Question

$\sin {750}^0.\cos {480}^0+\cos {120}^0.\cos {60}^0=\frac{\sqrt{3}-1}{4}$

• A. True
• B. False

Answer 1

The correct option is B False

$\cos {750}^{\circ }\cos {480}^{\circ }+\cos {120}^{\circ }\cos {60}^{\circ }$

$=\cos ({360}^{\circ }\times 2+{30}^{\circ })\cos ({360}^{\circ }+{120}^{\circ })+\cos ({180}^{\circ }-{60}^{\circ })\cos {60}^{\circ }$

$=-\cos {30}^{\circ }\cos {120}^{\circ }-\cos {60}^{\circ }\cos {60}^{\circ }$

$=-\cos {30}^{\circ }\cos ({180}^{\circ }-{60}^{\circ })-\cos {60}^{\circ }\cos {60}^{\circ }$

$=\cos {30}^{\circ }\cos {60}^{\circ }-\cos {60}^{\circ }\cos {60}^{\circ }$

$=\frac{\sqrt{3}}{2}\times \frac{1}{2}-\frac{1}{2}\times \frac{1}{2}$

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