Question

If \[\cos \theta =\dfrac{1}{2}\left( x+\dfrac{1}{x} \right)\], then \[\dfrac{1}{2}\left( {{x}^{2}}+\dfrac{1}{{{x}^{2}}} \right)\].
1.\[\sin 2\theta \]
2.\[\cos 2\theta \]
3.\[\tan 2\theta \]
4.\[\sec 2\theta \]

Answer 1

Hint: In order to find the value of \[\dfrac{1}{2}\left( {{x}^{2}}+\dfrac{1}{{{x}^{2}}} \right)\], we will be considering the given value i.e. \[\cos \theta =\dfrac{1}{2}\left( x+\dfrac{1}{x} \right)\]. Then we will be transposing the term \[\dfrac{1}{2}\] from RHS to the LHS. Then we will be squaring the obtained equation on both sides. And upon solving that we will be obtaining an expression. Then we must cross-check it with the given value. If they are the same, we must look at the value that produced the expression and that value will be our required answer.

Complete step by step answer:
Now let us have a brief regarding the trigonometric ratios. The counter-clockwise angle between the initial arm and the terminal arm of an angle in standard position is called the principal angle. Its value is between \[{{0}^{\circ }}\] and \[{{360}^{\circ }}\]. The relationship between the angles and sides of a triangle are given by the trigonometric functions. The basic trigonometric functions are sine, cosine, tangent, cotangent, secant and cosecant. These are the basic main trigonometric functions used.
Now let us start solving our problem.
So we are given with \[\cos \theta =\dfrac{1}{2}\left( x+\dfrac{1}{x} \right)\]
Firstly, let us transpose the term \[\dfrac{1}{2}\] from RHS to the LHS. We get,
\[\begin{align}
  & \cos \theta =\dfrac{1}{2}\left( x+\dfrac{1}{x} \right) \\
 & \Rightarrow 2\cos \theta =\left( x+\dfrac{1}{x} \right) \\
\end{align}\]
Now upon squaring on both sides, we get
\[\begin{align}
  & \Rightarrow {{\left( 2\cos \theta \right)}^{2}}={{\left( x+\dfrac{1}{x} \right)}^{2}} \\
 & \Rightarrow 4{{\cos }^{2}}\theta ={{x}^{2}}+\dfrac{1}{{{x}^{2}}}+2 \\
\end{align}\]
Upon solving further,
\[\begin{align}
  & \Rightarrow 4{{\cos }^{2}}\theta ={{x}^{2}}+\dfrac{1}{{{x}^{2}}}+2 \\
 & \Rightarrow 4{{\cos }^{2}}\theta -2={{x}^{2}}+\dfrac{1}{{{x}^{2}}} \\
 & \Rightarrow 2\left( 2{{\cos }^{2}}\theta -1 \right)={{x}^{2}}+\dfrac{1}{{{x}^{2}}} \\
\end{align}\]
We know that \[\cos 2\theta =2{{\cos }^{2}}\theta -1\], so upon substituting this we get,
\[\begin{align}
  & \Rightarrow 2\left( 2{{\cos }^{2}}\theta -1 \right)={{x}^{2}}+\dfrac{1}{{{x}^{2}}} \\
 & \Rightarrow 2\left( \cos 2\theta \right)={{x}^{2}}+\dfrac{1}{{{x}^{2}}} \\
 & \Rightarrow \cos 2\theta =\dfrac{1}{2}\left( {{x}^{2}}+\dfrac{1}{{{x}^{2}}} \right) \\
\end{align}\]
Now we can observe that the obtained RHS and the given expression in the problem statement are the same.
So the value that is equal to \[\dfrac{1}{2}\left( {{x}^{2}}+\dfrac{1}{{{x}^{2}}} \right)\] is \[\cos 2\theta \].

So, the correct answer is “Option 2”.

Note: While solving such problems, we must be analysing before proceeding in order to choose LHS or RHS for easy simplification. We must be aware of the trigonometric formulas and rules. We must be transposing the terms for our convenience.