Hello friends! Welcome to your very own guide to trigonometry. Whether you're preparing for your board exams or gearing up for competitive exams like JEE, trigonometry is a topic you just can't skip. It might seem a bit scary with all its formulas, but trust me, once you understand the logic, it's one of the most interesting topics in Maths.

This guide is designed to be your best friend on this journey. We'll start from the very basics (yes, SOH CAH TOA!) and slowly build our way up to the more advanced stuff, explaining everything in a simple, easy-to-understand way. So, let's begin!

Part 1: Starting with the Basics - The Right-Angled Triangle

Everything in trigonometry starts with a simple right-angled triangle. Think of it as the 'ABCD' of trigonometry.

The Six Magical Ratios

Imagine a right-angled triangle. For any angle `θ` (theta) in that triangle (that isn't the 90° one), we have three sides:

  • Hypotenuse:The longest side, always opposite the right angle.
  • Opposite Side:The side directly opposite to the angle `θ`.
  • Adjacent Side:The side that is next to the angle `θ` (but isn't the hypotenuse).

The Main Trio: Sine, Cosine, and Tangent

You've probably heard of "SOH CAH TOA". It's a simple way to remember the three main ratios:

  • Sine (sin):Opposite / Hypotenuse (SOH)
    $$\sin(\theta)=\frac{\text{Opposite}}{\text{Hypotenuse}}$$
  • Cosine (cos):Adjacent / Hypotenuse (CAH)
    $$\cos(\theta)=\frac{\text{Adjacent}}{\text{Hypotenuse}}$$
  • Tangent (tan):Opposite / Adjacent (TOA)
    $$\tan(\theta)=\frac{\text{Opposite}}{\text{Adjacent}}$$

What's the use? Finding a side.

Problem:Let's say you have a right triangle with an angle of 30° and the hypotenuse is 10 cm. You want to find the length of the opposite side (`x`).

  1. We have the Hypotenuse (H) and want the Opposite (O). The formula connecting O and H is Sine.
  2. So, \( \sin(30^\circ)=\frac{\text{Opposite}}{\text{Hypotenuse}}=\frac{x}{10} \).
  3. We know that \( \sin(30^\circ)=\frac{1}{2} \).
  4. So, \( \frac{1}{2}=\frac{x}{10} \). Solving this, we get \( x=5 \) cm. Simple!

The Other Three: Cosecant, Secant, and Cotangent

These three are just the "ulte" (reciprocals) of the main three.

  • Cosecant (csc):It's the reciprocal of sine.
    $$\csc(\theta)=\frac{1}{\sin(\theta)}=\frac{\text{Hypotenuse}}{\text{Opposite}}$$
  • Secant (sec):It's the reciprocal of cosine.
    $$\sec(\theta)=\frac{1}{\cos(\theta)}=\frac{\text{Hypotenuse}}{\text{Adjacent}}$$
  • Cotangent (cot):It's the reciprocal of tangent.
    $$\cot(\theta)=\frac{1}{\tan(\theta)}=\frac{\text{Adjacent}}{\text{Opposite}}$$

Example: Finding all ratios.

Imagine a triangle with Opposite=3, Adjacent=4. By Pythagoras theorem, the Hypotenuse will be \( \sqrt{3^2 + 4^2}=5 \).

  • \( \sin(\theta)=3/5 \implies \csc(\theta)=5/3 \)
  • \( \cos(\theta)=4/5 \implies \sec(\theta)=5/4 \)
  • \( \tan(\theta)=3/4 \implies \cot(\theta)=4/3 \)

Part 2: The Unit Circle - Trigonometry Beyond Triangles

Right-angled triangles are great, but they only have acute angles (less than 90°). What about angles like 120°, 270°, or even negative angles? For this, we use the Unit Circle.

The Unit Circle is just a circle with a radius of 1, drawn on a graph. The center is at (0,0).

Here's the key idea: For any angle \( \theta \), the point where the angle's line touches the circle has coordinates \( (x, y) \).

  • \( \cos(\theta)=x \)
  • \( \sin(\theta)=y \)

This simple definition works for anyangle!

