5.4 Exact trigonometric values — Practise with help

1. $\sqrt{2} \times \sqrt{2} = 2$ and $\sqrt{3} \times \sqrt{3} = 3$.
2. $1^2 + 1^2 = 2$, so the hypotenuse is $\sqrt{2}$.
3. $180 - 150 = 30$ and $\dfrac{180}{150} = \dfrac{6}{5}$.

If question 1 was hard, revise simplifying surds before you continue.

These come from the half-equilateral triangle with sides $1$, $\sqrt{3}$ and $2$.

For the $60^\circ$ angle, the side opposite it is $\sqrt{3}$ and the side next to it is $1$. The hypotenuse is $2$.

$\tan 30^\circ = \dfrac{\sin 30^\circ}{\cos 30^\circ}$, or read it from the half-equilateral triangle.

From the triangle, the side opposite $30^\circ$ is $1$ and the side next to it is $\sqrt{3}$, so $\tan 30^\circ = \dfrac{1}{\sqrt{3}}$. Rationalise the denominator.

$\dfrac{2\pi}{3}$ is $120^\circ$, in the second quadrant. Only sine is positive there.

The related acute angle is $\pi - \dfrac{2\pi}{3} = \dfrac{\pi}{3}$, which is $60^\circ$. So the answer is $+\sin 60^\circ$.

$\dfrac{5\pi}{4}$ is $225^\circ$, in the third quadrant. Only tangent is positive there.

The related acute angle is $\dfrac{5\pi}{4} - \pi = \dfrac{\pi}{4}$, which is $45^\circ$. Cosine is negative in the third quadrant.

$\dfrac{5\pi}{6}$ is $150^\circ$, in the second quadrant. Tangent is negative there.

The related acute angle is $\pi - \dfrac{5\pi}{6} = \dfrac{\pi}{6}$, which is $30^\circ$. So the answer is $-\tan 30^\circ$.

Work out each term on its own. $\dfrac{7\pi}{6}$ is $210^\circ$, in the third quadrant.

$\sin\dfrac{7\pi}{6} = -\sin\dfrac{\pi}{6} = -\dfrac{1}{2}$. Also $\cos\dfrac{\pi}{3} = \dfrac{1}{2}$. Add the two values.