Question

If A + B + C = $\frac{\pi}{2}$, then prove that cos 2A + cos 2B + cos 2C = 1 + 4 sin A sin B sin C.

Answer 1

Solution:

As we know that

$cos \: x + \: cos \: y = 2 \: cos( \frac{x + y}{2} )cos( \frac{x - y}{2} )$
so apply this formula in LHS in first two functions

$cos \: 2A + \: cos \: 2B = 2 \: cos( \frac{2A+ 2B}{2} )cos( \frac{2A - 2B}{2} ) \\ \\ = 2 \: cos \: (A + B)cos \: (A - B)$

So now LHS becomes

$2 \: cos \: (A + B)cos \: (A - B) + cos \: 2C\\ \\ as \\ cos \: 2C = 1 - 2 {sin}^{2} C \\ \\ so \\ \\ = 2 \: cos \: (A + B)cos \: (A - B) + 1 - 2 {sin}^{2}C \\ \\ = 2 \: cos \: ( \frac{\pi}{2} - C )cos \: (A - B) + 1 - 2 {sin}^{2}C \\ \\ = 2 \: sin \: C \: cos \: (A - B) + 1 - 2 {sin}^{2}C \\ \\ = 2sin \: C[cos \: (A - B) - sin \: C] + 1$

now from the given condition we can write
$C = \frac{\pi}{2} - A - B$
$2 \: sin \: C[cos \: (A - B) - sin(\frac{\pi}{2} - A - B)] + 1 \\ \\ = 2 \: sin \: C[cos \: (A - B) - cos( - A - B)] + 1 \\ \\ \\- 4 \: sin \: C [\: sin \frac{(A-B-A-B)}{2} sin \frac{(A-B+A+B)}{2}] + 1\\ \\ =- 4 \: [sin \: C \: sin \frac{(-2B)}{2} sin \frac{(2A)}{2}] + 1\\ \\=1+4 \: [sin \: A \: sin\:B\:sin\:C]\\ \\=R.H.S.$

since sin(-A) = -sin A