As we know that there are six Trigonometric functions of angles and their names are:

  1. Sine
  2. Cosine
  3. Tangent
  4. Cotangent
  5. Secant
  6. Cosecant

These functions are in relation to the right triangle in the following way:

In any right triangle ABC,

Sin A = Perpendicular/ Hypotenuse
Cos A = Base/ Hypotenuse
Tan A = Perpendicular/ Base
Cot A = Base/ Perpendicular
Cosec A = Hypotenuse/Perpendicular
Sec A = Hypotenuse/ Base

Formula of 2cosacosb

We know that,

cos (A + B) = cos A cos B – sin A sin B ….. (1)

cos (A – B) = cos A cos B + sin A sin B ….. (2)

Adding (1) and (2), we get

cos (A + B) + cos (A – B) = 2 cos A cos B

Solved Examples

Example 1: Prove that

cos 2x cos x2−cos 3x cos 9x2=sin 5x sin 5x2

Solution:
LHS=cos 2x cos x2−cos 3x cos 9x2=12[2cos 2x cos x2−2cos 3x cos 9x2]

Using the formula 2 cos A cos B = cos (A + B) + cos (A – B),

=12[cos(2x+x2)+cos(2x−x2)−cos(9x2+3x)−cos(9x2−3x)]
=12[cos5x2+cos3x2−cos15x2−cos3x2]=12[cos5x2−cos15x2]

Using the formula of cos x – cos y,

=12[−2 sin5x2+15x22 sin5x2−15x22]=−sin 5x sin(−5x2)=sin 5x sin 5x2=RHS

Hence, proved that

cos 2x cos x2−cos 3x cos 9x2=sin 5x sin 5x2

Example 2: Express 6 cos x cos 2x in terms of sum function.
Solution:
Consider,
6 cos x cos 2x
= 3 [2 cos x cos 2x]

Using the formula 2 cos A cos B = cos (A + B) + cos (A – B),
= 3[cos (x + 2x) + cos (x – 2x)] = 3[cos 3x + cos (-x)] = 3 [cos 3x + cos x]

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