Question

Math

If A+B+C=pi then prove that cos3A.cos3B+cos3B.cos3C+cos3C.cos3A=1

Answer 1

Explanation:

Given:

A+B+C= π

<=> 3A+3B+3C = 3π

<=> cos(3A+3B) = - cos3C

<=> cos3A.cos3B-sin3A.sin3B = - cos3C

<=> cos3A.cos3B = sin3A.sin3B - cos3C (1)

similarly apply for the other two angles, we have:

• cos3B.cos3C = sin3B.sin3C - cos3A (2)
• cos3C.cos3A = sin3C.sin3A - cos3B (3)

Grouping three equations, (1) + (2) + (3), we have:

<=> cos3A.cos3B+cos3B.cos3C+cos3C.cos3A = sin3A.sin3B + sin3B.sin3C + sin3C.sin3A - ( cos3A + cos3B + cos3C )

= 1

Hope it can find you well.