# Cos^2A+cos^2B-sin^2C=-2cosA.cosB.sinC?

Jul 15, 2018

If $A + B + C = \pi$ then

$L H S = {\cos}^{2} A + {\cos}^{2} B - {\sin}^{2} C$

$= \frac{1}{2} \left[2 {\cos}^{2} A + 2 {\cos}^{2} B - 2 {\sin}^{2} C\right]$

$= \frac{1}{2} \left[1 + \cos 2 A + 1 + \cos 2 B - \left(1 - \cos 2 C\right)\right]$

$= \frac{1}{2} \left[1 + \cos 2 A + 1 + \cos 2 B - 1 + \cos 2 C\right]$

$= \frac{1}{2} \left[1 + \cos 2 A + \cos 2 B + \cos 2 C\right]$

$= \frac{1}{2} \left[1 + 2 \cos \left(\frac{2 A + 2 B}{2}\right) \cos \left(\frac{2 A - 2 B}{2}\right) + \cos 2 \left(\pi - \left(A + B\right)\right)\right]$

$= \frac{1}{2} \left[1 + 2 \cos \left(A + B\right) \cos \left(A - B\right) + \cos \left(2 \pi - 2 \left(A + B\right)\right)\right]$

$= \frac{1}{2} \left[1 + 2 \cos \left(A + B\right) \cos \left(A - B\right) + 2 {\cos}^{2} \left(A + B\right) - 1\right]$

$= \frac{1}{2} \left[2 \cos \left(A + B\right) \left\{\cos \left(A - B\right) + \cos \left(A + B\right)\right\}\right]$

$= \cos \left(\pi - C\right) \cdot 2 \cos A \cos B = - 2 \cos A \cos B \cos C$

Jul 15, 2018

Below.

#### Explanation:

Let we take the values of $A + B + C = \pi$

$C = \pi - \left(A + B\right)$----------(i)

Multiplying LHS by $\frac{1}{2}$ we get,

$\frac{1}{2} \left(2 {\cos}^{2} A + 2 {\cos}^{2} B - 2 {\sin}^{2} C\right)$

$\implies$ $\frac{1}{2} \left(1 + \cos 2 A + 1 + \cos 2 B - \left\{1 - \cos 2 C\right\}\right)$

$\implies$ $\frac{1}{2} \left(1 + \cos 2 A + 1 + \cos 2 B - 1 + \cos 2 C\right)$

$\implies$ $\frac{1}{2} \left(1 + \cos 2 A + \cos 2 B + \cos 2 C\right)$

Using the formula of $\cos \left\{2 A + 2 B\right\}$ we get,

$\implies$ $\frac{1}{2} \left(1 + 2 \cos \left[\frac{2 A + 2 B}{2}\right] \cos \left[\frac{2 A - 2 B}{2}\right] + \cos 2 C\right)$

Putting the value of equation (i) in $\cos 2 C$ we get,

$\implies$ $\frac{1}{2} \left(1 + 2 \cos \left[\frac{2 A + 2 B}{2}\right] \cos \left[\frac{2 A - 2 B}{2}\right] + \cos 2 \left[\pi - \left\{A + B\right\}\right]\right)$

Resolving this equation further,

$\implies$ $\frac{1}{2} \left(1 + 2 \cos \left[A + B\right] \cos \left[A - B\right] + 2 {\cos}^{2} \left(A + B\right) - 1\right)$

$\implies$ $\left(\cos \left[\pi - C\right] \cdot 2 \cos A \cos B\right)$

This gives the final output as,

$\implies$ $- 2 \cos A \cdot \cos B \cdot \cos C$