Signs in Quadrants (The CAST Rule)

As the angle \( \theta \) moves around the circle, the signs of \( x \) and \( y \) change. This gives us a very important rule for the signs of the trig functions in the four quadrants.

A simple way to remember it is "All Silver Tea Cups"or "Add Sugar To Coffee". Starting from the top-right quadrant (Quadrant I) and going anti-clockwise:

  • Quadrant I (0° to 90°):All are positive.
  • Quadrant II (90° to 180°):Sine (and its reciprocal, csc) are positive.
  • Quadrant III (180° to 270°):Tangent (and its reciprocal, cot) are positive.
  • Quadrant IV (270° to 360°):Cosine (and its reciprocal, sec) are positive.

Example: Using the CAST rule.

If you are told that \( \tan(\theta)=-4/3 \) and \( \theta \) is in Quadrant IV, find \( \sin(\theta) \) and \( \cos(\theta) \).

  1. From \( \tan(\theta)=\frac{\text{Opposite}}{\text{Adjacent}}=4/3 \), let's imagine a triangle with Opposite=4, Adjacent=3. The Hypotenuse would be 5.
  2. So, the basic value for \( \sin(\theta) \) is 4/5 and for \( \cos(\theta) \) is 3/5.
  3. Now, we use the CAST rule. In Quadrant IV, Cosine is positive, and all others are negative.
  4. Therefore, \( \cos(\theta)=3/5 \) (positive) and \( \sin(\theta)=-4/5 \) (negative).

Part 3: The 'Rules of the Game' - Trigonometric Identities

Identities are formulas that are true for all angles. They are like the grammar rules of trigonometry and are super important for solving problems.

The Most Basic Identities

1. Quotient Identities

Use:To change everything into `sin` and `cos`, which makes simplifying easier.

$$\tan(\theta)=\frac{\sin(\theta)}{\cos(\theta)}$$
$$\cot(\theta)=\frac{\cos(\theta)}{\sin(\theta)}$$

2. Pythagorean Identities

These come from the Pythagoras theorem applied to the unit circle (\( x^2 + y^2=1 \)).

  • The most famous one:
    $$\sin^2\theta + \cos^2\theta=1$$
  • The other two (just divide the main one by \( \cos^2\theta \) and \( \sin^2\theta \)):
    $$1 + \tan^2\theta=\sec^2\theta$$
    $$1 + \cot^2\theta=\csc^2\theta$$

Example: Using a Pythagorean Identity.

If \( \sin(\theta)=3/5 \) and \( \theta \) is in Quadrant II, find \( \cos(\theta) \).

  1. We know \( \sin^2\theta + \cos^2\theta=1 \).
  2. \( (3/5)^2 + \cos^2\theta=1 \implies 9/25 + \cos^2\theta=1 \).
  3. \( \cos^2\theta=1 - 9/25=16/25 \).
  4. \( \cos(\theta)=\pm\sqrt{16/25}=\pm 4/5 \).
  5. Since \( \theta \) is in Quadrant II, \( \cos(\theta) \) must be negative.
  6. So, \( \cos(\theta)=-4/5 \).

Part 4: Advanced Formulas - Playing with Angles

These formulas help you handle sums, differences, and multiples of angles. They are very common in competitive exams.

Sum and Difference Formulas (The 'A+B' Formulas)

  • \( \sin(A \pm B)=\sin(A)\cos(B) \pm \cos(A)\sin(B) \)
  • \( \cos(A + B)=\cos(A)\cos(B) - \sin(A)\sin(B) \)
  • \( \cos(A - B)=\cos(A)\cos(B) + \sin(A)\sin(B) \)
  • \( \tan(A \pm B)=\frac{\tan(A) \pm \tan(B)}{1 \mp \tan(A)\tan(B)} \)

Use: Find \( \sin(75^\circ) \)

We can write 75° as (45° + 30°).

