Question

Find the value of ${\sin }^4\frac{\pi }{16}+{\sin }^4\frac{3\pi }{16}+{\sin }^4\frac{5\pi }{16}+{\sin }^4\frac{7\pi }{16}$

Answer 1

Given

$si{n}^4\frac{\pi }{16}+si{n}^4\frac{3\pi }{16}+si{n}^4\frac{5\pi }{16}+si{n}^4\frac{7\pi }{16}$

We know that $sin(\frac{\pi }{2}-\theta )=cos\theta$

and $sin(\pi -\theta )=sin\theta$

Now,

$\Rightarrow si{n}^4\frac{\pi }{16}+si{n}^4\frac{3\pi }{16}+si{n}^4\frac{5\pi }{16}+si{n}^4\frac{7\pi }{16}$

This can be written as

$\Rightarrow si{n}^4(\frac{\pi }{2}-\frac{7\pi }{16})+si{n}^4(\frac{\pi }{2}-\frac{5\pi }{16})+si{n}^4\frac{5\pi }{16}+si{n}^4\frac{7\pi }{16}$

$\Rightarrow co{s}^4\frac{7\pi }{16}+co{s}^4\frac{5\pi }{16}+si{n}^4\frac{5\pi }{16}+si{n}^4\frac{7\pi }{16}$

$\Rightarrow (co{s}^4\frac{7\pi }{16}+si{n}^4\frac{7\pi }{16})+(co{s}^4\frac{5\pi }{16}+si{n}^4\frac{5\pi }{16})$

We know that $\Rightarrow a^4+b^4=(a^2+b^2{)}^2-2a^2{b}^2$

Using this property we get,

$\Rightarrow \left[\right(co{s}^2\frac{7\pi }{16}+si{n}^2\frac{7\pi }{16}{)}^2-2si{n}^2\frac{7\pi }{16}co{s}^2\frac{7\pi }{16}]+[(co{s}^2\frac{5\pi }{16}+si{n}^2\frac{5\pi }{16}{)}^2-2si{n}^2\frac{5\pi }{16}co{s}^2\frac{5\pi }{16}]$

This can be written as

$\Rightarrow \left[\right(co{s}^2\frac{7\pi }{16}+si{n}^2\frac{7\pi }{16}{)}^2-\frac{4}{2}si{n}^2\frac{7\pi }{16}co{s}^2\frac{7\pi }{16}]+[(co{s}^2\frac{5\pi }{16}+si{n}^2\frac{5\pi }{16}{)}^2-\frac{4}{2}si{n}^2\frac{5\pi }{16}co{s}^2\frac{5\pi }{16}]$

$\Rightarrow \left[\right(co{s}^2\frac{7\pi }{16}+si{n}^2\frac{7\pi }{16}{)}^2-\frac{1}{2}\left(2sin\frac{7\pi }{16}cos\frac{7\pi }{16}{)}^2\right]+\left[\right(co{s}^2\frac{5\pi }{16}+si{n}^2\frac{5\pi }{16}{)}^2-\frac{1}{2}\left(2sin\frac{5\pi }{16}cos\frac{5\pi }{16}{)}^2\right]$

We know that $\Rightarrow si{n}^2\theta +co{s}^2\theta =1$ and $2sin\theta cos\theta =sin2\theta$

Using these properties we get,

$\Rightarrow [1-\frac{1}{2}si{n}^2(\frac{2\times 7\pi }{16}\left)\right]+[1-\frac{1}{2}si{n}^2(\frac{2\times 5\pi }{16}\left)\right]$

$\Rightarrow [1-\frac{1}{2}si{n}^2(\frac{7\pi }{8}\left)\right]+[1-\frac{1}{2}si{n}^2(\frac{5\pi }{8}\left)\right]$

$\Rightarrow 2-\frac{1}{2}si{n}^2\left(\frac{7\pi }{8}\right)-\frac{1}{2}si{n}^2\left(\frac{5\pi }{8}\right)$

$\Rightarrow 2-\frac{1}{2}\left[si{n}^2\right(\frac{7\pi }{8})+si{n}^2(\frac{5\pi }{8}\left)\right]$

This can be written as

$\Rightarrow 2-\frac{1}{2}\left[si{n}^2\right(\pi -\frac{\pi }{8})+si{n}^2(\pi -\frac{3\pi }{8}\left)\right]$

$\Rightarrow 2-\frac{1}{2}\left[si{n}^2\right(\frac{\pi }{8})+si{n}^2(\frac{3\pi }{8}\left)\right]$

$\Rightarrow 2-\frac{1}{2}\left[si{n}^2\right(\frac{\pi }{8})+si{n}^2(\frac{\pi }{2}-\frac{\pi }{8}\left)\right]$

$\Rightarrow 2-\frac{1}{2}\left[si{n}^2\right(\frac{\pi }{8})+co{s}^2(\frac{\pi }{8}\left)\right]$

$\Rightarrow 2-\frac{1}{2}\left[1\right]$

$\Rightarrow 2-\frac{1}{2}=\frac{3}{2}$

Therefore, $\Rightarrow si{n}^4\frac{\pi }{16}+si{n}^4\frac{3\pi }{16}+si{n}^4\frac{5\pi }{16}+si{n}^4\frac{7\pi }{16}=\frac{3}{2}$