Question

If A, B, C are angles in a triangle, then prove that cos² A + cos² B - cos² C = 1 - 2 sin A sinB cos C.

Answer 1

Answer:

Step-by-step explanation:

If A, B, C are angles in a triangle, then

A+B+C=π

so to prove that cos² A + cos² B - cos² C = 1 - 2 sin A sinB cos C

we have LHS

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

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

hence proved