Trigonometric ratios of standard angles

The angles $0^\circ, 30^\circ, 45^\circ, 60^\circ, 90^\circ$ appear constantly in geometry and trigonometry, and their sin/cos/tan have simple, exact values. Memorise these , they are the multiplication table of trigonometry.

The standard-angle table

(reciprocals: $\csc, \sec, \cot$ are the inverses of these.)

A quick mnemonic: write $\sin$ values as $\sqrt{0}/2, \sqrt{1}/2, \sqrt{2}/2, \sqrt{3}/2, \sqrt{4}/2$ for $0^\circ, 30^\circ, 45^\circ, 60^\circ, 90^\circ$. $\cos$ is the same list reversed.

Derivation

$45^\circ$. Consider an isoceles right triangle with legs $1, 1$. Hypotenuse $= \sqrt{2}$. So $\sin 45^\circ = 1/\sqrt{2}, \cos 45^\circ = 1/\sqrt{2}, \tan 45^\circ = 1$.

$30^\circ$ and $60^\circ$. Consider an equilateral triangle of side $2$. Drop the altitude , it splits the triangle into two right triangles with legs $1$ (half-base) and $\sqrt{3}$ (altitude), hypotenuse $2$. The angles in each right triangle are $30^\circ, 60^\circ, 90^\circ$.

For the $30^\circ$ angle (at the apex of the half-triangle): opp $= 1$, adj $= \sqrt{3}$, hyp $= 2$. $\sin 30^\circ = 1/2, \cos 30^\circ = \sqrt{3}/2, \tan 30^\circ = 1/\sqrt{3}$.

For the $60^\circ$ angle (at the base): opp $= \sqrt{3}$, adj $= 1$, hyp $= 2$. $\sin 60^\circ = \sqrt{3}/2, \cos 60^\circ = 1/2, \tan 60^\circ = \sqrt{3}$.

$0^\circ$ and $90^\circ$. These are limiting cases. As $\theta \to 0^\circ$, the opposite side shrinks to $0$ while the adjacent approaches the hypotenuse. So $\sin 0^\circ = 0, \cos 0^\circ = 1, \tan 0^\circ = 0$.

As $\theta \to 90^\circ$, the opposite approaches the hypotenuse and the adjacent shrinks to $0$. So $\sin 90^\circ = 1, \cos 90^\circ = 0$, $\tan 90^\circ$ is undefined (division by zero).

Patterns to remember

• $\sin$ increases from $0$ to $1$ as $\theta$ goes $0 \to 90^\circ$.
• $\cos$ decreases from $1$ to $0$ as $\theta$ goes $0 \to 90^\circ$.
• $\tan$ increases from $0$ and shoots to infinity as $\theta \to 90^\circ$.
• $\sin\theta = \cos(90^\circ - \theta)$ , a special-angle peek at the complementary identity.

Worked examples

Example 1. Compute $\sin 30^\circ + \cos 60^\circ$.

$\sin 30^\circ + \cos 60^\circ = 1/2 + 1/2 = 1$.

Example 2. Evaluate $\sin^2 30^\circ + \cos^2 30^\circ$.

$= 1/4 + 3/4 = 1$. (As predicted by the Pythagorean identity.)

Example 3. Evaluate $\tan 60^\circ - \cot 60^\circ$.

$= \sqrt{3} - 1/\sqrt{3} = (3 - 1)/\sqrt{3} = 2/\sqrt{3}$.

Example 4. Find $\theta$ if $\sin\theta = \sqrt{3}/2$, with $\theta \in (0^\circ, 90^\circ)$.

From the table, $\theta = 60^\circ$.

Example 5. Show that $\dfrac{\sin 30^\circ \cdot \cos 60^\circ + \cos 30^\circ \cdot \sin 60^\circ}{1} = \sin 90^\circ$.

LHS: $\frac{1}{2} \cdot \frac{1}{2} + \frac{\sqrt{3}}{2} \cdot \frac{\sqrt{3}}{2} = \frac{1}{4} + \frac{3}{4} = 1 = \sin 90^\circ$. ✓

(This previews the sine addition formula, which you'll learn formally in class XI.)

Try it yourself

1. Find: $\sin 45^\circ + \cos 45^\circ$.
2. Find: $\tan 45^\circ \cdot \cot 45^\circ$.
3. Evaluate $5 \sin 60^\circ + 4 \cos 30^\circ$.
4. Evaluate $\sin^2 60^\circ + \cos^2 60^\circ$.
5. Find $\theta$ if $\tan\theta = 1$.
6. Evaluate $\dfrac{4 \sin^2 30^\circ - \cos^2 60^\circ}{1 + \tan^2 45^\circ}$.
7. Compute $\dfrac{\sin 30^\circ + \tan 45^\circ - \cos 60^\circ}{\sec 60^\circ + \cot 45^\circ}$.
8. Verify: $\sin 30^\circ \cdot \cos 60^\circ + \cos 30^\circ \cdot \sin 60^\circ = 1$.
9. Find $\theta \in (0^\circ, 90^\circ)$ if $\cos\theta = 1/2$.
10. Show that $\sin 60^\circ = \cos 30^\circ$.

Pitfalls / Insight

• Memorise the table. It will pay off in every chapter that follows.
• $\tan 90^\circ$ is undefined, not $\infty$ (that distinction matters at higher levels).
• Watch for $\sin^2\theta$ vs $\sin\theta^2$. The first is $(\sin\theta)^2$; the second is $\sin(\theta^2)$. We always mean the first.

Insight. The standard angles aren't magic , they come from one isoceles right triangle and one equilateral triangle. Recreate them on paper a few times and the table will stick.