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 ways:

In a 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 x cos y = cos (x + y) + cos (x – y),
\(=\frac{1}{2}[cos(2x+\frac{x}{2})+cos(2x-\frac{x}{2})-cos(\frac{9x}{2}+3x)-cos(\frac{9x}{2}-3x)]\) \(=\frac{1}{2}[cos\frac{5x}{2}+cos\frac{3x}{2}-cos\frac{15x}{2}-cos\frac{3x}{2}]\\ =\frac{1}{2}[cos\frac{5x}{2}-cos\frac{15x}{2}]\) \(=\frac{1}{2}[-2\ sin{\frac{\frac{5x}{2}+\frac{15x}{2}}{2}}\ sin{\frac{\frac{5x}{2}-\frac{15x}{2}}{2}}]\\=-sin\ 5x\ sin(\frac{-5x}{2})\\=sin\ 5x\ sin\ \frac{5x}{2}\) =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 x cos y = cos (x + y) + cos (x – y),
= 3[cos (x + 2x) + cos (x – 2x)] = 3[cos 3x + cos (-x)] = 3 [cos 3x + cos x]

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