Question

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

Answer 1

Hint: 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.