Question

The value of $16\sin {144}^0\sin {108}^0\sin {72}^0\sin {36}^0$ is equal to

• A. $1$
• B. $3$
• C. $5$
• D. $\frac{1}{16}$

Answer 1

Answer: (C)

$5$

$\sin {144}^0=\sin ({180}^0-{36}^0)=\sin {36}^0$ and
$\sin {108}^0=\sin ({90}^0+{18}^0)=\cos {18}^0$
$\sin {72}^0=\sin ({90}^0-{18}^0)=\cos {18}^0$
$\therefore$ Given expression
$=16\times {\sin }^2{36}^0{\cos }^2{18}^0$
$=16\times {(\sqrt{\frac{10-2\sqrt{5}}{4}}\times \frac{\sqrt{10+2\sqrt{5}}}{4})}^2$
$=16\times {\left(\sqrt{\frac{100-20}{16}}\right)}^2=16{\left(\frac{4\sqrt{5}}{16}\right)}^2$
$=5$