Trigonometric Ratios of Specific Angles

The Standard Values Table

A handful of angles appear again and again: 0°, 30°, 45°, 60°, 90°. Their exact ratio values are worth memorising cold.

Ratio	0°	30°	45°	60°	90°
$\sin$	0	$\tfrac12$	$\tfrac{1}{\sqrt2}$	$\tfrac{\sqrt3}{2}$	1
$\cos$	1	$\tfrac{\sqrt3}{2}$	$\tfrac{1}{\sqrt2}$	$\tfrac12$	0
$\tan$	0	$\tfrac{1}{\sqrt3}$	1	$\sqrt3$	not defined

Key Point: Notice $\cos$ is just $\sin$ read backwards. And $\tan = \sin/\cos$.

[Board Important] $\tan 90°$ is not defined because $\cos 90° = 0$ and we cannot divide by zero. Similarly $\sec 90°$ and $\csc 0°$ are undefined.

A Memory Trick for sin

There's a neat pattern. Write $0, 1, 2, 3, 4$ under the angles $0°, 30°, 45°, 60°, 90°$, then take $\sqrt{\;}$ and divide by 2:

$$\sin\theta = \sqrt{\frac{0}{4}}, \sqrt{\frac{1}{4}}, \sqrt{\frac{2}{4}}, \sqrt{\frac{3}{4}}, \sqrt{\frac{4}{4}} = 0, \tfrac12, \tfrac{1}{\sqrt2}, \tfrac{\sqrt3}{2}, 1$$

Then $\cos$ is the same list reversed.

Key Point: $\sin\theta = \dfrac{\sqrt{n}}{2}$ where $n = 0,1,2,3,4$ for $0°,30°,45°,60°,90°$.

[JEE/NEET Tip] This $\sqrt{n}/2$ trick is the fastest way to reconstruct the table if you blank out in an exam.

Where the Values Come From

The values aren't magic — they come from two special triangles:

• A 45°-45°-90° triangle has legs equal, so if each leg is 1, the hypotenuse is $\sqrt2$, giving $\sin 45° = \cos 45° = \tfrac{1}{\sqrt2}$.
• A 30°-60°-90° triangle (half an equilateral triangle of side 2) has sides $1, \sqrt3, 2$, giving $\sin 30° = \tfrac12$, $\sin 60° = \tfrac{\sqrt3}{2}$, etc.

Key Point: Remember the two special triangles and you can re-derive every standard value.

[Board Important] Drawing the special triangle is a valid method to justify a value if you forget the table.

Reciprocal Ratios at Standard Angles

From the table, the reciprocal ratios follow at once:

• $\csc 30° = 2$, $\csc 45° = \sqrt2$, $\csc 60° = \tfrac{2}{\sqrt3}$.
• $\sec 30° = \tfrac{2}{\sqrt3}$, $\sec 45° = \sqrt2$, $\sec 60° = 2$.
• $\cot 30° = \sqrt3$, $\cot 45° = 1$, $\cot 60° = \tfrac{1}{\sqrt3}$.

Key Point: cosec, sec, cot at standard angles are just reciprocals of sin, cos, tan.

[Board Important] $\sec 0° = 1$ and $\csc 90° = 1$; but $\sec 90°$ and $\csc 0°$ are undefined (division by zero).

Solved Examples

Example 1: Direct evaluation

Evaluate $\sin 30° + \cos 60°$.

Solution:

1. $\sin 30° = \tfrac12$, $\cos 60° = \tfrac12$.
2. Sum $= \tfrac12 + \tfrac12 = 1$.

Final Answer: 1.

Takeaway: Read straight off the table.

Example 2: Product

Evaluate $\sin 60° \cos 30° + \cos 60° \sin 30°$.

Solution:

1. $= \tfrac{\sqrt3}{2}\cdot\tfrac{\sqrt3}{2} + \tfrac12\cdot\tfrac12 = \tfrac34 + \tfrac14 = 1$.

Final Answer: 1.

