tan30$$, $$\sin60$$, $$\cos60$$ and $$\tan60$$

First, sketch an equilateral triangle and mark the sides as length $$2$$. Because it's equilateral, we know that the angles are all $$60^{\circ}$$.

By bisecting the top (or any other) angle, we can create two right-angled trianlges. The base of each triangle will be of length $$1$$; half of the $$2$$ it was.

We only need one of these trianlges, so we take the shaded one as shown here. Once you know where this comes from, you can go straight to the last image. This method requires no memory of the final dimensions though and you can always start from the top!

Using Phythagoras' Theorem again, we can find that the vertical side is of length $$\sqrt{3}$$.

Now we just need our SOH CAH TOA;

$$\sin30=\frac{\text{opposite}}{\text{hypotenuse}}=\frac{1}{2}$$
$$\cos30=\frac{\text{adjacent}}{\text{hypotenuse}}=\frac{\sqrt{3}}{2}$$
$$\tan30=\frac{\text{opposite}}{\text{adjacent}}=\frac{1}{\sqrt{3}}$$
$$\sin60=\frac{\text{opposite}}{\text{hypotenuse}}=\frac{\sqrt{3}}{2}$$
$$\cos60=\frac{\text{adjacent}}{\text{hypotenuse}}=\frac{1}{2}$$
$$\tan60=\frac{\text{opposite}}{\text{adjacent}}=\frac{\sqrt{3}}{1}=\sqrt{3}$$