# Trigonometric Identities

## What is Trigonometry?

Trigonometry is an important branch of mathematics that is used to study the relationship between ratios of the sides of a right-angled triangle with its angles.  The word “trigonometry” is derived from the Greek words, “trigon” and “metron”. The word “trigon” means a triangle and the word ”metron” means a measure.  Hence, trigonometry means the science of measuring triangles. In broader cases it is that branch of mathematics which deals with the measurement of the sides and the angles of a triangle and the problems allied with the angles.

Before moving on to understand the concepts in trigonometry, it is important to recall what we mean by an angle.

## What are Trigonometric Ratios?

Consider an acute angle ∠YAX  = θ with initial side AX and terminal side AY. Let P be any point on the terminal side AY. Draw PM perpendicular from P on AX to get the right angled triangle AMP in which ∠PAM = θ.

Now, in the right angled triangle AMP, Base  = AM = x and Perpendicular  = PM = y and Hypotenuse  = AP = r

We define the following six trigonometric ratios

1. Sin θ = $\frac{Perpendiculur}{Hypotenuse}=\frac{y}{r}$, written as sin θ
2. Cos θ = $\frac{Base}{Hypotenuse}=\frac{x}{r}$, written as cos θ
3. Tan θ = $\frac{Perpendiculur}{Base}=\frac{y}{x}$, written as tan θ
4. Cosecant θ = $\frac{Hypotenuse}{Perpendicular}=\frac{r}{y}$, written as cosec θ
5. Secant θ = $\frac{Hypotenuse}{Base}=\frac{r}{x}$, written as sec θ
6. Cotangent θ = $\frac{Base}{Perpendicular}=\frac{x}{y}$, written as cot θ

Now, let us learn what are trigonometric identities.

## What are Trigonometric Identities?

We know that an equation is called an identity if it is true for all values of the variable (s) involved. For example,

x2 – 16 = ( x + 4 ) ( x – 4 )  is an algebraic identity as it is satisfied by every value of x.

Similarly, an equation involving trigonometric ratios of an angle θ is said to be a trigonometric identity if it is satisfied for all values of θ for which the given trigonometric ratios are defined.

For example,  cos2 θ – $\frac{1}{2}$ cos θ =  cos θ ( cos θ – $\frac{1}{2}$ ) is a trigonometric identity whereas,

cos θ ( cos θ – $\frac{1}{2}$ ) = 0 is an equation.

We shall now learn about various trigonometric Identities

## Reciprocal Trigonometric Identities

Reciprocal trigonometric identities are as follows –

1. cosec θ = $\frac{1}{sinsinθ}$
2. sec θ = $\frac{1}{coscosθ}$
3. cot θ = $\frac{1}{tantanθ}$

Let us understand these identities through an example.

Example If cos B = $\frac{1}{3}$, find the other trigonometric ratios

Solution We have been given that cos B = $\frac{1}{3}$. We need to find the values of the other trigonometric ratios.

So, we draw a triangle ABC, right angled at C such that AB = 3 units.

By Pythagoras theorem, we have,

AB2 = AC2 + BC2

⇒ 32 = 12 + AC2

⇒ AC2 = 9 – 1

⇒ AC = $\sqrt{8}$ = 2$\sqrt{2}$

So, we now have, Base = BC = 1

Perpendicular = AC = 2$\sqrt{2}$

Hypotenuse = AB = 3

Now, we can obtain the values of sin B and tan B.

We know that,

sin B =  $\frac{Perpendiculur}{Hypotenuse}=\frac{2\sqrt{2}}{3}$

Also, tan A = $\frac{Perpendiculur}{Base}=\frac{2\sqrt{2}}{1}$ = 2$\sqrt{2}$

To obtain the values of sec B, cosec B and cot B, we shall make use of the Reciprocal trigonometric identities

