Trigonometric Identities

Trigonometric Identities are useful whenever trigonometric functions are involved in an expression or an equation. Identity inequalities which are true for every value occurring on both sides of an equation. Geometrically, these identities involving certain functions of one or more angles. There are various distinct identities involving the side length as well as the angle of a triangle.

List of Trigonometric Identities

There are various identities in trigonometry which are used to many trigonometric problems. Let us see all the identities here.

Reciprocal Identities

  • Sin θ = 1/Csc θ or Csc θ = 1/Sin θ
  • Cos θ = 1/Sec θ or Sec θ = 1/Cos θ
  • Tan θ = 1/Cot θ or Cot θ = 1/Tan θ

Pythagorean Identities

  • sina + cosa = 1
  • 1+tan2 a  = sec2 a
  • coseca = 1 + cota

Ratio Identities

  • Tan θ = Sin θ/Cos θ
  • Cot θ = Cos θ/Sin θ

Opposite Angle Identities

  • Sin (-θ) = – Sin θ
  • Cos (-θ) = Cos θ
  • Tan (-θ) = – Tan θ
  • Cot (-θ) = – Cot θ
  • Sec (-θ) = Sec θ
  • Csc (-θ) = -Csc θ

Complementary Angles Identities

  • Sin (90 – θ) = Cos θ
  • Cos (90 – θ) = Sin θ
  • Tan (90 – θ) = Cot θ
  • Cot ( 90 – θ) = Tan θ
  • Sec (90 – θ) = Csc θ
  • Csc (90 – θ) = Sec θ

Angle Sum and Difference Identities

Consider two angles , α and β, the trigonometric sum and difference identities are as follows:

  • sin(α+β)=sin(α).cos(β)+cos(α).sin(β)
  • sin(α–β)=sinα.cosβ–cosα.sinβ
  • cos(α+β)=cosα.cosβ–sinα.sinβ
  • cos(α–β)=cosα.cosβ+sinα.sinβ
  • \(\tan (\alpha + \beta) = \frac{\tan \alpha + \tan \beta}{1 – \tan \alpha. \tan \beta}\)
  • \(\tan (\alpha – \beta) = \frac{\tan \alpha – \tan \beta}{1 + \tan \alpha. \tan \beta}\)

Trigonometric Identities Formula

Similarly, an equation which involves trigonometric ratios of an angle represents a trigonometric identity.
The upcoming discussion covers the fundamental trigonometric identities and their proofs.
Consider a right angle ∆ABC which is right-angled at B as shown in the given figure.

Trigonometry Identities Formula

Applying Pythagoras Theorem for the given triangle, we have

(hypotenuse)2 = (base)2 + (perpendicular)2

AC2 = AB2+BC2     ………………………..(1)

Let us prove the three Pythagoras trigonometric identities, which are commonly used.

Identity 1

Now, divide each term of equation (1) by AC2, we have

\(\frac{AC^2}{AC^2}\) = \(\frac{AB^2}{AC^2}~+~\frac{BC^2}{AC^2}\)

\(⇒\frac{AB^2}{AC^2}+\frac{BC^2}{AC^2}\) = \(1\)

\(⇒(\frac{AB}{AC})^2+(\frac{BC}{AC})^2\) = \(1\) ………………………….(2)

We know that,
\((\frac{AB}{AC})^2 = \cos a\) and \((\frac{BC}{AC})^2 = \sin a\), thus equation (2) can be written as-

sina + cosa = 1

Identity 1 is valid for angles 0 ≤ a ≤ 90.

Identity 2

Now Dividing the equation (1) by AB2, we get

\(\frac{AC^2}{AB^2}\) = \(\frac{AB^2}{AB^2}+\frac{BC^2}{AB^2}\)

\(⇒\frac{AC^2}{AB^2}\) = \(1+\frac{BC^2}{AB^2}\)

\(⇒ (\frac{AC}{AB})^2\) = \(1+(\frac{BC}{AB})^2 \) ……………………(3)

By referring trigonometric ratios, it can be seen that:

\(\frac{AC}{AB}\) = \(\frac{hypotenuse}{side~ adjacent~ to ~angle~ a}\) = \(\sec a\)

Similarly,

\(\frac{BC}{AB}\) = \(\frac{side~ opposite~ to~ angle~ a}{side ~adjacent~ to~ angle~ a}\) = \(\tan a\)

Replacing the values of \(\frac{AC}{AB}\) and \(\frac{BC}{AB}\) in the equation (3) gives,

1+tan2 a  = sec2 a

As it is known that tan a is not defined for a = 90° therefore  identity 2 obtained above is true for 0 ≤ A <90.

Identity 3

Dividing the equation (1) by BC2 , we get

\(\frac{AC^2}{BC^2}\) = \(\frac{AB^2}{BC^2}~ +~\frac{BC^2}{BC^2}\)

\(⇒\frac{AC^2}{BC^2}\) = \(\frac{AB^2}{BC^2}+1\)

\(⇒(\frac{AC}{BC})^2\) = \((\frac{AB}{BC})^2+1\) …………………..(iv) 

By referring trigonometric ratios, it can be seen that:

\(\frac{AC}{BC}\) = \(\frac{hypotenuse}{side ~opposite~ to~ angle~ a}\) = \(cosec a\)

Also,

\(\frac{AB}{BC}\) = \(\frac{side~ adjacent~ to~ angle~ a}{side~ opposite ~to ~angle~ a}\) = \(\cot a\)

Replacing the values of \(\frac{AC}{BC}\) and \(\frac{AB}{BC}\) in the equation (4) gives,

coseca = 1 + cota

Since cosec a and cot a are not defined for a = 0°, therefore the identity 3 is obtained is true for all the values of a except at a = 0°. Therefore, the identity is true for all a such that, 0° < a ≤ 90°.

Example

Consider a ∆ABC, right-angled at B. The length of the base, AB = 4 cm and length of perpendicular BC =3 cm. Find the value of sec A.

Solution:

As the length of the perpendicular and base is given; it can be concluded that,

tan A = 3/4

Now, using the trigonometric identity: 1+tan2 a  = sec2 a

sec2 A = 1 + (3/4)2

sec2 A = 25/16

sec A = ±5/4

Since, the ratio of lengths is positive, we can neglect sec A = 5/4.

Therefore, sec A = 5/4

Using these identities, we can solve various mathematical problems. All you need to know about trigonometry and its applications are just a click away,  visit BYJU’S to learn more.

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