Question

$\cos A.\cos \left({60}^{\circ }-A\right)\cos \left({60}^{\circ }+A\right)=$

• A. $(\frac{1}{4}{\cos }^{2}A-{\cos }^{3}A)$
• B. $(\frac{1}{4}\cos A-{\cos }^{3}A)$
• C. $(\frac{1}{2}\cos A-{\cos }^{2}A)$
• D. $(\frac{1}{4}\cos A-{\cos }^{2}A)$

Answer 1

Answer: (B)

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

We have,

$\cos A.\cos \left({60}^{\circ }-A\right)\cos \left({60}^{\circ }+A\right)$

$\Rightarrow \cos A(\cos {60}^{\circ }\sin A+\cos A\sin {60}^0)(\cos {60}^{\circ }\sin A-\cos A\sin {60}^0)$

We know that

$x^2-y^2=(x+y)(x-y)$

Therefore,

$\Rightarrow \cos A({\cos }^2{60}^{\circ }{\sin }^{2}A-{\cos }^{2}A{\sin }^2{60}^0)$

$\Rightarrow \cos A(\frac{1}{4}{\sin }^{2}A-{\cos }^{2}A\times \frac{3}{4})$

$\Rightarrow \frac{1}{4}\cos A({\sin }^{2}A-3{\cos }^{2}A)$

$\Rightarrow \frac{1}{4}\cos A(1-{\cos }^{2}A-3{\cos }^{2}A)$

$\Rightarrow \frac{1}{4}\cos A(1-4{\cos }^{2}A)$

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

Hence, this is the answer.