Question

${\sin }^2\theta +{\sin }^2({60}^{\circ }+\theta )+si{n}^2({60}^{\circ }-\theta )=$

• A. $\frac{3}{2}$
• B. $\frac{1}{2}$
• C. $\frac{3}{18}$
• D. $\frac{1}{4}$

Answer 1

The correct option is A $\frac{3}{2}$

${\sin }^2\theta +{\sin }^2({60}^{\circ }+\theta )+{\sin }^2({60}^{\circ }-\theta )={\sin }^2\theta +\frac{1}{2}(1-\cos ({120}^{\circ }+2\theta \left)\right)+\frac{1}{2}(1-\cos ({120}^{\circ }-2\theta \left)\right)$

$={\sin }^2\theta +1-\frac{1}{2}\left(\cos \right({120}^{\circ }+2\theta )+\cos ({120}^{\circ }-2\theta \left)\right)$

$=1+{\sin }^2\theta -\frac{1}{2}\left(2\cos {120}^{\circ }\cos 2\theta \right)$

$=1+{\sin }^2\theta +\frac{1}{2}(1-2{\sin }^2\theta )$

$=1+\frac{1}{2}=\frac{3}{2}$