Question

sin^4 π/16 + sin^4 3π/16 + sin^4 5π/16 + sin^4 7π/16 = 3/2

Answer 1

$\mathit{\large{GIVEN:- \;\;SIN^4\frac{\pi}{16} + SIN^4\frac{3\pi}{16} + SIN^4\frac{5\pi}{16} + SIN^4\frac{7\pi}{16}}}$

we know $\mathit{sin(\frac{\pi}{2} - \theta) = cos\theta}\\ \\ \mathit{sin(\pi - \theta) = sin\theta}$

Now,

$\mathit{\large{SIN^4(\frac{\pi}{2} - \frac{7\pi}{16}) + SIN^4(\frac{\pi}{2} - \frac{5\pi}{16}) + SIN^4\frac{5\pi}{16} + SIN^4\frac{7\pi}{16}}}$

$\mathit{\large{COS^4\frac{7\pi}{16} + COS^4\frac{5\pi}{16} + SIN^4\frac{5\pi}{16} + SIN^4\frac{7\pi}{16}}}$

$\mathit{\large{[COS^4\frac{7\pi}{16} + SIN^4\frac{7\pi}{16}] + [COS^4\frac{5\pi}{16} + SIN^4\frac{5\pi}{16}]}}$

$\mathit{\large{[(cos^2\frac{7\pi}{16} + sin^2\frac{7\pi}{16})^2 - 2sin^2\frac{7\pi}{16}*cos^2\frac{7\pi}{16}] + [(cos^2\frac{7/5\pi}{16} + sin^2\frac{5\pi}{16})^2 - 2sin^2\frac{5\pi}{16}*cos^2\frac{5\pi}{16}]}}$
$\mathit{\to\to\to\to\to\to\to\to\to\to\boxed{(a^4 + b^4) = (a^2 + b^2)^2 - 2ab}}$

$\mathit{\large{1 - \frac{1}{2}sin^2(2*\frac{7\pi}{16}) + 1 - \frac{1}{2}sin^2(2*\frac{5\pi}{16})}}$
$\mathit{\to\to\to\to\to\to\to\to\to\to\to\boxed{sin^2\theta + cos^2\theta = 1\;\;and\;\;\mathit2sin\theta*cos\theta = sin2\theta}}$

$\mathit{\large{2 - \frac{1}{2}sin^2\frac{7\pi}{8} - \frac{1}{2}sin^2\frac{5\pi}{8}}}$

$\mathit{\large{2 - \frac{1}{2}[sin^2(\pi - \frac{\pi}{8}) + sin^2(\pi - \frac{3\pi}{8})]}}$

$\mathit{\large{2 - \frac{1}{2}[sin^2\frac{\pi}{8} + sin^2\frac{3\pi}{8}]}}$

$\mathit{\large{2 - \frac{1}{2}[sin^2\frac{\pi}{8} + sin^2(\frac{\pi}{2} - \frac{\pi}{8})]}}$

$\mathit{\large{2 - \frac{1}{2}(sin^2\frac{\pi}{8} + cos^2\frac{\pi}{8})}}$

$\mathit{\large{2 - \frac{1}{2}}}$

$\mathit{\large{\frac{4 - 1}{2}}}$

$\mathit{\large{\frac{3}{2}}}$

Hence, $\mathit{\large{GIVEN:- \;\;SIN^4\frac{\pi}{16} + SIN^4\frac{3\pi}{16} + SIN^4\frac{5\pi}{16} + SIN^4\frac{7\pi}{16} = \frac{3}{2}}}$

Answer 2

Answer:

Step-by-step explanation: