Question

Value of ${\sin }^{6}7\frac{1}{2}+{\cos }^{6}7\frac{1}{2}$ is

• A. $\frac{3\sqrt{3}+2}{8}$
• B. $\frac{3\sqrt{3}-5}{16}$
• C. $\frac{10+3\sqrt{3}}{16}$
• D. $\frac{5+3\sqrt{3}}{16}$

Answer 1

Answer: (C)

$\frac{10+3\sqrt{3}}{16}$

${\sin }^{6}7\frac{1}{2}+{\cos }^{6}7\frac{1}{2}={\sin }^6\frac{15}{2}+{\cos }^6\frac{15}{2}=1-\frac{3}{4}{\sin }^{2}15=1-\frac{3}{4}{\left(\sin (45-30)\right)}^2=1-\frac{3}{4}{(\sin 45\cos 30-\cos 45\sin 30)}^2=1-\frac{3}{4}{(\frac{1}{\sqrt{2}}\frac{\sqrt{3}}{2}-\frac{1}{\sqrt{2}}.\frac{1}{2})}^2=1-\frac{3}{4.8}{(\sqrt{3}-1)}^2=1-\frac{3}{32}(3+1-2\sqrt{3})=\frac{32-12+6\sqrt{3}}{32}=\frac{20+6\sqrt{3}}{32}=\frac{10+3\sqrt{3}}{16}$