Takeaway: This is $\sin(60°+30°) = \sin 90° = 1$ — a nice check.

Example 3: tan values

Evaluate $\tan^2 60° + \tan^2 45°$.

Solution:

1. $\tan 60° = \sqrt3$, so $\tan^2 60° = 3$.
2. $\tan 45° = 1$, so $\tan^2 45° = 1$.
3. Sum $= 3 + 1 = 4$.

Final Answer: 4.

Takeaway: Square after substituting the value.

Example 4: Mixed expression

Evaluate $\dfrac{\tan 30°}{\cot 60°}$.

Solution:

1. $\tan 30° = \tfrac{1}{\sqrt3}$, $\cot 60° = \tfrac{1}{\sqrt3}$.
2. Ratio $= \dfrac{1/\sqrt3}{1/\sqrt3} = 1$.

Final Answer: 1.

Takeaway: $\tan 30° = \cot 60°$ — a complementary-angle hint.

Example 5: Find the angle

If $\sin\theta = \tfrac{\sqrt3}{2}$ and $\theta$ is acute, find $\theta$.

Solution:

1. From the table, $\sin 60° = \tfrac{\sqrt3}{2}$.

Final Answer: $\theta = 60°$.

Takeaway: Reading the table backwards finds the angle.

Example 6: Evaluate with squares

Evaluate $4\sin^2 60° - 3\tan^2 30°$.

Solution:

1. $\sin^2 60° = \tfrac34$, so $4\cdot\tfrac34 = 3$.
2. $\tan^2 30° = \tfrac13$, so $3\cdot\tfrac13 = 1$.
3. $3 - 1 = 2$.

Final Answer: 2.

Takeaway: Square the value first, then multiply by the coefficient.

Example 7: Fraction

Evaluate $\dfrac{\cos 45°}{\sec 30° + \csc 30°}$.

Solution:

1. $\cos 45° = \tfrac{1}{\sqrt2}$; $\sec 30° = \tfrac{2}{\sqrt3}$; $\csc 30° = 2$.
2. Denominator $= \tfrac{2}{\sqrt3} + 2 = \tfrac{2 + 2\sqrt3}{\sqrt3}$.
3. $\dfrac{1/\sqrt2}{(2+2\sqrt3)/\sqrt3} = \dfrac{\sqrt3}{\sqrt2 \,(2 + 2\sqrt3)} = \dfrac{\sqrt3}{2\sqrt2(1+\sqrt3)}$.
4. Rationalise: $= \dfrac{\sqrt3(\sqrt3 - 1)}{2\sqrt2(3-1)} = \dfrac{3 - \sqrt3}{4\sqrt2}$.

Final Answer: $\dfrac{3 - \sqrt3}{4\sqrt2}$.

Takeaway: Substitute, combine, then rationalise the surd.

Example 8: Verify an identity numerically

Verify $\sin^2 30° + \cos^2 30° = 1$.

Solution:

1. $\sin^2 30° = \tfrac14$, $\cos^2 30° = \tfrac34$.
2. Sum $= \tfrac14 + \tfrac34 = 1$.

Final Answer: 1 — verified.

Takeaway: $\sin^2\theta + \cos^2\theta = 1$ holds for every angle.

Example 9: Solve for angle in an equation

Find acute $\theta$ if $2\cos\theta = 1$.

Solution:

1. $\cos\theta = \tfrac12$.
2. From the table, $\cos 60° = \tfrac12$.

Final Answer: $\theta = 60°$.

Takeaway: Isolate the ratio, then read the angle.

Example 10: Combine angles

Evaluate $\sin 90° - 2\cos^2 45° + \tan^2 60°$.

Solution:

1. $\sin 90° = 1$.
2. $\cos^2 45° = \tfrac12$, so $2\cdot\tfrac12 = 1$.
3. $\tan^2 60° = 3$.
4. $1 - 1 + 3 = 3$.

Final Answer: 3.

Takeaway: Evaluate each term separately, then combine.