Conditional Identities – II

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.