Question

$\sin {36}^o\sin {72}^o.\sin {108}^o.\sin {144}^o=\frac{5}{16}$.

• A. True
• B. False

Answer 1

The correct option is A True

$\sin {72}^{\circ }\sin {108}^{\circ }=\sin ({90}^{\circ }-{18}^{\circ })\sin ({90}^{\circ }+{18}^{\circ })={\cos }^2{18}^{\circ }=1-{\sin }^2{18}^{\circ }=1-(\frac{\sqrt{5}-1}{4}{)}^2=\frac{16-(5+1-2\sqrt{5})}{16}$

$=\frac{10+2\sqrt{5}}{16}$

$\sin {36}^{\circ }\sin {144}^{\circ }=\sin {36}^{\circ }\sin ({180}^{\circ }-{36}^{\circ })={\sin }^2{36}^{\circ }=1-{\cos }^2{36}^{\circ }=1-(\frac{\sqrt{5}+1}{4}{)}^2=\frac{16-(5+1+2\sqrt{5})}{16}$

$=\frac{10-2\sqrt{5}}{16}$

So $\sin {36}^{\circ }\sin {72}^{\circ }\sin {108}^{\circ }\sin {144}^{\circ }=\left(\sin {36}^{\circ }\sin {144}^{\circ }\right)\left(\sin {72}^{\circ }\sin {108}^{\circ }\right)$

$=\frac{10-2\sqrt{5}}{16}\times \frac{10+2\sqrt{5}}{16}=\frac{{10}^2-4\times 5}{{16}^2}=\frac{5}{16}$

So the given relation is $\text{True}$