We know that

cosec θ = $\frac{1}{sinsin θ}$

therefore,

cosec B = $\frac{1}{sinsin B}=\frac{2\sqrt{2}}{3}$

We also know that,

sec B = $\frac{1}{coscos B} = \frac{3}{1}$ = 3

cot θ = $\frac{1}{tantan θ}=\frac{1}{2\sqrt{2}}$

Hence, if cos B = $\frac{1}{3}$, the remaining values of trigonometric ratios will be –

sin B = $\frac{2\sqrt{2}}{3}$

tan B = 2$\sqrt{2}$

cosec B =$\frac{2\sqrt{2}}{3}$

sec B = 3

cot θ = $\frac{1}{2\sqrt{2}}$

## Pythagorean Trigonometric Identities

Pythagorean Trigonometric identities are some of the fundamental Trigonometric identities that have been derived using the Pythagoras theorem. These identities are –

1. sin2 θ  + cos2 θ = 1 or cos2 θ = 1 – sin2 θ or sin2 θ  = 1 – cos2 θ
2. sec2 θ  = 1 + tan2 θ or sec2 θ  – tan2 θ = 1 or sec2 θ  – 1 = tan2 θ
3. cosec2 θ  = 1 + cot2 θ  or cosec2 θ  – cot2 θ = 1 or cosec2 θ  – 1 = cot2 θ
4. tan θ = $\frac{sinsinθ}{coscosθ}$
5. cot θ = $\frac{coscosθ}{sinsinθ}$
6. sec θ + tan θ = $\frac{1}{secsecθ-tantanθ}$
7. cosec θ + cot θ = $\frac{1}{cosecθ-cotcotθ}$

## Complementary and Supplementary Trigonometric Identities

Let us recall what we mean by complementary angles. Two angles are said to be complementary if their sum is 90o. It follows from this definition that θ and ( 90o – θ ) are complementary angles for acute angles θ.

So, we have, complementary trigonometry identities are –

1. sin ( 90o – θ ) = cos θ
2. cos ( 90o – θ ) = sin θ
3. tan ( 90o – θ ) = cot θ
4. cot ( 90o – θ ) = tan θ
5. sec ( 90o – θ ) = cosec θ
6. cosec ( 90o – θ ) = sec θ

Supplementary Trigonometry Identities are –

1. sin (180°- θ) = sin θ
2. cos (180°- θ) = – cos θ
3. cosec (180°- θ) = cosec θ
4. sec (180°- θ)= – sec θ
5. tan (180°- θ) = – tan θ
6. cot (180°- θ) = – cot θ

Let us now learn about some trigonometric identities that involve the sum and difference of trigonometric functions.

1. sin ( A + B ) = sin A cos B + cos A sin B
2. sin ( A – B ) = sin A cos B – cos A sin B
3. cos ( A + B ) = cos A cos B – sin A sin B
4. cos ( A – B ) = cos A cos B + sin A sin B
5. tan ( A + B ) = $\frac{(tan A + tan B)}{(1 – tan A tan B)}$
6. tan ( A – B ) = $\frac{(tan A- tan B)}{(1+ tan A tan B)}$
7. sin ( A + B ) sin ( A – B ) = sin2 A  – sin2 B = cos2 B – cos2 A
8. cos ( A + B ) cos ( A  – B ) = cos2 A  – sin2 B = cos2 B – sin2 A
9. sin ( A + B + C ) = sin A cos B cos C + cos A sin B cos C + cos A cos B sin C –  sin A sin B sin C
10. cos ( A + B + C ) = cos A cos B cos C – cos A sin B sin C – sin A cos B sin C – sin A sin B cos C
11. tan ( A  + B + C ) = $\frac{tan A+tan B+tan C-tan A tan B tan C}{1-tan A tan b-tan B tan C-tan C tan A}$

## Formulae to Transform the Product into Sum or Difference

We have just learnt the formulae involving the identities, sin ( A  + B ), sin ( A – B ) and so on. Now we shall discuss about the identities that help convert the product of two sines or two cosines or one sine and one cosine into the sum or difference of two sines or two cosines.

