Question

Find the values of other five trigonometric functions if $\cos (x)=\dfrac{-1}{2}$, x lies in the third quadrant.

Answer 1

Hint: In this question, we are given the value of the cosine of the angle and the quadrant in which the angle x lies. Therefore, using the definition of cos(x), and other trigonometric formulas, we can obtain the values of other trigonometric ratios by solving the corresponding equations.

Complete step by step solution:

We are given the value of cot(x). We can use the trigonometric relation

${{\sin }^{2}}\left( x \right)=1-{{\cos }^{2}}\left( x \right)$

With the given value of cos(x) to find

$ {{\sin }^{2}}\left( x \right)=1-{{\left( \dfrac{-1}{2} \right)}^{2}}=1-\dfrac{1}{4}=\dfrac{4-1}{4}=\dfrac{3}{4} $

$ \Rightarrow \sin (x)=\pm \sqrt{\dfrac{3}{4}}=\pm \dfrac{\sqrt{3}}{2} $

However, in the third quadrant, the value of sin(x) is negative, so we should take only the negative value in the above equation to obtain

 $\sin (x)=\dfrac{-\sqrt{3}}{2}............(1.1)$

Now, the expression for tan(x) is given by,

$\tan (x)=\dfrac{\sin (x)}{\cos (x)}$

Therefore, using the values of cos(x) and sin(x) from the values given in question and (1.1),

$\tan (x)=\dfrac{\sin (x)}{\cos(x)}=\dfrac{\dfrac{-1}{2}}{\dfrac{-\sqrt{3}}{2}}=\dfrac{1}{\sqrt{3}}............(1.2)$

Also tan(x) and cot(x) are related by

$\tan (x)=\dfrac{1}{\cot (x)}$

Therefore, using the value of tan(x) from (1.2) in the above equation, we find

$\cot (x)=\dfrac{1}{\dfrac{1}{\sqrt{3}}}=\sqrt{3}...............(1.3)$

We know that cosec(x) and sec(x) are the multiplicative inverses of sin(x) and cos(x) as,

therefore using the given value of cos(x) and by using equation (1.1), we obtain

$\text{cosec}(x)=\dfrac{1}{\sin(x)}=\dfrac{1}{\dfrac{-\sqrt{3}}{2}}=\dfrac{-2}{\sqrt{3}}..........(1.4)$

And

$\text{sec}(x)=\dfrac{1}{\cos (x)}=\dfrac{1}{\dfrac{-1}{2}}=-2...........(1.5)$

Therefore, from equations (1.1), (1.2), (1.3), (1.4) and (1.5), we have found the other trigonometric ratios as

$\sin (x)=\dfrac{-\sqrt{3}}{2}$, $\tan (x)=\dfrac{1}{\sqrt{3}}$, $\cot (x)=\sqrt{3}$ , $\sec(x)=-2$, $\text{cosec}(x)=\dfrac{-2}{\sqrt{3}}$

Which is the required answer to this question.

Note: We could also have found sec(x) by using $\sec (x)=\dfrac{1}{\cos (x)}$ and then tan(x) from sec(x) by using the identity ${{\sec }^{2}}\left( x \right)=1+{{\tan }^{2}}\left( x \right)$ and from there we could have found sin(x) and cos(x). However, the answer would have remained the same as found out in the solution above.