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```Trigonometric Identities & Equations

Trigonometric Identities & Equations is a vital topic of IIT JEE Trigonometry syllabus. As the name implies, 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 topic of Trigonometric identities is also important as it fetches some direct questions in JEE on the basis of formulae. Hence, it is extremely essential to learn the identities besides knowing the application of these identities.

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)

For more, refer the video

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)

Besides these identities, trigonometry is flooded with various trigonometric formulas and conditional identities as well. Students are advised to learn all the formulas of trigonometry in order to remain competitive in the JEE.

Trigonometric Equations:

The equations involving trigonometric functions of unknown angles are known as trigonometric equations. For e.g. sin2 A + sin A = 2 is a trigonometric equation.

Period of a function: A function f(x) is said to be periodic if there exists a T > 0 such that f(x+T) = f(x) for all x in the domain. If ‘T’ is the smallest positive real number satisfying this condition, then it is called the period of f(x).

As shown above, sin x is periodic with period 2π, i.e. the graph repeats itself afetr every intervla of 2π.

These concepts have been discussed in detail in the coming sections. For more, please refer the following pages.

Illustration: In any triangle, prove that

cot A/2 + cot B/2 + cot C/2 = cot A/2 cot B/2 cot C/2     (2000)

Solution: We know that if A, B and C are angles of a triangle, then A + B + C = π.

This means that A/2 + B/2 + C/2 = π/2.

or, we have A/2 + B/2 = π/2 - C/2.

Then, cot (A/2 + B/2) = cot (π/2 - C/2)

Using the identity cot (A + B) = (cot A cot B - 1)/(cot A + cot B), we have

[cot A/2.cot B/2 – 1]/ [cot (A/2 + B/2)] = tan C/2

This gives cot A/2 .cot B/2. cot C/2 - cot C/2 = cot A/2 + cot B/2

Hence, we have cot A/2 + cot B/2 + cot C/2 = cot A/2 .cot B/2. cot C/2

Illustration: The number of integral values of k for which the equation 7 cos x + 5 sin x = 2k +1 has a solution is (2002)

(1) 4                                                               (2) 8

(3) 10                                                             (4) 12

Solution: We know that -√a2 + b2 ≤ a sin x + b cos x ≤ √a2 + b2

-√74 ≤ 5 sin x + 7 cos x ≤ √74

-√74 ≤ 2k + 1 ≤ √74

Now, since k is an integer, we have

-9 ≤ 2k + 1 ≤ 9

This gives -10 < 2k < 8.

This further yields that -5 < k < 4.

Hence, since there are 8 integers lying in this interval, so the number of positive integral values of k is 8.

You might like to refer some of the related resources listed below:

Look into the Revision Notes on Trigonometry for a quick revision.

Various Recommended Books of Mathematics are just a click away.

To read more, Buy study materials of Trigonometry comprising study notes, revision notes, video lectures, previous year solved questions etc. Also browse for more study materials on Mathematics here.
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