Trigonometric functions are basically defined for the acute angles of the triangle. We know that the trigonometric functions of an acute angle is the ratio of the pair of sides of the right triangle as per the property. Using the basic trigonometric identities, we can have identities like, angle sum property, triple angle formula, half angle formulae etc. These formulas are helpful in solving any types of triangles for the angles and the side. In real life situations, it is helpful to find the distance and the direction of of ships and the planes from a particular point. In this section let us see some of these identities and the problems based on it.

## Angle Sum Formula

### Basic Identities :

The basic identities of trigonometry are,
1.   Tan $\theta$ = $\frac{sin\theta}{cos\theta}$

2.   Cot $\theta$ = $\frac{cos\theta}{sin\theta}$

3.   Sin2 $\theta$ + Cos2 $\theta$ = 1

4.  1 + tan2 $\theta$ = sec2 $\theta$

5. 1 + cot2 $\theta$ = cosec2 $\theta$

These identities are used to derive other identities.

### Angle Sum Formula:

Diagram to find the trigonometric ratios           Diagram to find the trigonometric ratios
of the angle A + B
of the angle A - B

Let us assume that A and B are the two angles as shown in the above diagram.
Then the angle sum formula for sine of the sum of the angles A and B is,

Sin ( A + B )        = Sin A. Cos B + Cos A. Sin B
This can be proved geometrically by considering the trigonometric ratios of the angles A, B and A + B from the above diagram.

Similarly,
Cos ( A + B )      = Cos A. Cos B - Sin A. Sin B
Replacing B as - B we get,
Sin ( A + ( - B ) ) = Sin A. Cos ( -B ) + Cos A. Sin ( - B )
Therefore,
Sin ( A - B )        = Sin A. Cos B - Cos A. sin B [ since Sin ( - B) =  - Sin B  and Cos ( - B ) = Cos B ]

Cos ( A - B ) = Cos A. Cos B + Sin A . Sin B

### Angle Sum Formula:

1. Sin ( A + B ) = Sin A. Cos B + Cos A. Sin B

2. Cos ( A + B ) = Cos A. Cos B - Sin A. Sin B

3. Tan ( A + B ) = $\frac{tanA\;+\;tanB}{1\;-\;tanA\;tanB}$

### Angle Difference Formula

4.   Sin ( A - B ) = Sin A. Cos B - Cos A. Sin B

5. Cos ( A - B ) = Cos A. Cos B + Sin A . Sin B

6. Tan ( A - B ) = $\frac{tanA\;-\;tanB}{1\;+\;tanA\;tanB}$

## Trigonometry Formulas

### Reciprocal Identities:

1. Sin $\theta$ = $\frac{1}{Csc\;\theta}$

2. Cos $\theta$ = $\frac{1}{Sec\;\theta}$

3. Tan $\theta$ = $\frac{1}{Cot\;\theta}$

4. Csc $\theta$ = $\frac{1}{Sin\;\theta}$

5. Sec $\theta$ = $\frac{1}{Cos\;\theta}$

6. Cot $\theta$ = $\frac{1}{Tan\;\theta}$

### Quotient Identities:

1. Tan $\theta$ = $\frac{Sin\;\theta}{Cos\;\theta}$

2. Cot $\theta$ = $\frac{Cos\;\theta}{Sin\;\theta}$

### Negative Angle Identities:

1. Sin ( - $\theta$ ) = - Sin $\theta$
2. Cos ( - $\theta$ ) = Cos $\theta$
3. Tan ( - $\theta$ ) = - Tan $\theta$
4. Csc ( - $\theta$ ) = - Csc $\theta$
5. Sec ( - $\theta$ ) = Cos $\theta$
6. Cot ( - $\theta$ ) = - Tan $\theta$

