^4 4 + ^4 3 8 + ^4 5 8 + ^4 7 8 = - Vidyadip

MCQ

${\sin ^4}\frac{\pi }{4} + {\sin ^4}\frac{{3\pi }}{8} + {\sin ^4}\frac{{5\pi }}{8} + {\sin ^4}\frac{{7\pi }}{8} = $

A
$\frac{1}{2}$
B
$\frac{1}{4}$
✓
$\frac{3}{2}$
D
$\frac{3}{4}$

✓

Answer

Correct option: C.

$\frac{3}{2}$

c
(c) ${\sin ^4}\frac{\pi }{8} + {\sin ^4}\frac{{3\pi }}{8} + {\sin ^4}\frac{{5\pi }}{8} + {\sin ^4}\frac{{7\pi }}{8}$

$= \frac{1}{4}\,\left[ {{{\left( {2{{\sin }^2}\frac{\pi }{8}} \right)}^2} + {{\left( {2{{\sin }^2}\frac{{3\pi }}{8}} \right)}^2}} \right]$

$ + \frac{1}{4}\,\left[ {{{\left( {2{{\sin }^2}\frac{\pi }{8}} \right)}^2} + {{\left( {2{{\sin }^2}\frac{{3\pi }}{8}} \right)}^2}} \right]$

= $\frac{1}{4}\,\left[ {{{\left( {1 - \cos \frac{\pi }{4}} \right)}^2} + {{\left( {1 - \cos \frac{{3\pi }}{4}} \right)}^2}} \right]$

$ + \frac{1}{4}\,\left[ {{{\left( {1 - \cos \frac{\pi }{4}} \right)}^2} + {{\left( {1 - \cos \frac{{3\pi }}{4}} \right)}^2}} \right]$

$= \frac{1}{4}\,\left[ {{{\left( {1 - \frac{1}{{\sqrt 2 }}} \right)}^2} + {{\left( {1 + \frac{1}{{\sqrt 2 }}} \right)}^2}} \right] + \frac{1}{4}\,\left[ {{{\left( {1 - \frac{1}{{\sqrt 2 }}} \right)}^2} + {{\left( {1 + \frac{1}{{\sqrt 2 }}} \right)}^2}} \right]$

$= \frac{1}{4}(3)\, + \frac{1}{4}(3) = \frac{3}{2}$.

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