Trigonometric ratios of 30°, 45°, 60° — derive and build the table

Question

Derive the trigonometric ratios for $30°$ , $45°$ , and $60°$ and present them in a table. Show the derivation using special triangles.

(NCERT Class 10, Chapter 8 — Introduction to Trigonometry)

Solution — Step by Step

Take a right triangle with both legs equal to 1. By Pythagoras, the hypotenuse is $\sqrt{1^2 + 1^2} = \sqrt{2}$ .

The angles opposite equal sides are both $45°$ . So:

\sin 45° = \frac{1}{\sqrt{2}}, \quad \cos 45° = \frac{1}{\sqrt{2}}, \quad \tan 45° = \frac{1}{1} = 1

Start with an equilateral triangle of side 2. Drop a perpendicular from one vertex to the opposite side — this bisects the base into two segments of length 1 and creates a $30°$ - $60°$ - $90°$ triangle.

The perpendicular height $= \sqrt{2^2 - 1^2} = \sqrt{3}$ .

Now we have sides: 1 (short), $\sqrt{3}$ (medium), 2 (hypotenuse).

In the $30°$ - $60°$ - $90°$ triangle, the side opposite $30°$ is 1:

\sin 30° = \frac{1}{2}, \quad \cos 30° = \frac{\sqrt{3}}{2}, \quad \tan 30° = \frac{1}{\sqrt{3}}

The side opposite $60°$ is $\sqrt{3}$ :

\sin 60° = \frac{\sqrt{3}}{2}, \quad \cos 60° = \frac{1}{2}, \quad \tan 60° = \frac{\sqrt{3}}{1} = \sqrt{3}

Angle	$\sin$	$\cos$	$\tan$	$\csc$	$\sec$	$\cot$
$30°$	$\frac{1}{2}$	$\frac{\sqrt{3}}{2}$	$\frac{1}{\sqrt{3}}$	$2$	$\frac{2}{\sqrt{3}}$	$\sqrt{3}$
$45°$	$\frac{1}{\sqrt{2}}$	$\frac{1}{\sqrt{2}}$	$1$	$\sqrt{2}$	$\sqrt{2}$	$1$
$60°$	$\frac{\sqrt{3}}{2}$	$\frac{1}{2}$	$\sqrt{3}$	$\frac{2}{\sqrt{3}}$	$2$	$\frac{1}{\sqrt{3}}$

Why This Works

We’re not pulling numbers out of thin air — every value comes from the sides of two special triangles: the $45°$ - $45°$ - $90°$ (isosceles right triangle) and the $30°$ - $60°$ - $90°$ (half of an equilateral triangle). These are the only triangles where the side ratios involve clean surds.

Notice the beautiful pattern: $\sin 30° = \cos 60°$ and $\sin 60° = \cos 30°$ . This happens because $30° + 60° = 90°$ , and for complementary angles, sine and cosine swap.

Alternative Method — The Memory Pattern

For quick recall, use the “0-1-2-3-4” pattern for sine values of $0°, 30°, 45°, 60°, 90°$ :

\sin \theta = \frac{\sqrt{0}}{2}, \frac{\sqrt{1}}{2}, \frac{\sqrt{2}}{2}, \frac{\sqrt{3}}{2}, \frac{\sqrt{4}}{2}

That gives $0, \frac{1}{2}, \frac{1}{\sqrt{2}}, \frac{\sqrt{3}}{2}, 1$ . For cosine, reverse the order. This trick saves precious time in board exams.

Common Mistake

Students swap $\sin 30°$ and $\sin 60°$ all the time. Here’s a quick check: a bigger angle should have a bigger sine (at least between $0°$ and $90°$ ). Since $60° > 30°$ , we need $\sin 60° > \sin 30°$ . Indeed, $\frac{\sqrt{3}}{2} \approx 0.87 > 0.5 = \frac{1}{2}$ . If your values don’t follow this, you’ve swapped them.

Question

Solution — Step by Step

Why This Works

Alternative Method — The Memory Pattern

Common Mistake

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