Question

Please someone help me to prove this!!​

Answer 1

The answer : 4 sin c sin b sin a


Step by step
Formula = A+b+c = 180

B+c = 180-A

C+A= 180-B

A+B= 180-c




2 sin c • 2 cos a-b+c/ 2 cos a-b-c / 2

4 sin c cos 180-2b/2 cos a-(180-a)/ 2

4 sin c cos (90-b ) cos ( A-90)

4 sin c sin b sin A

Answer 2

Answer: see proof below

Explanation:

Given: A + B + C = π → A + B = π - C

A + C = π - B

B + C = π - A

Use the Cofunction Identity: sin A = cos (π/2 - A)

Use the following Sum to Product Identity:

sin A + sin B = 2 sin [(A + B)/2] · cos [(A + B)/2]

Use the Double Angle Identity: sin 2A = 2 sin A · cos A

Proof LHS → RHS

LHS: sin (B + C - A) + sin (C + A - B) + sin (A + B - C)

Given: sin[(π - A) - A) + sin [(π - B) - B] + sin [(π - C) - C]

= sin (π - 2A) + sin (π - 2B) + sin (π - 2C)

= sin 2A + sin 2B + sin 2C

= (sin 2A + sin 2B) + sin 2C

`\text{Sum to Product:}\qquad 2\sin \bigg((2A+2B)/(2)\bigg)\cdot \cos \bigg((2A-2B)/(2)\bigg)+\sin 2C\\\\.\qquad \qquad \qquad \qquad =2\sin (A+B)\cdot \cos (A-B)+\sin 2C`

`\text{Double Angle:}\qquad 2\sin (A+B)\cdot \cos (A-B)+2\sin C \cdot \cos C`

Given: 2 sin C · cos (A - B) + 2 sin C · cos C

Factor: 2 sin C [cos (A - B) + cos C]

`\text{Sum Product:}\qquad 2\sin C\cdot 2\cos \bigg((A-B+C)/(2)\bigg)\cdot \cos \bigg((A-B-C)/(2)\bigg)\\\\.\qquad \qquad \qquad =2\sin C\cdot 2\cos \bigg(((A+C)-B)/(2)\bigg)\cdot \cos \bigg((A-(B+C))/(2)\bigg)`

`\text{Given:}\qquad \qquad 4\sin C\cdot \cos \bigg(((\pi -B)-B)/(2)\bigg)\cdot \cos \bigg((A-(\pi -A))/(2)\bigg)\\\\.\qquad \qquad \qquad =4\sin C\cdot \cos \bigg((\pi -2B)/(2)\bigg)\cdot \cos \bigg((2A-\pi)/(2)\bigg)\\\\.\qquad \qquad \qquad =4\sin C\cdot \cos \bigg((\pi)/(2)-B\bigg)\cdot \cos \bigg((\pi)/(2)-A\bigg)`

Cofunction: 4 sin A · sin B · sin C

LHS = RHS: 4 sin A · sin B · sin C = 4 sin A · sin B · sin C
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