As we know an identity that ;
- [tex]{\boxed{\bf{\cos{(2\theta)}=2\cos^{2}(\theta)-1}}}[/tex]
Setting, [tex]{\bf{{\theta}=\footnotesize \dfrac{\pi}{8}}}[/tex] will give us ;
[tex]{:\implies \quad \sf \cos \left(2\times \dfrac{\pi}{8}\right)=2\cos^{2}\left(\dfrac{\pi}{8}\right)-1}[/tex]
[tex]{:\implies \quad \sf 2\cos^{2}\left(\dfrac{\pi}{8}\right)=\dfrac{1}{\sqrt{2}}+1}[/tex]
[tex]{:\implies \quad \sf 2\cos^{2}\left(\dfrac{\pi}{8}\right)=\dfrac{1+\sqrt{2}}{\sqrt{2}}}[/tex]
[tex]{:\implies \quad \sf \cos^{2}\left(\dfrac{\pi}{8}\right)=\dfrac{1+\sqrt{2}}{2\sqrt{2}}}[/tex]
Rationalizing the denominator of RHS, will yield ;
[tex]{:\implies \quad \sf \cos^{2}\left(\dfrac{\pi}{8}\right)=\dfrac{1+\sqrt{2}}{2\sqrt{2}}\times \dfrac{2\sqrt{2}}{2\sqrt{2}}}[/tex]
[tex]{:\implies \quad \sf \cos^{2}\left(\dfrac{\pi}{8}\right)=\dfrac{2\sqrt{2}+4}{8}}[/tex]
[tex]{:\implies \quad \sf \cos^{2}\left(\dfrac{\pi}{8}\right)=\dfrac{2\sqrt{2}+4}{8}}[/tex]
[tex]{:\implies \quad \sf \cos \left(\dfrac{\pi}{8}\right)=\pm \sqrt{\dfrac{2\sqrt{2}+4}{8}}}[/tex]
Now, as we know that ;
- [tex]{\boxed{\bf{\sin (2\theta)=2\sin (\theta)\cos (\theta)}}}[/tex]
Now, setting the same [tex]{\bf{{\theta}=\footnotesize \dfrac{\pi}{8}}}[/tex]
[tex]{:\implies \quad \sf 2\sin \left(\dfrac{\pi}{8}\right)\cos \left(\dfrac{\pi}{8}\right)=\sin \left(2\times \dfrac{\pi}{8}\right)}[/tex]
[tex]{:\implies \quad \sf 2\sin \left(\dfrac{\pi}{8}\right)\cos \left(\dfrac{\pi}{8}\right)=\dfrac{1}{\sqrt{2}}}[/tex]
[tex]{:\implies \quad \sf \sin \left(\dfrac{\pi}{8}\right)\left(\pm \sqrt{\dfrac{2\sqrt{2}+4}{8}}\right)=\dfrac{1}{2\sqrt{2}}}[/tex]
[tex]{:\implies \quad \sf \sin \left(\dfrac{\pi}{8}\right)=\dfrac{\sqrt{8}}{2\sqrt{2}(\pm \sqrt{2\sqrt{2}+4})}}[/tex]
[tex]{:\implies \quad \sf \sin \left(\dfrac{\pi}{8}\right)=\pm \dfrac{1}{\sqrt{2\sqrt{2}+4}}}[/tex]
This is the required answer
