Question

The value of $(4{\cos }^2{9}^0-1)(4{\cos }^2{27}^0-1)(4{\cos }^2{81}^0-1)(4{\cos }^2{243}^0-1)$ is

• A. $1$
• B. $-1$
• C. $2$
• D. None of these

Answer 1

The correct option is A $1$

$(4{\cos }^2{9}^{\circ }-1)(4{\cos }^2{27}^{\circ }-1)(4{\cos }^2{81}^{\circ }-1)(4{\cos }^2{243}^{\circ }-1)$

$=(4{\cos }^2{9}^{\circ }-1)(4{\cos }^2{27}^{\circ }-1)(4{\sin }^2{9}^{\circ }-1)(4{\sin }^2{27}^{\circ }-1)$

$=(16{\cos }^2{9}^{\circ }{\sin }^2{9}^{\circ }-4{\cos }^2{9}^{\circ }-4{\sin }^2{9}^{\circ }+1)(16{\cos }^2{27}^{\circ }{\sin }^2{27}^{\circ }-4{\cos }^2{27}^{\circ }-4{\sin }^2{27}^{\circ }+1)$

$=(4\sin 2{18}^{\circ }-3)(4{\sin }^2{54}^{\circ }-3)$

$=(3-4{\sin }^2{18}^{\circ })(3-4{\sin }^2{54}^{\circ })$
multiplying $\sin {18}^o$ and $\sin {54}^o$ in numerator and denominator

$=\frac{(3\sin {18}^{\circ }-4{\sin }^3{18}^{\circ })}{\sin {18}^{\circ }}\frac{(3\sin {54}^{\circ }-4{\sin }^3{54}^{\circ })}{\sin {54}^{\circ }}$

$=\left(\frac{\sin {54}^{\circ }}{\sin {18}^{\circ }}\right)\left(\frac{\sin {162}^{\circ }}{\sin {54}^{\circ }}\right)$

$=1$ (As $\sin {162}^{\circ }=\sin {18}^{\circ }$)