Question

Please help me to prove this!​

Answer 1

Answer: see proof below

Explanation:

Given: A + B = C → A = C - B

→ B = C - A

Use the Double Angle Identity: cos 2A = 2 cos² A - 1

→ (cos 2A + 1)/2 = cos² A

Use Sum to Product Identity: cos A + cos B = 2 cos [(A + B)/2] · 2 cos [(A - B)/2]

Use Even/Odd Identity: cos (-A) = cos (A)

Proof LHS → RHS:

LHS: cos² A + cos² B + cos² C

`\text{Double Angle:}\qquad (\cos 2A+1)/(2)+(\cos 2B+1)/(2)+\cos^2 C\\\\\\.\qquad \qquad \qquad =(1)/(2)\bigg(2+\cos 2A+\cos 2B\bigg)+\cos^2 C\\\\\\.\qquad \qquad \qquad =1+(1)/(2)\bigg(\cos 2A+\cos 2B\bigg)+\cos^2 C`

`\text{Sum to Product:}\quad 1+(1)/(2)\bigg[2\cos \bigg((2A+2B)/(2)\bigg)\cdot \cos \bigg((2A-2B)/(2)\bigg)\bigg]+\cos^2 C\\\\\\.\qquad \qquad \qquad =1+\cos (A+B)\cdot \cos (A-B)+\cos^2 C`

`\text{Given:}\qquad \qquad 1+\cos C\cdot \cos (A-B)+\cos^2C`

`\text{Factor:}\qquad \qquad 1+\cos C[\cos (A-B)+\cos C]`

`\text{Sum to Product:}\quad 1+\cos C\bigg[2\cos \bigg((A-B+C)/(2)\bigg)\cdot \cos \bigg((A-B-C)/(2)\bigg)\bigg]\\\\\\.\qquad \qquad \qquad =1+2\cos C\cdot \cos \bigg((A+(C-B))/(2)\bigg)\cdot \cos \bigg((-B-(C-A))/(2)\bigg)`

`\text{Given:}\qquad \qquad =1+2\cos C\cdot \cos \bigg((A+A)/(2)\bigg)\cdot \cos \bigg((-B-B)/(2)\bigg)\\\\\\.\qquad \qquad \qquad =1+2\cos C \cdot \cos A\cdot \cos (-B)`

`\text{Even/Odd:}\qquad \qquad 1+2\cos C \cdot \cos A\cdot \cos B\\\\\\.\qquad \qquad \qquad \quad =1+2\cos A \cdot \cos B\cdot \cos C`

LHS = RHS: 1 + 2 cos A · cos B · cos C = 1 + 2 cos A · cos B · cos C
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