Precalculus - Trigonometric Identities


  • We know that equations involving trigonometric functions of a variable (angle) are termed Trigonometric equations.
  • Trigonometric Identities are special trigonometric equations that hold true for all the values of the angles(s) involved.
  • For example, 2sinθ-1=0 is a trigonometric equation and not an identity, as it doesn't hold true for all the values of θ. It is only true for some specific values of θ.

Trigonometric Identities

Trigonometric Identities can be divided into several groups, as mentioned below: 

Reciprocal Identities

  • sin θ=1cosec θ  or,  cosec θ=1sin θ
  • cos θ=1sec θ  or,  sec θ=1cos θ
  • tan θ=1cot θ  or,  cot θ=1tan θ


Quotient Identities

  • tan θ=sin θcos θ
  • cot θ=cos θsin θ


Pythagorean Identities

  • sin2 θ+cos2 θ=1
  • sec2 θ=1+tan2 θ
  • cosec2 θ=1+cot2 θ


Sign Identities / Opposite Angle Identities

  • sin-θ=-sin θ
  • cos-θ=cos θ
  • tan-θ=-tan θ


Cofunction Identities

  • sinπ2-θ=cos θ
  • cosπ2-θ=sin θ
  • tanπ2-θ=cot θ
  • cotπ2-θ=tan θ
  • secπ2-θ=cosec θ
  • cosecπ2-θ=sec θ


Double Angle Identities

  • sin 2θ=2sin θ·cos θ=2tan θ1+tan2 θ
  • cos 2θ=cos2 θ-sin2 θ=2cos2 θ-1=1-2sin2 θ
  • tan 2θ=2tan θ1-tan 2θ


Triple Angle Identities

  • sin 3θ=3sin θ-4sin3 θ
  • cos 3θ=4cos3 θ-3cos θ


Sum and Difference of Angle Identities

  • sin x+sin y=2sin x+y2cos x-y2
  • sin x-sin y=2cos x+y2sin x-y2
  • cos x+cos y=2cos x+y2cos x-y2
  • cos x-cos y=2sin x+y2sin y-x2

Solved Examples

Example 1: Find the value of 1-cos2θ cosec2θ.

Solution: 1-cos2θ cosec2θ

=sin2θ cosec2θ  [since, 1-cos2θ=sin2θ]

=1cosec2θ cosec2θ = 1

Example 2: Find the value of sinθ+sinθtan2θ 1+cosecθ.

Solution: sinθ+sinθtan2θ 1+cosecθ

=sinθ 1+1tan2θ 1+cosecθ

=sinθ 1+cot2θ 1+cosecθ

=sinθ 1+cosec2θ-1 1+cosecθ  [since, cosec2θ=1+cot2θ cot2θ=cosec2θ-1]

=sinθ 1+cosecθ-1 cosecθ-1 1+cosecθ

=sinθ 1+cosecθ-1

=sinθ 1+cosecθ-1

=sinθ cosecθ

=sinθ 1sinθ  [since cosecθ=1sinθ]


Example 3: Find the value of cos2θ-2cos2θ+1.

Solution: cos2θ-2cos2θ+1

=2cos2θ-1-2cos2θ+1 [since, cos2θ=2cos2θ-1]



Example 4: Find the value of 1-sinθ 1+sinθ 1+tan2θ.

Solution: 1-sinθ 1+sinθ 1+tan2θ

=1-sin2θ sec2θ [since sec2θ=1+tan2θ ]

=cos2θ 1cos2θ [since, secθ=1cosθ]


Example 5: Verify the identity 1+cot2θ tanθsec2θ=cotθ.

Solution: 1+cot2θ tanθsec2θ=cotθ

Simplifying LHS

1+cot2θ tanθsec2θ

=cosec2θ tanθsec2θ [since, cosec2θ=1+cot2θ]

=cos2θ tanθsin2θ [since, cosecθ=1sinθ and secθ=1cosθ]

=cot2θ tanθ [since, cotθ=cosθsinθ]

=cot2θ cotθ [since, cotθ=1tanθ]



Cheat Sheet

  • sin θ=1cosec θ  or  cosec θ=1sin θ
  • cos θ=1sec θ  or  sec θ=1cos θ
  • tan θ=1cot θ  or  cot θ=1tan θ
  • tan θ=sin θcos θ and cot θ=cos θsin θ
  • sin2 θ+cos2 θ=1
  • sec2 θ=1+tan2 θ
  • cosec2 θ=1+cot2 θ
  • sin-θ=-sin θ
  • cos-θ=cos θ
  • tan-θ=-tan θ
  • sin 2θ=2sin θ·cos θ
  • cos 2θ=cos2 θ-sin2 θ=2cos2 θ-1=1-2sin2 θ
  • tan 2θ=2tan θ1-tan2 θ
  • sinπ2-θ=cos θ
  • cosπ2-θ=sin θ
  • tanπ2-θ=cot θ

Blunder Areas

  • All the trigonometric identities are trigonometric equations but not all trigonometric equations are trigonometric identities.