Question

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

Answer 1

Hint: 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.