Ratios of standard angles

The angles $0^\circ$, $30^\circ$, $45^\circ$, $60^\circ$ and $90^\circ$ appear in prob­lems so often that their ratios should be known by heart. They do not need to be mem­o­rised blindly, though: each one comes from a sim­ple fig­ure, a square cut by its diag­o­nal or an equi­lat­eral tri­an­gle cut in half. This les­son derives the table of val­ues, shows a pat­tern that makes it easy to remem­ber, and uses it to eval­u­ate expres­sions and find unknown angles.

The angles 0° and 90°#

Take a point at dis­tance $r$ on the pos­i­tive x-axis: $P(r, 0)$. Its angle is $0^\circ$, so

$$\displaystyle \sin 0^\circ = \frac{0}{r} = 0, \qquad \cos 0^\circ = \frac{r}{r} = 1, \qquad \tan 0^\circ = 0$$

Take $P(0, r)$ on the pos­i­tive y-axis. Its angle is $90^\circ$, so

$$\displaystyle \sin 90^\circ = 1, \qquad \cos 90^\circ = 0, \qquad \tan 90^\circ = \frac{r}{0} \text{ (not defined)}$$

The angle 45°#

In a right tri­an­gle with a $45^\circ$ angle, the other acute angle is also $45^\circ$, so the two legs are equal; call each $a$. The hypotenuse is $\sqrt{a^2 + a^2} = \sqrt{2}a$.

$$\displaystyle \sin 45^\circ = \cos 45^\circ = \frac{a}{\sqrt{2}a} = \frac{1}{\sqrt{2}}, \qquad \tan 45^\circ = \frac{a}{a} = 1$$

The angles 30° and 60°#

Take an equi­lat­eral tri­an­gle of side $a$; each angle is $60^\circ$. The per­pen­dic­u­lar from one ver­tex to the oppo­site side bisects that side and the ver­tex angle, mak­ing a right tri­an­gle with angles $30^\circ$, $60^\circ$, $90^\circ$, hypotenuse $a$, short side $\displaystyle \frac{a}{2}$ and height $\displaystyle \sqrt{a^2 - \frac{a^2}{4}} = \frac{\sqrt{3}}{2}a$.

The short side is oppo­site $30^\circ$ and the height is oppo­site $60^\circ$, so

$$\displaystyle \sin 30^\circ = \frac{1}{2}, \quad \cos 30^\circ = \frac{\sqrt{3}}{2}, \quad \sin 60^\circ = \frac{\sqrt{3}}{2}, \quad \cos 60^\circ = \frac{1}{2}$$

and $\displaystyle \tan 30^\circ = \frac{1}{\sqrt{3}}$, $\tan 60^\circ = \sqrt{3}$.

The table#

A pat­tern for sine. The sines of $0^\circ, 30^\circ, 45^\circ, 60^\circ, 90^\circ$ are $\displaystyle \sqrt{\frac{0}{4}}, \sqrt{\frac{1}{4}}, \sqrt{\frac{2}{4}}, \sqrt{\frac{3}{4}}, \sqrt{\frac{4}{4}}$, that is $\displaystyle 0, \frac{1}{2}, \frac{1}{\sqrt{2}}, \frac{\sqrt{3}}{2}, 1$. The cosines are the same list in reverse order. Every other ratio fol­lows from these two.

As $\theta$ grows from $0^\circ$ to $90^\circ$, $\sin\theta$ rises from $0$ to $1$ and $\cos\theta$ falls from $1$ to $0$.

Exam­ple 1#

Eval­u­ate $\sin 30^\circ \cos 60^\circ + \cos 30^\circ \sin 60^\circ$.

$$\displaystyle \frac{1}{2} \times \frac{1}{2} + \frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{2} = \frac{1}{4} + \frac{3}{4} = 1$$

Exam­ple 2#

Eval­u­ate $\displaystyle \frac{\tan^2 60^\circ + 4\cos^2 45^\circ + 3\sec^2 30^\circ + 5\cos^2 90^\circ}{\csc 30^\circ + \sec 60^\circ - \cot^2 30^\circ}$.

Numer­a­tor: $\displaystyle 3 + 4 \times \frac{1}{2} + 3 \times \frac{4}{3} + 0 = 3 + 2 + 4 = 9$.

Denom­i­na­tor: $2 + 2 - 3 = 1$.

The value is $9$.

Exam­ple 3#

Find the acute angle $\theta$ if $2\sin\theta = \sqrt{3}$.

$\displaystyle \sin\theta = \frac{\sqrt{3}}{2}$, and from the table $\displaystyle \theta = 60^\circ = \frac{\pi}{3}$.

Exam­ple 4#

If $\sin(A + B) = 1$ and $\displaystyle \cos(A - B) = \frac{\sqrt{3}}{2}$, where $A$ and $B$ are acute and $A > B$, find $A$ and $B$.

$\sin(A + B) = 1$ gives $A + B = 90^\circ$. $\displaystyle \cos(A - B) = \frac{\sqrt{3}}{2}$ gives $A - B = 30^\circ$. Adding, $2A = 120^\circ$, so $A = 60^\circ$ and $B = 30^\circ$.

Prac­tice#

1. Eval­u­ate $\sin 60^\circ \cos 30^\circ - \cos 60^\circ \sin 30^\circ$.
2. Eval­u­ate $2\tan^2 45^\circ + \cos^2 30^\circ - \sin^2 60^\circ$.
3. Eval­u­ate $\displaystyle \frac{\cos 60^\circ + \sin 30^\circ}{\tan 45^\circ + \cot 45^\circ}$.
4. Eval­u­ate $4\cot^2 45^\circ - \sec^2 60^\circ + \sin^2 60^\circ + \cos^2 90^\circ$.
5. Eval­u­ate $\displaystyle \sin\frac{\pi}{4}\cos\frac{\pi}{4} + \tan\frac{\pi}{3}\cot\frac{\pi}{3}$.
6. Find the acute angle $\theta$ if $\sqrt{3}\tan\theta = 1$.
7. Find the acute angle $\theta$ if $2\cos 2\theta = 1$.
8. If $\displaystyle \sin(A - B) = \frac{1}{2}$ and $\displaystyle \cos(A + B) = \frac{1}{2}$, where $A$ and $B$ are acute and $A > B$, find $A$ and $B$.
9. Ver­ify that $\cos 60^\circ = 1 - 2\sin^2 30^\circ$.

Answers#