Trigonometric Ratios and Standard Values | JEE Main Trigonometry & Inverse Trigonometry Crash Course | Mathematicon

Trigonometric Ratios and Standard Values: Your JEE Foundation

Welcome, future engineers! Trigonometry is not just about triangles; it's a fundamental tool that pops up everywhere in JEE Main, from calculus to coordinate geometry and even physics. Mastering trigonometric ratios and their values at standard angles is crucial. This lesson provides a comprehensive guide to these essential concepts, equipping you with the knowledge and intuition to tackle a wide range of problems. Let's begin!

Understanding Trigonometric Ratios

Imagine a right-angled triangle. The trigonometric ratios relate the angles of this triangle to the ratios of its sides. There are six fundamental trigonometric ratios:

Sine ( $\sin$ )
Cosine ( $\cos$ )
Tangent ( $\tan$ )
Cosecant ( $\csc$ )
Secant ( $\sec$ )
Cotangent ( $\cot$ )

Consider a right-angled triangle $ABC$ , where angle $B$ is $90^\circ$ . Let's consider the angle $\theta$ at vertex $A$ . Then, $BC$ is the side opposite to $\theta$ , $AB$ is the side adjacent to $\theta$ , and $AC$ is the hypotenuse.

1. Sine: $\sin\theta = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{BC}{AC}$

The sine of an angle is the ratio of the length of the side opposite to the angle to the length of the hypotenuse. As $\theta$ increases from 0 to 90 degrees, the opposite side approaches the length of the hypotenuse, so $\sin\theta$ increases.

2. Cosine: $\cos\theta = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{AB}{AC}$

The cosine of an angle is the ratio of the length of the side adjacent to the angle to the length of the hypotenuse. As $\theta$ increases from 0 to 90 degrees, the adjacent side shrinks towards zero, so $\cos\theta$ decreases.

3. Tangent: $\tan\theta = \frac{\sin\theta}{\cos\theta} = \frac{\text{opposite}}{\text{adjacent}} = \frac{BC}{AB}$

The tangent of an angle is the ratio of the length of the side opposite to the angle to the length of the side adjacent to the angle. You can also think of it as the slope of the line that forms the angle $\theta$ with the x-axis. The formula $\tan\theta = \frac{\sin\theta}{\cos\theta}$ is directly derived from the definitions of sine and cosine.

4. Cosecant, Secant, and Cotangent:

$\csc\theta = \frac{1}{\sin\theta} = \frac{\text{hypotenuse}}{\text{opposite}} = \frac{AC}{BC}$
$\sec\theta = \frac{1}{\cos\theta} = \frac{\text{hypotenuse}}{\text{adjacent}} = \frac{AC}{AB}$
$\cot\theta = \frac{1}{\tan\theta} = \frac{\cos\theta}{\sin\theta} = \frac{\text{adjacent}}{\text{opposite}} = \frac{AB}{BC}$

These are the reciprocals of sine, cosine, and tangent, respectively. Knowing these relationships simplifies calculations.

Standard Angles and Their Trigonometric Values

Certain angles appear frequently in trigonometric problems. Knowing the values of trigonometric ratios for these standard angles will save you valuable time during the JEE Main exam.

These standard angles are: 0°, 30°, 45°, 60°, and 90°.

Angle ( $\theta$ )	$\sin\theta$	$\cos\theta$	$\tan\theta$	$\csc\theta$	$\sec\theta$	$\cot\theta$
0°	0	1	0	Undefined	1	Undefined
30° ( $\frac{\pi}{6}$ )	$\frac{1}{2}$	$\frac{\sqrt{3}}{2}$	$\frac{1}{\sqrt{3}}$	2	$\frac{2}{\sqrt{3}}$	$\sqrt{3}$
45° ( $\frac{\pi}{4}$ )	$\frac{1}{\sqrt{2}}$	$\frac{1}{\sqrt{2}}$	1	$\sqrt{2}$	$\sqrt{2}$	1
60° ( $\frac{\pi}{3}$ )	$\frac{\sqrt{3}}{2}$	$\frac{1}{2}$	$\sqrt{3}$	$\frac{2}{\sqrt{3}}$	2	$\frac{1}{\sqrt{3}}$
90° ( $\frac{\pi}{2}$ )	1	0	Undefined	1	Undefined	0

How to remember this table?

