PROVE THAT cos3A + cos3(120+A) + cos3(240+A) = 3/4 cos3A - Maths - Trigonometric Functions

Question

PROVE THAT
cos3A + cos3(120+A) +
cos3(240+A) = 3/4 cos3A

Gursheen kaur. · Accepted Answer

As we know that,

$\cos 3 A = 4 \cos^{3} A - 3 \cos A \Rightarrow \cos 3 A + 3 \cos A = 4 \cos^{3} A \Rightarrow \cos^{3} A = \frac{1}{4} (\cos 3 A + 3 \cos A)$

Now, Taking L.H.S

$\cos^{3} A + \cos^{3} (120 + A) + \cos^{3} (240 + A)$

$= \frac{1}{4} (\cos 3 A + 3 \cos A) + \frac{1}{4} [\cos (360 + 3 A) + 3 \cos (120 + A)] + \frac{1}{4} [\cos (720 + 3 A) + 3 \cos (240 + A)] = \frac{1}{4} (\cos 3 A + 3 \cos A) + \frac{1}{4} [\cos 3 A + 3 \cos (120 + A)] + \frac{1}{4} [\cos 3 A + 3 \cos (240 + A)] = \frac{3}{4} \cos 3 A + \frac{3}{4} [\cos A + \cos (120 + A) + \cos (240 + A)] = \frac{3}{4} \cos 3 A + \frac{3}{4} [\cos A + 2 \cos (180 + A) . \cos 60] = \frac{3}{4} \cos 3 A + \frac{3}{4} [\cos A - 2 \cos A .. \frac{1}{2}] = \frac{3}{4} \cos 3 A + \frac{3}{4} [\cos A - \cos A .] = \frac{3}{4} \cos 3 A = R H S$

Hence proved.