Question

The value of the expression ${{\cos }^{2}}\dfrac{\pi }{12}+{{\cos }^{2}}\dfrac{\pi }{4}+{{\cos }^{2}}\dfrac{5\pi }{12}$ is
(a) $\dfrac{2}{3}$
(b) $\dfrac{2}{3+\sqrt{3}}$
(c) $\dfrac{3}{2}$
(d) $\dfrac{3+\sqrt{3}}{2}$

Answer 1

Hint: In the trigonometric expression given to us, we know the value of $\cos \dfrac{\pi }{4}$ which is equal to $\dfrac{1}{\sqrt{2}}$. For simplifying the reset of the two terms, we can add their arguments to find a relation between the two. The relation will be obtained as \[\dfrac{\pi }{12}=\dfrac{\pi }{2}-\dfrac{5\pi }{12}\], on substituting which into the given expression, and using the trigonometric identity given by ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$, we will be able to find the final value of the given expression.

Complete step by step solution:
Let us consider the trigonometric expression given in the above question as
$\Rightarrow E={{\cos }^{2}}\dfrac{\pi }{12}+{{\cos }^{2}}\dfrac{\pi }{4}+{{\cos }^{2}}\dfrac{5\pi }{12}$
We know that
$\Rightarrow \cos \dfrac{\pi }{4}=\dfrac{1}{\sqrt{2}}$
Taking squares on both the sides, we get
$\begin{align}
  & \Rightarrow {{\cos }^{2}}\dfrac{\pi }{4}={{\left( \dfrac{1}{\sqrt{2}} \right)}^{2}} \\
 & \Rightarrow {{\cos }^{2}}\dfrac{\pi }{4}=\dfrac{1}{2} \\
\end{align}$
Putting the above equation in the given expression, we get
\[\begin{align}
  & \Rightarrow E={{\cos }^{2}}\dfrac{\pi }{12}+\dfrac{1}{2}+{{\cos }^{2}}\dfrac{5\pi }{12} \\
 & \Rightarrow E=\dfrac{1}{2}+{{\cos }^{2}}\dfrac{\pi }{12}+{{\cos }^{2}}\dfrac{5\pi }{12}......\left( i \right) \\
\end{align}\]
Now, let us add the two arguments of the cosine terms to get
\[\begin{align}
  & \Rightarrow \dfrac{\pi }{12}+\dfrac{5\pi }{12} \\
 & \Rightarrow \dfrac{6\pi }{12} \\
 & \Rightarrow \dfrac{\pi }{2} \\
\end{align}\]
This means we can write
$\Rightarrow \dfrac{\pi }{12}+\dfrac{5\pi }{12}=\dfrac{\pi }{2}$
Subtracting $\dfrac{5\pi }{12}$ from both the sides, we get
\[\begin{align}
  & \Rightarrow \dfrac{\pi }{12}+\dfrac{5\pi }{12}-\dfrac{5\pi }{12}=\dfrac{\pi }{2}-\dfrac{5\pi }{12} \\
 & \Rightarrow \dfrac{\pi }{12}=\dfrac{\pi }{2}-\dfrac{5\pi }{12} \\
\end{align}\]
Now, on substituting the above relation in (i) we get
\[\begin{align}
  & \Rightarrow E=\dfrac{1}{2}+{{\cos }^{2}}\left( \dfrac{\pi }{2}-\dfrac{5\pi }{12} \right)+{{\cos }^{2}}\dfrac{5\pi }{12} \\
 & \Rightarrow E=\dfrac{1}{2}+{{\left[ \cos \left( \dfrac{\pi }{2}-\dfrac{5\pi }{12} \right) \right]}^{2}}+{{\cos }^{2}}\dfrac{5\pi }{12}........\left( ii \right) \\
\end{align}\]
Now, we know that
\[\Rightarrow \cos \left( \dfrac{\pi }{2}-\theta \right)=\sin \theta \]
Substituting \[\theta =\dfrac{5\pi }{12}\] in the above identity, we get
\[\Rightarrow \cos \left( \dfrac{\pi }{2}-\dfrac{5\pi }{12} \right)=\sin \dfrac{5\pi }{12}\]
Putting the above identity in (ii) we get
\[\begin{align}
  & \Rightarrow E=\dfrac{1}{2}+{{\left[ \sin \dfrac{5\pi }{12} \right]}^{2}}+{{\cos }^{2}}\dfrac{5\pi }{12} \\
 & \Rightarrow E=\dfrac{1}{2}+{{\sin }^{2}}\dfrac{5\pi }{12}+{{\cos }^{2}}\dfrac{5\pi }{12} \\
\end{align}\]
Now, using the trigonometric identity given by ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$, we can write the above expression as
\[\begin{align}
  & \Rightarrow E=\dfrac{1}{2}+1 \\
 & \Rightarrow E=\dfrac{3}{2} \\
\end{align}\]
Thus, the value of the given trigonometric expression is found to be equal to \[\dfrac{3}{2}\].
Hence, the correct answer is option (c).

Note: We must remember all the important trigonometric identities for solving these types of questions. We can also use the trigonometric identity ${{\cos }^{2}}\theta =\dfrac{1+\cos 2\theta }{2}$ on all the three terms to simplify the given expression as \[\dfrac{1}{2}\left[ \left( 1+\cos \dfrac{\pi }{6} \right)+\left( 1+\cos \dfrac{\pi }{2} \right)+\left( 1+\cos \dfrac{5\pi }{6} \right) \right]\]. Then, on substituting $\cos \left( \dfrac{\pi }{6} \right)=\dfrac{\sqrt{3}}{2}$, \[\cos \left( \dfrac{5\pi }{6} \right)=\dfrac{-\sqrt{3}}{2}\], and \[\cos \left( \dfrac{\pi }{2} \right)=0\], we will obtain the required value.