Question

If A = ${30}^{\circ }$, then $cos2A=co{s}^{4}A-si{n}^{4}A$

If the above statement is true then mention answer as 1, else mention 0 .

Answer 1

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

= $\frac{8}{16}$

= $\frac{1}{2}$

The given statement is true.