5.4 Exact trigonometric values — Practise

1. The value

Answer: From the half-equilateral triangle with sides $1$, $\sqrt{3}$ and $2$, the side opposite $60^\circ$ is $\sqrt{3}$ and the hypotenuse is $2$. So $\sin 60^\circ = \dfrac{\sqrt{3}}{2}$.

2. The value

Answer: From the half-equilateral triangle, $\tan 30^\circ = \dfrac{1}{\sqrt{3}}$. Rationalising gives $\dfrac{1}{\sqrt{3}} \times \dfrac{\sqrt{3}}{\sqrt{3}} = \dfrac{\sqrt{3}}{3}$.

3. The value

Answer: $\dfrac{2\pi}{3}$ is $120^\circ$, in the second quadrant where sine is positive. The related acute angle is $\dfrac{\pi}{3}$, so $\sin\dfrac{2\pi}{3} = \sin\dfrac{\pi}{3} = \dfrac{\sqrt{3}}{2}$.

4. The value

Answer: $\dfrac{5\pi}{4}$ is $225^\circ$, in the third quadrant where cosine is negative. The related acute angle is $\dfrac{\pi}{4}$, so $\cos\dfrac{5\pi}{4} = -\cos\dfrac{\pi}{4} = -\dfrac{\sqrt{2}}{2}$.

5. The value

Answer: $\dfrac{5\pi}{6}$ is $150^\circ$, in the second quadrant where tangent is negative. The related acute angle is $\dfrac{\pi}{6}$, so $\tan\dfrac{5\pi}{6} = -\tan\dfrac{\pi}{6} = -\dfrac{\sqrt{3}}{3}$.

6. The value

Answer: $\dfrac{7\pi}{6}$ is $210^\circ$, in the third quadrant, so $\sin\dfrac{7\pi}{6} = -\dfrac{1}{2}$. Also $\cos\dfrac{\pi}{3} = \dfrac{1}{2}$. The sum is $-\dfrac{1}{2} + \dfrac{1}{2} = 0$.

7. The value

Answer: $\dfrac{2\pi}{3}$ is $120^\circ$, in the second quadrant where cosine is negative. The related acute angle is $\dfrac{\pi}{3}$, so $\cos\dfrac{2\pi}{3} = -\dfrac{1}{2}$.

8. The value

Answer: Both factors equal $\dfrac{\sqrt{2}}{2}$, so their product is $\dfrac{\sqrt{2}}{2} \times \dfrac{\sqrt{2}}{2} = \dfrac{2}{4} = \dfrac{1}{2}$.

9. The value

Answer: Each term is $\dfrac{\sqrt{2}}{2}$, so the sum is $2 \times \dfrac{\sqrt{2}}{2} = \sqrt{2}$.

10. The value

Answer: From the half-equilateral triangle, the side opposite $60^\circ$ is $\sqrt{3}$ and the side next to it is $1$, so $\tan 60^\circ = \sqrt{3}$.