These identities are –

1. 2 sin A cos B =  sin ( A  + B ) + sin ( A  – B )
2. 2 cos A cos B = cos ( A  + B ) + cos ( A  –  B )
3. 2 cos A sin B = sin ( A + B ) – sin ( A  – B )
4. 2 sin A sin B = cos ( A – B ) – cos ( A + B )

Similarly, if we substitute A + B = C and A – B = D in the above formulae, we will get,

A = $\frac{C+D}{2}$ and B = $\frac{C-D}{2}$

Therefore, new identities can be derived as –

1. sin C + sin D = 2 sin $\frac{C+D}{2} cos \frac{C-D}{2}$
2. sin C – sin D = 2 sin $\frac{C-D}{2} cos \frac{C+D}{2}$
3. cos C + cos D = 2 cos $\frac{C+D}{2} cos \frac{C-D}{2}$
4. cos D – cos C =  2 sin $\frac{C+D}{2} cos \frac{C-D}{2}$

## Trigonometric Identities as values of functions at 2x in terms of vales at x

Now we shall introduce identities expressing the trigonometric functions as multiples of x, i.e. 2x, 3s, 4x etc. in terms of the values of x. some of the commonly used identities are –

1. sin 2x = 2 sin x cos x
2. cos 2x = cos2 x – sin2 x
3. cos 2x = 2 cos2 x – 1  or 1 + cos 2x = 1 + 2 cos2 x
4. cos 2x = 1 – 2 sin2 x or 1 – cos 2x = 2 sin2 x
5. tan 2x = $\frac{2tanx}{1-tan^{2}x}$
6. sin 2x = $\frac{2tanx}{1-tan^{2}x}$
7. cos 2x = $\frac{1-tanx}{1+tan^{2}x}$

## Solved Examples

Example 1 Find the value of $\frac{coscos 37^{0}}{sinsin 53^{0}}$

Solution We have,

$\frac{coscos 37^{0}}{sinsin 53^{0}}$

= $\frac{coscos 90^{0} – 53^{0}}{sinsin 53^{0}}$

= $\frac{sinsin 53^{0}}{sinsin 53^{0}}$ = 1

Hence, $\frac{coscos 37^{0}}{sinsin 53^{0}}$ = 1

Example 2 Prove that tan 3x tan 2x tan x = tan 3x – tan 2x – tan x

Solution Let us consider the L.H.S

Now, 3x can be written as 3x = 2 x + x

Therefore,

tan 3x = tan ( 2x + x)

Now, we know that

tan ( A + B ) = $\frac{(tan A + tan B)}{(1 – tan A tan B)}$

Hence

⇒ tan 3x = $\frac{tantan 2x + tantan x }{1 – tantan 2x tanx}$

⇒ tan 3x (1-tan 2x tanx ) = tan 2x + tan x

⇒ tan 3x – tan 3x tan 2x tan x = tan 2x + tan x

⇒ tan 3x – tan 2x – tan x = tan 3x tan 2x tan x

Hence,

tan 3x tan 2x tan x = tan 3x – tan 2x – tan x

Example 3 If x sin3 A + y cos3 A = sin A cos A and x sin A = y cos A, prove that x2 + y2 = 1

Solution We have been given that

x sin3 A + y cos3 A = sin A cos A and x sin A = y cos A

We need to prove that x2 + y2 = 1

Starting from L.H.S we have,

x sin3 A + y cos3 A = sin A cos A

⇒ ( x sin A ) sin2 A + (y cos A ) cos2 A = sin A cos A

Replacing y cos A by x sin A ( as we have been given that   x sin A = y cos A), we get,

