Question

If $\cos x = \dfrac{5}{{13}}$ and $x$ is reflex, how do you find the exact value of $\cos 2x$?

Answer 1

We will put the given value of cosine in the formula and then we will solve the question. On doing some simplification we get the required answer.

Formula used: $\cos 2x$ have four different kinds of formulas.
$(1)\;\cos 2x = {\cos ^2}x - {\sin ^2}x$
Now, we know that ${\cos ^2}x + {\sin ^2}x = 1$.
Using this formula we can derive the following formula also:
$(2)\;\;\operatorname{Cos} 2x = 1 - 2{\operatorname{Sin} ^2}x$
Now, using the ${\cos ^2}x + {\sin ^2}x = 1$, we can convert the $\cos 2x$in following way:
$(3)\;\;\cos 2x = 2{\cos ^2}x - 1$
$(4)\;\;\cos 2x = \dfrac{{1 - {{\tan }^2}x}}{{1 + {{\tan }^2}x}}$.

Complete Step by Step Solution:
In the above question, the value of ‘$\cos x$’ is given.
So, we will put the value of ‘$\cos x$’ directly in the formula to derive the value of ‘$\cos 2x$’.
So, we will put the value of $\cos x = \dfrac{5}{{13}}$ in the third formula.
So, we will put $\cos x = \dfrac{5}{{13}}$in $\cos 2x = 2{\cos ^2}x - 1$.
So, after putting this value, we can rewrite the equation in following way:
$\Rightarrow \cos 2x = 2 \times {\left( {\dfrac{5}{{13}}} \right)^2} - 1$.
Now, squaring the constant term, we get the following value:
$\Rightarrow \cos 2x = 2 \times \left( {\dfrac{{25}}{{169}}} \right) - 1$.
Now, multiply the first two terms, we get the following expression:
$\Rightarrow \cos 2x = \dfrac{{50}}{{169}} - 1$.
Now, subtract the above terms, we get the following value:
$\Rightarrow \cos 2x = \dfrac{{50 - 169}}{{169}}$.
Now, subtract the terms in numerator, we get the following value:
$\Rightarrow \cos 2x = \dfrac{{ - 119}}{{\;\;169}}$.
Now, we can rewrite the above expression in following way:
$\Rightarrow \cos 2x = - \dfrac{{119}}{{169}}$.

Therefore, the exact value of $\cos 2x$ is $- \dfrac{{119}}{{169}}$.

Note: Points to remember:
A reflex angle is one that lies between ${180^ \circ }$ and ${360^ \circ }$.
It means that it will be either in the $III$quadrant or in $IV$ quadrant.
Recall that for ${\text{cos}}\theta = \dfrac{{{\text{adjacent}}}}{{{\text{hypotenuse}}}}$.
Hypotenuse is never negative, so the adjacent must be the negative one.
If the adjacent is the negative one, it must be in the $III$ quadrant.
This also means that the opposite must be negative as well.