Question

The value of ${\cos }^3({60}^{\circ }-A)-{\cos }^3({60}^{\circ }+A)$ is

• A. $3\sqrt{3}$
• B. $\frac{3\sqrt{3}}{4}\sin A$
• C. $\frac{3\sqrt{3}}{4}$
• D. $3\sqrt{3}\sin A$

Answer 1

The correct option is B $\frac{3\sqrt{3}}{4}\sin A$

${\cos }^3({60}^{\circ }-A)-{\cos }^3({60}^{\circ }+A)$

We Know,
${\cos }^{3}A=\frac{1}{4}[\cos 3A+3\cos A]$
Similarly,
${\cos }^3({60}^{\circ }-A)=\frac{1}{4}\left[\cos \right({180}^{\circ }-3A)+3\cos ({60}^{\circ }-A\left)\right]=\frac{1}{4}[-\cos (3A)+3\cos ({60}^{\circ }-A\left)\right]\cdots \left(1\right)$
${\cos }^3({60}^{\circ }+A)=\frac{1}{4}\left[\cos \right({180}^{\circ }+3A)+3\cos ({60}^{\circ }+A\left)\right]=\frac{1}{4}[-\cos (3A)+3\cos ({60}^{\circ }+A\left)\right]\cdots \left(2\right)$

Using equation $\left(1\right)$ and $\left(2\right)$,
${\cos }^3({60}^{\circ }-A)-{\cos }^3({60}^{\circ }+A)=\frac{3}{4}\left[\cos \right({60}^{\circ }-A)-\cos ({60}^{\circ }+A\left)\right]=\frac{3}{4}\times 2\sin {60}^{\circ }\sin A=\frac{3\sqrt{3}}{4}\sin A$