Question

Please answer this question​

Answer 1

As we know an identity that ;

• 
  `{\boxed{\bf{cos((2\theta))=2\cos^(2)(\theta)-1}}}`

Setting,
`{\bf{{\theta}=\footnotesize (\pi)/(8)}}` will give us ;

`{:\implies \quad \sf \cos \left(2* (\pi)/(8)\right)=2\cos^(2)\left((\pi)/(8)\right)-1}`

`{:\implies \quad \sf 2\cos^(2)\left((\pi)/(8)\right)=(1)/(√(2))+1}`

`{:\implies \quad \sf 2\cos^(2)\left((\pi)/(8)\right)=(1+√(2))/(√(2))}`

`{:\implies \quad \sf \cos^(2)\left((\pi)/(8)\right)=(1+√(2))/(2√(2))}`

Rationalizing the denominator of RHS, will yield ;

`{:\implies \quad \sf \cos^(2)\left((\pi)/(8)\right)=(1+√(2))/(2√(2))* (2√(2))/(2√(2))}`

`{:\implies \quad \sf \cos^(2)\left((\pi)/(8)\right)=(2√(2)+4)/(8)}`

`{:\implies \quad \sf \cos^(2)\left((\pi)/(8)\right)=(2√(2)+4)/(8)}`

`{:\implies \quad \sf \cos \left((\pi)/(8)\right)=\pm \sqrt{(2√(2)+4)/(8)}}`

Now, as we know that ;

• 
  `{\boxed{\bf{\sin (2\theta)=2\sin (\theta)\cos (\theta)}}}`

Now, setting the same
`{\bf{{\theta}=\footnotesize (\pi)/(8)}}`

`{:\implies \quad \sf 2\sin \left((\pi)/(8)\right)\cos \left((\pi)/(8)\right)=\sin \left(2* (\pi)/(8)\right)}`

`{:\implies \quad \sf 2\sin \left((\pi)/(8)\right)\cos \left((\pi)/(8)\right)=(1)/(√(2))}`

`{:\implies \quad \sf \sin \left((\pi)/(8)\right)\left(\pm \sqrt{(2√(2)+4)/(8)}\right)=(1)/(2√(2))}`

`{:\implies \quad \sf \sin \left((\pi)/(8)\right)=\frac{√(8)}{2√(2)(\pm \sqrt{2√(2)+4})}}`

`{:\implies \quad \sf \sin \left((\pi)/(8)\right)=\pm \frac{1}{\sqrt{2√(2)+4}}}`

This is the required answer