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cos a + cos b + cos c = sin a +sin b+ sin c =0 then value of cos 3a + cos 3b+ cos 3c is..

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2 years ago

Arun
24738 Points
```							Dear student we have cos A + cos B + cos C = 0 ...1sin A + sin B + sin C = 0 ....2multiply 2nd by i and add (cos A + i sin A ) + ( cos B + i sin B) + (cos C + i sin C) = 0or e^(iA) + e^(iB) + e^(iC)= 0 as e^ix = cos x+ i sin xand as if (x+y+z) = 0 then x^3+y^3 + z^3 = 3xyzso e^(i3A) + e^(i3B ) + e^(i3C) = 3 e^i(A+B+C)so ( cos 3A + i sin 3 A) + (cos 3B + i sin 3B ) + ( cos 3C + i sin 3C) = 3 cos (A+B+C) + 3i sin (A+B+ C)or (cos 3A + cos 3 B + cos 3C) + i( sin 3A + sin 3B + sin 3C) = 3 cos (A+B+C) + 3i sin (A+B+ C)equating real and imaginary parts on both sides we getcos 3A + cos 3 B + cos 3C = 3 cos (A+B+C) sin 3A + sin 3B + sin 3C = 3 sin (A+B+ C)  RegardsArun  (askIITians forum expert)
```
2 years ago
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### Course Features

• 101 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions