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Please answer this question​

Please answer this question​-example-1

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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

User Matt Rowland
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