\( \sin(75^\circ)=\sin(45^\circ + 30^\circ)=\sin(45^\circ)\cos(30^\circ) + \cos(45^\circ)\sin(30^\circ) \)

\(=(\frac{1}{\sqrt{2}})(\frac{\sqrt{3}}{2}) + (\frac{1}{\sqrt{2}})(\frac{1}{2}) \)

\(=\frac{\sqrt{3} + 1}{2\sqrt{2}} \)

Double-Angle and Half-Angle Formulas

These are special cases of the sum formulas.

Double-Angle:

  • \( \sin(2A)=2\sin(A)\cos(A) \)
  • \( \cos(2A)=\cos^2(A) - \sin^2(A) \)
  • \( \cos(2A)=2\cos^2(A) - 1 \)
  • \( \cos(2A)=1 - 2\sin^2(A) \)
  • \( \tan(2A)=\frac{2\tan(A)}{1 - \tan^2(A)} \)

Half-Angle:

(These come from rearranging the \( \cos(2A) \) formulas)

  • \( \sin(A/2)=\pm\sqrt{\frac{1 - \cos A}{2}} \)
  • \( \cos(A/2)=\pm\sqrt{\frac{1 + \cos A}{2}} \)

Remember:For half-angle formulas, the `+` or `-` sign depends on the quadrant of the angle \( A/2 \).

Part 5: For Any Triangle - The Sine and Cosine Rules

SOH CAH TOA only works for right-angled triangles. What about other triangles? We have two powerful rules for them.

Let a triangle have angles A, B, C and sides opposite to them be a, b, c.

The Sine Rule

$$\frac{a}{\sin(A)}=\frac{b}{\sin(B)}=\frac{c}{\sin(C)}$$

When to use:When you know:

  • Two angles and one side (AAS or ASA).
  • Two sides and a non-included angle (SSA - the "ambiguous case").

The Cosine Rule

$$c^2=a^2 + b^2 - 2ab\cos(C)$$

Or to find an angle:

$$\cos(C)=\frac{a^2 + b^2 - c^2}{2ab}$$

When to use:When you know:

  • Two sides and the angle between them (SAS).
  • All three sides (SSS).

Part 6: More Advanced Formulas for Competitive Exams

Let's dive a little deeper. These formulas are very helpful for JEE and other competitive exams for simplifying complex problems.

Product-to-Sum Formulas

These formulas change a product of sines and cosines into a sum or difference. They are great for integration problems in calculus!

  • \( 2\sin(A)\cos(B)=\sin(A+B) + \sin(A-B) \)
  • \( 2\cos(A)\sin(B)=\sin(A+B) - \sin(A-B) \)
  • \( 2\cos(A)\cos(B)=\cos(A+B) + \cos(A-B) \)
  • \( -2\sin(A)\sin(B)=\cos(A+B) - \cos(A-B) \)

Example: Product-to-Sum

Problem:Express \( \sin(5x)\cos(3x) \) as a sum.

We use the formula \( 2\sin(A)\cos(B)=\sin(A+B) + \sin(A-B) \). So, \( \sin(A)\cos(B)=\frac{1}{2}[\sin(A+B) + \sin(A-B)] \).

Here, A=5x and B=3x.

\( \sin(5x)\cos(3x)=\frac{1}{2}[\sin(5x+3x) + \sin(5x-3x)]=\frac{1}{2}[\sin(8x) + \sin(2x)] \)

Sum-to-Product Formulas

These are the reverse of the above. They change a sum or difference into a product. Very useful for solving trigonometric equations.

  • \( \sin(A) + \sin(B)=2\sin(\frac{A+B}{2})\cos(\frac{A-B}{2}) \)
  • \( \sin(A) - \sin(B)=2\cos(\frac{A+B}{2})\sin(\frac{A-B}{2}) \)
  • \( \cos(A) + \cos(B)=2\cos(\frac{A+B}{2})\cos(\frac{A-B}{2}) \)
  • \( \cos(A) - \cos(B)=-2\sin(\frac{A+B}{2})\sin(\frac{A-B}{2}) \)

Example: Sum-to-Product

Problem:Simplify \( \cos(75^\circ) + \cos(15^\circ) \).

We use the formula \( \cos(A) + \cos(B)=2\cos(\frac{A+B}{2})\cos(\frac{A-B}{2}) \).

