Question

If $\tan \alpha = \dfrac{1}{7}$ and $\tan \beta = \dfrac{1}{3}$, then $\cos 2\alpha$ is equal to:
(A) $\sin 2\beta$
(B) $\sin 4\beta$
(C) $\sin 3\beta$
(D) None of these

Answer 1

The given question deals with finding the value of trigonometric expression doing basic simplification of trigonometric functions by using some of the simple trigonometric formulae such as double angle formula for cosine in terms of tangent as $\cos 2x = \dfrac{{1 - {{\tan }^2}x}}{{1 + {{\tan }^2}x}}$. Basic algebraic rules and trigonometric identities are to be kept in mind while doing simplification in the given problem. We must know the simplification rules to solve the problem with ease. We first find the value of $\cos 2\alpha$ using $\tan \alpha = \dfrac{1}{7}$ and then find the values of all the options one by one.

Complete step by step answer:
In the given problem, we are given $\tan \alpha = \dfrac{1}{7}$ and $\tan \beta = \dfrac{1}{3}$.
Now, we have to find the expression whose value is equal to $\cos 2\alpha$.
So, we have, $\cos 2\alpha$.
We know the double angle formula for cosine $\cos 2x = \dfrac{{1 - {{\tan }^2}x}}{{1 + {{\tan }^2}x}}$.
So, we get, $\cos 2\alpha = \dfrac{{1 - {{\tan }^2}\alpha }}{{1 + {{\tan }^2}\alpha }}$.
We are given the value of $\tan \alpha$ as $\left( {\dfrac{1}{7}} \right)$. So, we get,
$\Rightarrow \cos 2\alpha = \dfrac{{1 - {{\left( {\dfrac{1}{7}} \right)}^2}}}{{1 + {{\left( {\dfrac{1}{7}} \right)}^2}}}$
Computing the squares of the terms, we get,
$\Rightarrow \cos 2\alpha = \dfrac{{1 - \left( {\dfrac{1}{{49}}} \right)}}{{1 + \left( {\dfrac{1}{{49}}} \right)}}$
Taking LCM in numerator and denominator, we get,
$\Rightarrow \cos 2\alpha = \dfrac{{\left( {\dfrac{{49 - 1}}{{49}}} \right)}}{{\left( {\dfrac{{49 + 1}}{{49}}} \right)}}$
$\Rightarrow \cos 2\alpha = \dfrac{{48}}{{50}}$
Cancelling the common factors in numerator and denominator, we get,
$\Rightarrow \cos 2\alpha = \dfrac{{24}}{{25}}$
Now, we analyze the options and find the values of expression in the options. So, we get,
Option (A):
We have, $\sin 2\beta$.
We are given the value of $\tan \beta$ as $\dfrac{1}{3}$. We also know the double angle formula for sine as $\sin 2x = \dfrac{{2\tan x}}{{1 + {{\tan }^2}x}}$. So, we get,
$\Rightarrow \sin 2\beta = \dfrac{{2\tan \beta }}{{1 + {{\tan }^2}\beta }}$
So, putting value of $\tan \beta$, we get,
$\Rightarrow \sin 2\beta = \dfrac{{2 \times \dfrac{1}{3}}}{{1 + {{\left( {\dfrac{1}{3}} \right)}^2}}}$
Simplifying the terms, we get,
$\Rightarrow \sin 2\beta = \dfrac{{\dfrac{2}{3}}}{{1 + \left( {\dfrac{1}{9}} \right)}}$
$\Rightarrow \sin 2\beta = \dfrac{{\left( {\dfrac{2}{3}} \right)}}{{\left( {\dfrac{{10}}{9}} \right)}}$
$\Rightarrow \sin 2\beta = \left( {\dfrac{2}{3}} \right) \times \left( {\dfrac{9}{{10}}} \right)$
$\Rightarrow \sin 2\beta = \left( {\dfrac{3}{5}} \right)$
Now, Option (B):
We have $\sin 4\beta$.
So, we first evaluate the value of $\cos 2\beta$ as well.
So, we have, $\tan \beta = \dfrac{1}{3}$.
$\cos 2\beta = \dfrac{{1 - {{\tan }^2}\beta }}{{1 + {{\tan }^2}\beta }}$
$\Rightarrow \cos 2\beta = \dfrac{{1 - {{\left( {\dfrac{1}{3}} \right)}^2}}}{{1 + {{\left( {\dfrac{1}{3}} \right)}^2}}}$
On further simplifications
$\Rightarrow \cos 2\beta = \dfrac{{1 - \left( {\dfrac{1}{9}} \right)}}{{1 + \left( {\dfrac{1}{9}} \right)}}$
$\Rightarrow \cos 2\beta = \dfrac{{\left( {\dfrac{8}{9}} \right)}}{{\left( {\dfrac{{10}}{9}} \right)}}$
Simplifying the expression further, we get,
$\Rightarrow \cos 2\beta = \dfrac{4}{5}$
Now, we can find the value of $\sin 4\beta$ using $\sin 2x = 2\sin x\cos x$.
So, $\sin 4\beta = 2\left( {\sin 2\beta } \right)\left( {\cos 2\beta } \right)$
$\Rightarrow \sin 4\beta = 2\left( {\dfrac{3}{5}} \right)\left( {\dfrac{4}{5}} \right)$
$\Rightarrow \sin 4\beta = \dfrac{{24}}{{25}}$
So, the value of $\sin 4\beta$ is $\dfrac{{24}}{{25}}$. Hence, $\cos 2\alpha = \sin 4\beta$. Thus, Option (B) is the correct answer.

Note:
We must have a strong grip over the concepts of trigonometry, related formulae and rules to ace these types of questions. Besides these simple trigonometric formulae, trigonometric identities are also of significant use in such types of questions where we have to simplify trigonometric expressions with help of basic knowledge of algebraic rules and operations.