### Double Angle Formula:

1. Sin ( 2 $\theta$ ) = 2 Sin $\theta$. Cos $\theta$

Sin ( 2 $\theta$ ) = $\frac{2tan\:\theta}{1+tan^{2}\:\theta}$

2. Cos ( 2 $\theta$ ) = Cos2 $\theta$ - Sin2 $\theta$

Cos ( 2 $\theta$ ) = 1 - 2 Sin2 $\theta$

Cos ( 2 $\theta$ ) = 2 Cos2 $\theta$ - 1

Cos ( 2 $\theta$ ) = $\frac{1-tan^{2}\:\theta}{1+tan^{2}\:\theta}$

3. Tan ( 2 $\theta$ ) = $\frac{2tan\:\theta}{1-tan^{2}\:\theta}$

4. Cot ( 2 $\theta$ ) = $\frac{cot^{2}\theta\;-\;1}{2cot\;\theta}$

### Triple Angle formula:

1. Sin ( 3 $\theta$ ) = 3 Sin $\theta$ - 4 Sin3 $\theta$

2. Cos ( 3 $\theta$ ) = 4 Cos3 $\theta$ - 3 Cos $\theta$

3. Tan ( 3 $\theta$ ) = $\frac{3tan\;\theta-tan^{3}\theta}{1-3tan^{2}\theta}$

4. Cot ( 3 $\theta$ ) = $\frac{cot^{3}\;\theta-3cot\;\theta}{3cot^{2}\;\theta-1}$

### Half Angle Formula:

1. Sin ( $\theta$ ) = 2 Sin $\frac{\theta}{2}$. Cos $\frac{\theta}{2}$

Sin ( $\theta$ ) = $\frac{2tan\;\frac{\theta}{2}}{1-tan^{2}\;\frac{\theta}{2}}$

2. Cos ( $\theta$ ) = Cos2 $\frac{\theta}{2}$ - Sin2 $\frac{\theta}{2}$

Cos ( $\theta$ ) = 1 - 2 Sin2 $\frac{\theta}{2}$

Cos ( $\theta$ ) = 2 Cos2 $\frac{\theta}{2}$ - 1

Cos ( $\theta$ ) = $\frac{1-tan^{2}\;\frac{\theta}{2}}{1+tan^{2}\;\frac{\theta}{2}}$

3. Tan ( $\theta$ ) = $\frac{2tan\;\frac{\theta}{2}}{1-tan^{2}\;\frac{\theta}{2}}$

4. Cot ( $\theta$ ) = $\frac{cot^{2}\frac{\theta }{2}-1}{2cot\; \frac{\theta }{2}}$

### One Third Angle Formula:

1. Sin $\theta$ = 3 sin ( $\frac{\theta}{3}$ ) - 4 sin3 ( $\frac{\theta}{3}$)

2. Cos $\theta$ = 4 cos3 ( $\frac{\theta}{3}$ ) - 3 cos ($\frac{\theta}{3}$ )

3. Tan $\theta$ = $\frac{3tan\;\frac{\theta }{3}-tan^{3}\;\frac{\theta }{3}}{1-3tan^{2}\;\frac{\theta }{3}}$

## Sum to Product Identity

For the two angles A and B,
1. Sin A + Sin B = 2 Sin $\frac{A+B}{2}$ . Cos $\frac{A-B}{2}$

2. Sin A - Sin B = 2 Cos $\frac{A+B}{2}$ . Sin $\frac{A-B}{2}$

3. Cos A + Cos B = 2 Cos $\frac{A+B}{2}$ . Cos $\frac{A-B}{2}$

4. Cos A + Cos B = - 2 Sin $\frac{A+B}{2}$ . Sin $\frac{A-B}{2}$

### Product to Sum Identities:

1. Sin A . Cos B = $\frac{1}{2}$ [ Sin ( A + B ) + sin ( A - B ) ]

2. Cos A. sin B = $\frac{1}{2}$ [ Sin ( A + B ) - sin ( A - B ) ]

3. Cos A. Cos B = $\frac{1}{2}$ [ Cos ( A + B ) + Cos ( A - B ) ]

4. Sin A . sin B = - $\frac{1}{2}$ [ Cos ( A + B ) - Cos ( A - B ) ]

= $\frac{1}{2}$ [ Cos ( A - B ) - Cos ( A + B ) ]

## Trigonometric Identities Problems

Question 1:

### Find the value of sin 15o = sin ( $\frac{\pi}{12}$ )

Solution:

Since 15 = 45 - 30
Therefore, we have $\frac{\pi}{12}$ = $\frac{\pi}{4}$ - $\frac{\pi}{6}$

sin $\frac{\pi}{12}$ = sin ( $\frac{\pi}{4}$ - $\frac{\pi}{6}$ )
Using the identity on Angle difference, Sin ( A - B ) = sin A . Cos B - Cos A . Sin B,
we get,
sin ( $\frac{\pi}{4}$ - $\frac{\pi}{6}$ ) = sin $\frac{\pi}{4}$ . cos $\frac{\pi}{6}$  - cos $\frac{\pi}{4}$   sin $\frac{\pi}{6}$

= $\frac{1}{\sqrt{2}}$ . $\frac{\sqrt{3}}{2}$ - $\frac{1}{\sqrt{2}}$ . $\frac{1}{2}$

= $\frac{\sqrt{3}}{2\sqrt{2}}$ - $\frac{1}{2\sqrt{2}}$

= $\frac{\sqrt{3}\;-\;1}{2\sqrt{2}}$

Therefore we have Sin 15o = sin $\frac{\pi}{12}$ = $\frac{\sqrt{3}\;-\;1}{2\sqrt{2}}$

Question 2:

### If tan $\theta$ = 3, find tan 3 $\theta$

Solution:

From the list of identities, we have

Tan ( 3 $\theta$ )   = $\frac{3tan\;\theta-tan^{3}\theta}{1-3tan^{2}\theta}$

Substituting tan $\theta$ = 3, we get,

tan (3 $\theta$ )   = $\frac{3(3)\;-\;3^{3}}{1\;-\;3(3^{2})}$

= $\frac{3(3)\;-\;27}{1-27}$

= $\frac{-18}{-26}$

= $\frac{9}{13}$