\(=2\cos(\frac{75+15}{2})\cos(\frac{75-15}{2})=2\cos(45^\circ)\cos(30^\circ) \)

\(=2(\frac{1}{\sqrt{2}})(\frac{\sqrt{3}}{2})=\frac{\sqrt{3}}{\sqrt{2}}=\frac{\sqrt{6}}{2} \)

Triple-Angle Formulas

Just like double-angle formulas, these help with triple angles.

  • \( \sin(3x)=3\sin(x) - 4\sin^3(x) \)
  • \( \cos(3x)=4\cos^3(x) - 3\cos(x) \)
  • \( \tan(3x)=\frac{3\tan(x) - \tan^3(x)}{1 - 3\tan^2(x)} \)

Example: Triple-Angle

Problem:If \( \cos(x)=1/2 \), find \( \cos(3x) \).

We use the formula \( \cos(3x)=4\cos^3(x) - 3\cos(x) \).

\( \cos(3x)=4(\frac{1}{2})^3 - 3(\frac{1}{2})=4(\frac{1}{8}) - \frac{3}{2}=\frac{1}{2} - \frac{3}{2}=-1 \)

(This makes sense, because if \( \cos(x)=1/2 \), then \( x=60^\circ \), and \( 3x=180^\circ \), and \( \cos(180^\circ)=-1 \).)

Cofunction Identities (Formulas of 90° - θ)

These show the relationship between a function and its "co-" function (like sine and co-sine). They apply to complementary angles (angles that add up to 90°).

  • \( \sin(90^\circ - \theta)=\cos(\theta) \)
  • \( \cos(90^\circ - \theta)=\sin(\theta) \)
  • \( \tan(90^\circ - \theta)=\cot(\theta) \)

Example: Cofunction Identity

Problem:Simplify \( \frac{\sin(20^\circ)}{\cos(70^\circ)} \).

We know that \( \cos(70^\circ)=\cos(90^\circ - 20^\circ) \).

Using the cofunction identity, \( \cos(90^\circ - \theta)=\sin(\theta) \), we get \( \cos(70^\circ)=\sin(20^\circ) \).

So, the expression becomes \( \frac{\sin(20^\circ)}{\sin(20^\circ)}=1 \).

Formulas in Terms of Semi-Perimeter (s)

These are very useful in solving triangle problems, especially in geometry. Here, 's' is the semi-perimeter of the triangle: \( s=\frac{a+b+c}{2} \).

  • \( \sin(A/2)=\sqrt{\frac{(s-b)(s-c)}{bc}} \)
  • \( \cos(A/2)=\sqrt{\frac{s(s-a)}{bc}} \)
  • \( \tan(A/2)=\sqrt{\frac{(s-b)(s-c)}{s(s-a)}} \)

These formulas are also used to derive Heron's Formulafor the area of a triangle: \( \text{Area}=\sqrt{s(s-a)(s-b)(s-c)} \).

Example: Area using Heron's Formula

Problem:Find the area of a triangle with sides a=13, b=14, c=15.

  1. First, find the semi-perimeter, s:
    \( s=\frac{13+14+15}{2}=\frac{42}{2}=21 \).
  2. Now, use Heron's Formula:
    \( \text{Area}=\sqrt{s(s-a)(s-b)(s-c)} \)
  3. \( \text{Area}=\sqrt{21(21-13)(21-14)(21-15)} \)
  4. \( \text{Area}=\sqrt{21 \times 8 \times 7 \times 6}=\sqrt{7056}=84 \) square units.

Conclusion: You Can Do It!

And that's a wrap! We've gone from the simple right-angled triangle to rules that work for any triangle, and even covered some advanced formulas for your competitive exams. It might seem like a lot of formulas, but the key is to practice. The more you solve problems, the more comfortable you'll become.

Don't just memorize the formulas; try to understand where they come from. See how they are all connected. The Pythagorean identities come from the unit circle, the double-angle formulas come from the sum formulas, and so on.

Keep this guide handy, practice regularly, and you'll master trigonometry in no time. All the best for your studies!