This subsection is not completed yet, but in the meantime here are some practice questions.

Remember:

The special angles that you need to know, with their sines, cosines, and tangents are:

$\theta$ | $\sin\theta$ | $\cos\theta$ | $\tan\theta$ |

$30^\circ=\pi/6$ | $\dfrac{\sqrt1}{2}$ | $\dfrac{\sqrt3}{2}$ | $\dfrac{\sqrt1}{\sqrt3}=\dfrac{1}{\sqrt3}$ |

$45^\circ=\pi/4$ | $\dfrac{\sqrt2}{2}$ | $\dfrac{\sqrt2}{2}$ | $\dfrac{\sqrt2}{\sqrt2} =1$ |

$60^\circ=\pi/3$ | $\dfrac{\sqrt3}{2}$ | $\dfrac{\sqrt1}{2}$ | $\dfrac{\sqrt3}{\sqrt1}=\sqrt3$ |

When written in this way, the sines make a lovely pattern of $\sqrt1/2$, $\sqrt2/2$, $\sqrt3/2$, with the cosines in the opposite order. You can work out the tangents by dividing sine by cosine. (I’m using $\sqrt1/2$ to make the pattern clear, but we would of course use $1/2$ instead.)

Alternatively, you can work it out by sketching the special triangles:

Or if you want to use a calculator, you can learn to recognise $0.866\cdots$ as $\sqrt3/2$ and $0.707\cdots$ as $\sqrt2/2$, or you can get the exact value by squaring the answer, for example $\sin60^\circ$ will give $0.866\cdots$, and if you square this, you’ll get $3/4$, whose square root is $\sqrt3/2$.