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Prove that cos3A + cos3(120+A) + cos3(240+A) = 3/4 cos3A |
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Answer» cos3A + cos3(120+A) + cos3(240 + A) = cos3A + (cos 120 cos A - sin 120 sin A)3 + (cos 240 cos A - sin 240 sin A)3 (\(\because\) cos(A+B) = cos A cos B - sin A sin B)3 = cos3A + (-sin 30 cos A - cos 30 sin A)3 + (-sin 30 cos A + cos 30 sin A)3 \(\big(\)\(\because\) cos (90 + \(\theta\)) = -sin\(\theta\), cos (270 - \(\theta\)) = - sin\(\theta\) sin(90 + \(\theta\)) = cos\(\theta\), sin(270 - \(\theta\)) = -cos \(\theta\)\(\big)\) = cos3A - \(\big( \frac{1}{2} cos A + \frac{\sqrt{3}}{2} sin A\big)^3 + (\frac{\sqrt{3}}{2} sin A - \frac{1}{2} cos A\big)^3\) \(\big(\)\(\because\) sin 30 = \(\frac{1}{2}\) & cos 30 = \(\frac{\sqrt{3}}{2}\)\(\big)\) = cos3A - (\(\frac{1}{8}\) cos3A + \(\frac{3\sqrt{3}}{8}\) cos2A sin A + \(\frac{9}{8} \) cosAsin2A + \(\frac{3\sqrt{3}}{8} sin^3A\)) + \(\frac{3\sqrt{3}}{8} \) sin3A - \(\frac{9}{8}\) sin2A cos A + \(\frac{3\sqrt{3}}{8} \) sin A cos2 A - \(\frac{1}{8}\) cos3A = cos3A - \(\frac{1}{4}\) cos3A - \(\frac{9}{4}\) cos A sin2A = \(\frac{3}{4}\) cos3A - \(\frac{9}{4} \) cos A (1 - cos2A) (\(\because\) sin2A = 1 - cos2A) = \(\frac{3}{4}\) cos3A - \(\frac{9}{4} \) cos A + \(\frac{9}{4} \) cos3A = 3 cos3A - \(\frac{9}{4}\) cos A = \(\frac{3}{4}\) (4 cos3A - 3 cos A) = \(\frac{3}{4}\) cos 3A |
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