Question

How do you find $\cos 2x$, given $\sin x=\dfrac{1}{5}$ and $\cos x<0$?

Answer 1

This question is from the topic of trigonometry. In this question, we will first use the formula ${{\cos }^{2}}x=1-{{\sin }^{2}}x$ and put here the value of sinx in the equation and find the value of cosx. After that, using the formula $\cos 2x=2{{\cos }^{2}}x-1$, we will find the value of $\cos 2x$. After that, we will see the alternate method to solve this question.

Complete step by step solution:
Let us solve this question.
In this question, it is asked us to find the value of $\cos 2x$ and it is given that $\sin x=\dfrac{1}{5}$ and $\cos x<0$.
So, we can see that $\cos x<0$ and it is also given that $\sin x=\dfrac{1}{5}$. Hence, the value of sin is positive and cos is negative. So, we can say that the value of x lies between 90 and 180 degrees. Therefore, using the formula ${{\sin }^{2}}x+{{\cos }^{2}}x=1$, we can write this formula as
${{\cos }^{2}}x=1-{{\sin }^{2}}x$
Now, putting the value of $\sin x$ as $\dfrac{1}{5}$, we can write the above equation as
$\Rightarrow {{\cos }^{2}}x=1-{{\left( \dfrac{1}{5} \right)}^{2}}$
The above equation can also be written as
$\Rightarrow {{\cos }^{2}}x=1-\dfrac{1}{25}$
The above equation can also be written as
$\Rightarrow {{\cos }^{2}}x=\dfrac{24}{25}$
Now, squaring root both the side of the equation, we can write the above equation as
$\Rightarrow \sqrt{{{\cos }^{2}}x}=\pm \sqrt{\dfrac{24}{25}}$
$\Rightarrow \cos x=\pm \dfrac{\sqrt{24}}{\sqrt{25}}$
As we know that the square of 25 is 5, so we can write
$\Rightarrow \cos x=\pm \dfrac{\sqrt{24}}{5}$
Now, we know that 24 is a multiple of 6 and 4, so we can write
$\Rightarrow \cos x=\pm \dfrac{\sqrt{4\times 6}}{5}=\pm \dfrac{2\sqrt{6}}{5}$
So, we have got the value of $\cos x$ as $\dfrac{2\sqrt{6}}{5}$ and $\left( -\dfrac{2\sqrt{6}}{5} \right)$.
But, it is given in the question that $\cos x<0$, so we can say that $\cos x=-\dfrac{2\sqrt{6}}{5}$
Now, we will use the formula here. The formula is
$\cos 2x=2{{\cos }^{2}}x-1$
Now, putting the value of $\cos x$ as $-\dfrac{2\sqrt{6}}{5}$ in the above equation, we can write
$\Rightarrow \cos 2x=2{{\left( -\dfrac{2\sqrt{6}}{5} \right)}^{2}}-1$
The above equation can also be written as
$\Rightarrow \cos 2x=2\left( \dfrac{4\times 6}{25} \right)-1$
The above equation can also be written as
$\Rightarrow \cos 2x=2\left( \dfrac{24}{25} \right)-1=\dfrac{48}{25}-1=\dfrac{48-25}{25}$
The above equation can also be written as
$\Rightarrow \cos 2x=\dfrac{23}{25}$
So, we have found the value of $\cos 2x$. The value of $\cos 2x$ is $\dfrac{23}{25}$.

Note:
We should have a better knowledge in the topic of trigonometry. We should remember the following formulas to solve this type of question easily:
${{\sin }^{2}}x+{{\cos }^{2}}x=1$
$\cos 2x=2{{\cos }^{2}}x-1$
$\cos 2x=1-2{{\sin }^{2}}x$
We can solve this question by alternate method.
It is given that $\sin x=\dfrac{1}{5}$ and $\cos x<0$, and we have to find $\cos 2x$.
So, using the formula $\cos 2x=1-2{{\sin }^{2}}x$, by putting the value of $\sin x$ as $\dfrac{1}{5}$ we can write this formula as
$\cos 2x=1-2{{\left( \dfrac{1}{5} \right)}^{2}}=1-\left( \dfrac{2}{25} \right)=\dfrac{25-2}{25}=\dfrac{23}{25}$
Hence, we get the same answer. So, we can use this method too.