2cosacosb Formula

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 \(cos\ 2x\ cos\ \frac{x}{2}-cos\ 3x\ cos\ \frac{9x}{2}=sin\ 5x\ sin\ \frac{5x}{2}\)
Solution:
\(LHS=cos\ 2x\ cos\ \frac{x}{2}-cos\ 3x\ cos\ \frac{9x}{2}\\ =\frac{1}{2}[2cos\ 2x\ cos\ \frac{x}{2}-2cos\ 3x\ cos\ \frac{9x}{2}]\)

Using the formula 2 cos A cos B = cos (A + B) + cos (A – B),
\(=\frac{1}{2}[cos(2x+\frac{x}{2})+cos(2x-\frac{x}{2})-cos(\frac{9x}{2}+3x)-cos(\frac{9x}{2}-3x)]\) \(=\frac{1}{2}[cos\frac{5x}{2}+cos\frac{3x}{2}-cos\frac{15x}{2}-cos\frac{3x}{2}]\\ =\frac{1}{2}[cos\frac{5x}{2}-cos\frac{15x}{2}]\)

Using the formula of cos x – cos y,

\(=\frac{1}{2}[-2\ sin{\frac{\frac{5x}{2}+\frac{15x}{2}}{2}}\ sin{\frac{\frac{5x}{2}-\frac{15x}{2}}{2}}]\\=-sin\ 5x\ sin(\frac{-5x}{2})\\=sin\ 5x\ sin\ \frac{5x}{2}\\=RHS\)

Hence, proved that \(cos\ 2x\ cos\ \frac{x}{2}-cos\ 3x\ cos\ \frac{9x}{2}=sin\ 5x\ sin\ \frac{5x}{2}\)

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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