Question

If $A={30}^0$, then
$\cos 2A={\cos }^{4}A-{\sin }^{4}A$

If true enter 1 else 0

Answer 1

Given $A={30}^{\circ }$
To Prove: $\cos 2A={\cos }^{4}A-{\sin }^{4}A$
L.H.S : $\cos 2A=\cos {60}^{\circ }=\frac{1}{2}$
R.H.S: ${\cos }^{4}A-si{n}^{4}A$
= ${\cos }^4{30}^0-si{n}^4{30}^0$
= $(\frac{\sqrt{3}}{2}{)}^4-(\frac{1}{2}{)}^4$
= $\frac{9}{16}-\frac{1}{16}$
= $\frac{8}{16}$
= $\frac{1}{2}$