
Find the value of $\cos \dfrac{\pi }{8}$.
A) $\sqrt{\dfrac{\sqrt{2}-1}{2\sqrt{2}}}$
B) $\sqrt{\dfrac{\sqrt{2}+1}{2\sqrt{2}}}$
C) $\sqrt{\dfrac{\sqrt{2}+1}{2\sqrt{3}}}$
D) $\sqrt{\dfrac{\sqrt{2}+1}{\sqrt{2}}}$
Answer
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Hint: In this question, we have to bring the given expression in the form of the cosine of a standard angle, whose cosine value is known. Thereafter, we can obtain the value of the given expression in terms of the cosine value of the standard angle (which is ${{45}^{\circ }}$ in this case).
Complete step-by-step answer:
In this case, the expression to be evaluated is $\cos \dfrac{\pi }{8}$. We should try to bring it in terms of the trigonometric ratios of one of the standard angles i.e. $\dfrac{\pi }{2},\dfrac{\pi }{3},\dfrac{\pi }{4},\dfrac{\pi }{6}$ whose trigonometric ratios are known. In this case, we notice that $\dfrac{\pi }{8}$ can be written as $\dfrac{\pi }{8}=\dfrac{\pi }{2\times 4}=\dfrac{1}{2}\times \dfrac{\pi }{4}$.
We can now use the cosine of sum of angles rule
\[\cos (x+y)=\cos (x)\cos (y)-\sin (x)\sin (y)\]
We can take y=x in the above formula which gives the cosine of twice an angle x as
$\begin{align}
& \cos (2x)={{\cos }^{2}}(x)-{{\sin }^{2}}(x)={{\cos }^{2}}(x)-(1-{{\cos }^{2}}x)=2{{\cos }^{2}}x-1 \\
& \Rightarrow \cos (x)=\sqrt{\dfrac{1+\cos (2x)}{2}}..................................(1.1) \\
\end{align}$
Thus, in this case we can take $x=\dfrac{\pi }{8}$. Then, $2x=\dfrac{\pi }{4}$, using these values in equation(1.1), we obtain
\[\cos (\dfrac{\pi }{8})=\sqrt{\dfrac{1+\cos \left( \dfrac{\pi }{4} \right)}{2}}.......................(1.2)\]
Now, we also know that \[\cos \left( \dfrac{\pi }{4} \right)=\cos ({{45}^{\circ }})=\dfrac{1}{\sqrt{2}}\]. Using these values in equation (1.2), we get
\[\cos (\dfrac{\pi }{8})=\sqrt{\dfrac{1+\cos \left( {{45}^{\circ }} \right)}{2}}=\sqrt{\dfrac{1+\dfrac{1}{\sqrt{2}}}{2}}=\sqrt{\dfrac{\sqrt{2}+1}{2\sqrt{2}}}\]
Therefore, the answer to this question should be option(B).
Note: In this case we should be careful to use the cosine of twice an angle formula because there are three formulas for cos of twice an angle and note that there should be an overall square root in the answer.
Complete step-by-step answer:
In this case, the expression to be evaluated is $\cos \dfrac{\pi }{8}$. We should try to bring it in terms of the trigonometric ratios of one of the standard angles i.e. $\dfrac{\pi }{2},\dfrac{\pi }{3},\dfrac{\pi }{4},\dfrac{\pi }{6}$ whose trigonometric ratios are known. In this case, we notice that $\dfrac{\pi }{8}$ can be written as $\dfrac{\pi }{8}=\dfrac{\pi }{2\times 4}=\dfrac{1}{2}\times \dfrac{\pi }{4}$.
We can now use the cosine of sum of angles rule
\[\cos (x+y)=\cos (x)\cos (y)-\sin (x)\sin (y)\]
We can take y=x in the above formula which gives the cosine of twice an angle x as
$\begin{align}
& \cos (2x)={{\cos }^{2}}(x)-{{\sin }^{2}}(x)={{\cos }^{2}}(x)-(1-{{\cos }^{2}}x)=2{{\cos }^{2}}x-1 \\
& \Rightarrow \cos (x)=\sqrt{\dfrac{1+\cos (2x)}{2}}..................................(1.1) \\
\end{align}$
Thus, in this case we can take $x=\dfrac{\pi }{8}$. Then, $2x=\dfrac{\pi }{4}$, using these values in equation(1.1), we obtain
\[\cos (\dfrac{\pi }{8})=\sqrt{\dfrac{1+\cos \left( \dfrac{\pi }{4} \right)}{2}}.......................(1.2)\]
Now, we also know that \[\cos \left( \dfrac{\pi }{4} \right)=\cos ({{45}^{\circ }})=\dfrac{1}{\sqrt{2}}\]. Using these values in equation (1.2), we get
\[\cos (\dfrac{\pi }{8})=\sqrt{\dfrac{1+\cos \left( {{45}^{\circ }} \right)}{2}}=\sqrt{\dfrac{1+\dfrac{1}{\sqrt{2}}}{2}}=\sqrt{\dfrac{\sqrt{2}+1}{2\sqrt{2}}}\]
Therefore, the answer to this question should be option(B).
Note: In this case we should be careful to use the cosine of twice an angle formula because there are three formulas for cos of twice an angle and note that there should be an overall square root in the answer.
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