⇒ ( x sin A ) sin2 A + ( x sin A ) cos2 A = sin A cos A

⇒ x sin A  ( sin2 A + cos2 A ) = sin A cos A

⇒ x sin A  = sin A cos A

⇒ x = cos A

Now, x sin A  = y cos A

⇒ cos A sin A = y cos A

⇒ y = cos A

Now, we have, x = cos A and y = sin A

Therefore, we can say that, x2 + y2 = 1

## Remember

1. An equation involving trigonometric ratios of an angle θ is said to be a trigonometric identity if it is satisfied for all values of θ for which the given trigonometric ratios are defined.
2. cosec θ = $\frac{1}{sinsinθ}$
3. sec θ = $\frac{1}{coscosθ}$
4. cot θ = $\frac{1}{tantanθ}$
5. sin2 θ + cos2 θ = 1 or cos2 θ = 1 – sin2 θ or sin2 θ = 1 – cos2 θ
6. sec2 θ = 1 + tan2 θ or sec2 θ – tan2 θ = 1 or sec2 θ – 1 = tan2 θ
7. cosec2 θ = 1 + cot2 θ or cosec2 θ – cot2 θ = 1 or cosec2 θ – 1 = cot2 θ
8. tan θ = $\frac{sinsinθ}{coscosθ}$
9. cot θ = $\frac{coscosθ}{sinsinθ}$
10. sec θ + tan θ = $\frac{1}{secsecθ-tantanθ}$
11. cosec θ + cot θ = $\frac{1}{cosecθ-cotcotθ}$
12. sin ( 90o – θ ) = cos θ
13. cos ( 90o – θ ) = sin θ
14. tan ( 90o – θ ) = cot θ
15. cot ( 90o – θ ) = tan θ
16. sec ( 90o – θ ) = cosec θ
17. cosec ( 90o – θ ) = sec θ
18. sin (180°- θ) = sin θ
19. cos (180°- θ) = – cos θ
20. cosec (180°- θ) = cosec θ
21. sec (180°- θ)= – sec θ
22. tan (180°- θ) = – tan θ
23. cot (180°- θ) = – cot θ
24. sin ( A + B ) = sin A cos B + cos A sin B
25. sin ( A – B ) = sin A cos B – cos A sin B
26. cos ( A + B ) = cos A cos B – sin A sin B
27. cos ( A – B ) = cos A cos B + sin A sin B
28. tan ( A + B ) = $\frac{(tan A + tan B)}{(1 – tan A tan B)}$
29. tan ( A – B ) = $\frac{(tan A- tan B)}{(1+ tan A tan B)}$
30. sin ( A + B ) sin ( A – B ) = sin2 A  – sin2 B = cos2 B – cos2 A
31. cos ( A + B ) cos ( A  – B ) = cos2 A  – sin2 B = cos2 B – sin2 A
32. sin ( A + B + C ) = sin A cos B cos C + cos A sin B cos C + cos A cos B sin C –  sin A sin B sin C
33. cos ( A + B + C ) = cos A cos B cos C – cos A sin B sin C – sin A cos B sin C – sin A sin B cos C
34. tan ( A  + B + C ) = $\frac{tan A+tan B+tan C-tan A tan B tan C}{1-tan A tan b-tan B tan C-tan C tan A}$
35. 2 sin A cos B =  sin ( A  + B ) + sin ( A  – B )
36. 2 cos A cos B = cos ( A  + B ) + cos ( A  –  B )
37. 2 cos A sin B = sin ( A + B ) – sin ( A  – B )
38. 2 sin A sin B = cos ( A – B ) – cos ( A + B )
39. sin C + sin D = 2 sin ($\frac{C+D}{2}$) cos ($\frac{C-D}{2}$)
40. sin C – sin D = 2 sin ($\frac{C-D}{2}$) cos ($\frac{C+D}{2}$)
41. cos C + cos D = 2 cos ($\frac{C+D}{2}$) cos ($\frac{C-D}{2}$)
42. cos D – cos C =  2 sin ($\frac{C+D}{2}$) cos ($\frac{C-D}{2}$)
43. sin 2x = 2 sin x cos x
44. cos 2x = cos2 x – sin2 x
45. cos 2x = 2 cos2 x – 1  or 1 + cos 2x = 1 + 2 cos2 x
46. cos 2x = 1 – 2 sin2 x or 1 – cos 2x = 2 sin2 x
47. tan 2x = $\frac{2tanx}{1-tan^{2}x}$
48. sin 2x = $\frac{2tanx}{1-tan^{2}x}$
49. cos 2x = $\frac{1-tanx}{1+tan^{2}x}$