Question

The value of ${\cos }^4\frac{\pi }{4}-{\cos }^4\frac{\pi }{6}+{\sin }^4\frac{\pi }{6}+{\sin }^4\frac{4\pi }{3}$ is

• A. $\frac{1}{16}$
• B. $\frac{1}{8}$
• C. $\frac{-13}{16}$
• D. $\frac{3}{16}$

Answer 1

Answer: (C)

$\frac{-13}{16}$

Given that,

$\Rightarrow {\cos }^4\frac{\pi }{4}-{\cos }^4\frac{\pi }{6}+{\sin }^4\frac{\pi }{6}+{\sin }^4\frac{4\pi }{3}$

$\Rightarrow {\cos }^4\frac{\pi }{4}-{\cos }^4\frac{\pi }{6}+{\sin }^4\frac{\pi }{6}+{\sin }^4(\pi +\frac{\pi }{3})$

$\Rightarrow {\cos }^4\frac{\pi }{4}-{\cos }^4\frac{\pi }{6}+{\sin }^4\frac{\pi }{6}+(-{\sin }^4\left(\frac{\pi }{3}\right))$

$\Rightarrow {\cos }^4\frac{\pi }{4}-{\cos }^4\frac{\pi }{6}+{\sin }^4\frac{\pi }{6}-{\sin }^4\left(\frac{\pi }{3}\right)$

$\Rightarrow {\left(\frac{1}{\sqrt{2}}\right)}^4-{\left(\frac{\sqrt{3}}{2}\right)}^4+{\left(\frac{1}{2}\right)}^4-{\left(\frac{\sqrt{3}}{2}\right)}^4$

$\Rightarrow \frac{1}{4}+\frac{1}{16}-\frac{2.{\sqrt{3}}^4}{2^4}$

$\Rightarrow \frac{4+1}{16}-\frac{3^2}{2^3}$

$\Rightarrow \frac{5}{16}-\frac{9}{8}$

$\Rightarrow \frac{-13}{16}$