For $\sin\theta$ : Write down $\sqrt{0/4}, \sqrt{1/4}, \sqrt{2/4}, \sqrt{3/4}, \sqrt{4/4}$ . Simplify them to get values for 0°, 30°, 45°, 60°, 90° respectively.
For $\cos\theta$ : Reverse the $\sin\theta$ sequence.
For $\tan\theta$ : Divide $\sin\theta$ by $\cos\theta$ .
The rest ( $\csc$ , $\sec$ , $\cot$ ) are reciprocals.

Trigonometric Ratios and the Unit Circle

The unit circle is a circle with a radius of 1 centered at the origin of a coordinate plane. It offers a beautiful way to visualize trigonometric ratios for any angle, not just those between 0° and 90°.

Consider a point $P(x, y)$ on the unit circle. Let $\theta$ be the angle formed by the positive x-axis and the line segment $OP$ . Then:

$x = \cos\theta$
$y = \sin\theta$

Therefore, the coordinates of point $P$ are $(\cos\theta, \sin\theta)$ . Also, $\tan\theta = \frac{y}{x}$ .

Extending beyond 90°: The unit circle allows us to define trigonometric ratios for angles greater than 90° and even negative angles. The sign of the trigonometric ratios depends on the quadrant in which the point $P$ lies.

All Students Take Calculus (ASTC): This mnemonic helps remember which trigonometric ratios are positive in each quadrant:

1st Quadrant (0° - 90°): All ratios are positive.
2nd Quadrant (90° - 180°): Sine and cosecant are positive.
3rd Quadrant (180° - 270°): Tangent and cotangent are positive.
4th Quadrant (270° - 360°): Cosine and secant are positive.

Tip: To find the trigonometric ratios of an angle greater than 90°, find its reference angle (the acute angle formed with the x-axis) and use the ASTC rule to determine the sign.

Relationships Between Trigonometric Ratios

There are fundamental identities that relate trigonometric ratios to each other. These identities are essential for simplifying expressions and solving trigonometric equations.

$\sin^2\theta + \cos^2\theta = 1$
$1 + \tan^2\theta = \sec^2\theta$
$1 + \cot^2\theta = \csc^2\theta$

Derivation of $\sin^2\theta + \cos^2\theta = 1$ :

Consider the right-angled triangle $ABC$ again. According to the Pythagorean theorem:

$AB^2 + BC^2 = AC^2$

Dividing both sides by $AC^2$ :

$\frac{AB^2}{AC^2} + \frac{BC^2}{AC^2} = 1$

Since $\cos\theta = \frac{AB}{AC}$ and $\sin\theta = \frac{BC}{AC}$ :

$\cos^2\theta + \sin^2\theta = 1$

The other two identities can be derived from this one by dividing by $\cos^2\theta$ and $\sin^2\theta$ respectively.

Common Mistake: Forgetting the quadrant! Always consider the quadrant of the angle when determining the sign of the trigonometric ratio. For example, $\sin(210^\circ)$ is negative because 210° lies in the third quadrant.

JEE Trick: When faced with a trigonometric problem, try to express everything in terms of $\sin\theta$ and $\cos\theta$ . This often simplifies the problem and makes it easier to solve.

Now that you have a solid foundation in trigonometric ratios and standard values, you're well-equipped to tackle more complex trigonometric problems in JEE Main. Remember to practice regularly and apply these concepts in different scenarios to solidify your understanding. Good luck!