Question

$4\left({\sin }^{4}30°+{\cos }^{4}60°\right)-\frac{2}{3}\left({\sin }^{2}60°-{\cos }^{2}45°\right)+\frac{1}{2}{\tan }^{2}60°=?$
(a) $\frac{8}{3}$
(b) $\frac{11}{6}$
(c) $\frac{13}{6}$
(d) $\frac{13}{8}$

Answer 1

(b) $\frac{11}{6}$

$$\begin{gathered} 4\left({\sin }^4{30}^o+{\cos }^4{60}^o\right)-\frac{2}{3}\left({\sin }^2{60}^o-{\cos }^2{45}^o\right)+\frac{1}{2}{\tan }^2{60}^o \\ =4\left[{\left(\frac{1}{2}\right)}^4+{\left(\frac{1}{2}\right)}^4\right]-\frac{2}{3}\left[{\left(\frac{\sqrt{3}}{2}\right)}^2-{\left(\frac{1}{\sqrt{2}}\right)}^2\right]+\frac{1}{2}\times {\left(\sqrt{3}\right)}^2 \\ =4\left(\frac{1}{16}+\frac{1}{16}\right)-\frac{2}{3}\left(\frac{3}{4}-\frac{1}{2}\right)+\frac{3}{2} \\ =4\times \frac{1}{8}-\frac{2}{3}\times \frac{1}{4}+\frac{3}{2} \\ =\frac{1}{2}-\frac{1}{6}+\frac{3}{2} \\ =\frac{3-1+9}{6} \\ =\frac{11}{6} \end{gathered}$$