Question

If A, B, C are angles in a triangle, then prove that sin² A + sin² B - sin² C = 2 sin A sinB cos C.

Answer 1

Answer:

Step-by-step explanation:

If A, B, C are angles in a triangle, then prove that sin² A + sin² B - sin² C = 2 sin A sinB cos C

so

A+B+C=π

As we have

$sin^{2}A+sin^{2}B-sin^{2}C\\\\=\frac{1}{2}(1-cos2A )+\frac{1}{2}(1-cos2B )-\frac{1}{2}(1-cos2C )\\\\\\=\frac{1}{2}-\frac{1}{2}[cos2A +cos\:2B-cos\:2C]\\\\\\=\frac{1}{2}-\frac{1}{2}[2cos(A+B)cos(A-B)-2cos^{2}+1]\\\\=\frac{1}{2}-\frac{1}{2}[2cos(\pi-C)cos(A-B)-2cos^{2}+1]\\\\=\frac{1}{2}-\frac{1}{2}[-2cosC\:cos(A-B)-2cos^{2}+1]\\\\\\=cosC\:cos(A-B)-cos^{2}C\\\\=cosC[cos(A-B)-cos\:C]$

$=cosC[cos(A-B)-cos\:(\pi-(A+B)]\\\\=cosC[cos(A-B)-cos\:((A+B)]\\\\\\=2sin\:A\:sin\:B\:cos\:C$

=RHS

hence proved