Conditional Identities – II
1. sin A + sin B + sin C = 4 cos(A/2) cos(B/2) cos(C/2)
Proof: Sin A + sin B + sin C
= \(2\sin \left( \frac{A+B}{2} \right).\cos \left( \frac{A-B}{2}\right)+\sin C\).
\(\frac{A+B+C}{2}=\frac{\pi }{2}\).
\(\frac{A+B}{2}=\frac{\pi }{2}-\frac{C}{2}\).
\(=2\sin \left( \frac{\pi }{2}-\frac{C}{2} \right).\cos \left( \frac{A-B}{2} \right)+\sin C\).
\(=2\cos \left( \frac{C}{2} \right).\cos \left( \frac{A-B}{2} \right)+\sin C\).
\(=2\cos \left( \frac{C}{2} \right).\cos \left( \frac{A-B}{2} \right)+2\sin \left( \frac{C}{2} \right)\cos \left( \frac{C}{2} \right)\).
\(=2\cos \left( \frac{C}{2} \right)\left( \cos \left( \frac{A-B}{2} \right)+\sin \left( \frac{C}{2} \right) \right)\).
\(=2\cos \left( \frac{C}{2} \right)\left( \cos \left( \frac{A-B}{2} \right)+\sin \left( \frac{\pi }{2}-\left( \frac{A+B}{2} \right) \right) \right)\).
\(=2\cos \left( \frac{C}{2} \right)\left( \cos \left( \frac{A-B}{2} \right)+\cos \left( \left( \frac{A+B}{2} \right) \right) \right)\).
\(=2\cos \left( \frac{C}{2} \right)\left( 2\cos \left( \frac{A}{2} \right)\cos \left( \frac{B}{2} \right) \right)\).
x\(=2\cos \left( \frac{A}{2} \right)\cos \left( \frac{B}{2} \right)\cos \left( \frac{C}{2} \right)\).
Hence proved.
sinA + sinB + sinC = 4 cos(A/2) cos(B/2) cos(C/2).
2. If A + B + C = 180, prove that cos²A + cos²B + cos²C = 1 – 2cosA cos B cos C.
Proof: cos²A + cos²B + cos²C = 1 – 2cosA cos B cos C.
cos²A + cos²B + cos²C
\(=\left( \frac{1+\cos A}{2} \right)+\left( \frac{1+\cos B}{2} \right)+\left( \frac{1+\cos C}{2} \right)\).
\(=\frac{1}{2}\left( \cos 2A+\cos 2B+\cos 2C \right)+\frac{3}{2}\).
\(=\frac{1}{2}\left( -1-4\cos A\cos B\cos C \right)+\frac{3}{2}\).
= 1 – 2 cosA cosB cosC
Hence proved.
cos²A + cos²B + cos²C = 1 – 2cosA cos B cos C
3. Prove that in triangle ABC, cos²A + cos²B – cos²C = 1 – 2 sin A sin B cos C.
Proof: cos²A + cos²B – cos²C = 1 – 2 sin A sin B cos C.
cos²A + cos²B – cos²C = cos²A + sin²C – sin²B
= cos²A + sin(C+B) sin (C – B)
= 1 – sin²A + sin (C + B) sin (C – B)
A + B + C = π
C + B = π – A
= 1 – sin²A + sin (π – A) sin (C – B)
= 1 – sin²A + sin A sin (C – B)
= 1 – sin A (sin A + sin(C – B))
= 1 – sin A (sin (C + B) + sin (C – B))
= 1 – 2 sin A sin B sin C.