Trigonometric Identities For IIT JEE

Can you please tell me the Trigonometric Identities for preparation of IIT Joint Entrarnce Exam JEE ?

Trigonometric Identities & Equations is a vital topic of IIT JEE Trigonometry syllabus.

Trigonometric identities consist of various formulae which are equalities that involve trigonometric functions and are true for every value of the occurring variable.

Geometrically, these identities involve functions of one or more angles.

They are particularly useful for simplification of trigonometric problems.

The various topics that have been covered in this section include:

Measurement of Angles

Trigonometric Functions

Identities

Trigonometric Equations

Trigonometric Identity:

A trigonometric equation that holds good for every angle is called a trigonometric identity. Some of the important trigonometric identities are listed below:
Angle-Sum and Difference Identities:

sin (α + β) = sin (α)cos (β) + cos (α)sin (β)

sin (α – β) = sin (α)cos (β) – cos (α)sin (β)

cos (α + β) = cos (α)cos (β) – sin (α)sin (β)

cos (α – β) = cos (α)cos (β) + sin (α)sin (β)

tan (A + B) = (tan A + tan B)/(1 - tan A tan B)

tan (A - B) = (tan A - tan B)/(1 + tan A tan B)

cot (A + B) = (cot A cot B - 1)/(cot A + cot B)

cot (A - B) = (cot A cot B + 1)/(cot B - cot A)

Multiple angle identities:

sin 2A = 2 sin A cos A = 2 tan A/ (1 + tan2A)

cos 2A = (1 - tan2A)/(1 + tan2A)

tan 2A = 2 tan A/(1 - tan2A)

sin 3A = 3 sin A – 4 sin3A

sin 3A = 4 sin (60° - A) sin A sin (60° + A)

cos 3A = 4 cos3A – 3 cos A

cos 3A = 4 cos (60° - A) cos A cos (60° + A)

tan 3A = tan (60° - A) tan A tan (60° + A)

tan 3A = (3tan A – tan3A)/(1 - 3tan2A) (provided A ≠ nπ + π/6)

Half-Angle Identities:

sin A/2 = ± √(1 - cos A)/ 2

cos A/2 = ± √(1 + cos A)/ 2

tan A/2 = ± √(1 - cos A)/(1 + cos A)

Other important formulae:

sin A + sin B = 2 sin (A+B)/2 . cos (A-B)/2

sin A - sin B = 2 cos (A+B)/2 . sin (A-B)/2

cos A + cos B = 2 cos (A+B)/2 . cos (A-B)/2

cos A - cos B = 2 sin (A+B)/2 . sin (B-A)/2

tan A ± tan B = sin (A ± B)/ cos A cos B, provided A ≠ nπ + π/2, B ≠ mπ

cot A ± cot B = sin (B ± A)/ sin A sin B, provided A ≠ nπ, B ≠ mπ+ π/2

1 + tan A tan B = cos (A-B)/ cos A cos B

1 - tan A tan B = cos (A+B)/ cos A cos B

Product Identities:

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)

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