Question

The value of: ${\cos }^2{30}^{\circ }-{\sin }^2{30}^{\circ }$ is $\cos {60}^o$ or $\frac{1}{2}$.( Enter 1 if true or 0 otherwise)

Answer 1

To prove: ${\cos }^2{30}^{\circ }-{\sin }^2{30}^{\circ }=\cos {60}^{\circ }$

L.H.S: ${\cos }^2{30}^{\circ }-{\sin }^2{30}^{\circ }$

= $(\frac{\sqrt{3}}{2}{)}^2-(\frac{1}{2}{)}^2$

= $\frac{3}{4}-\frac{1}{4}$

= $\frac{2}{4}$

= $\frac{1}{2}$

= $\cos {60}^{\circ }$

Thus, $L.H.S.=R.H.S$