Question

Find the sines and cosines of all angles in the first four quadrants whose tangents are equal to $\cos \left( {{135}^{\circ }} \right)$ .

Answer 1

Hint:In this case, the value of the tangent is given to be $\cos \left( {{135}^{\circ }} \right)$, therefore, we should try to find out the value of this quantity first. Now, we can use the trigonometric identity of the sum of sine and cosine and use the relation from the relation obtained above from the tangent to obtain the required answer.

Complete step-by-step answer:
Let the angle be $\theta $. It is given that tangent of the angle is equal to $\cos \left( {{135}^{\circ }} \right)$ i.e.
$\tan \left( \theta \right)=\cos \left( {{135}^{\circ }} \right)....................(1.1)$
Now, we can use the following identity
$\cos \left( a+b \right)=\cos \left( a \right)\cos \left( a \right)-\sin \left( a \right)\sin \left( b \right)$
In equation (1.1) to obtain
$\begin{align}
  & \tan \left( \theta \right)=\cos \left( {{135}^{\circ }} \right)=\cos \left( {{90}^{\circ }}+{{45}^{\circ }} \right)=\cos \left( {{90}^{\circ }} \right)\cos \left( {{45}^{\circ }} \right)-\sin \left( {{90}^{\circ }} \right)\sin \left( {{45}^{\circ }} \right) \\
 & =0\times \dfrac{1}{\sqrt{2}}-1\times \dfrac{1}{\sqrt{2}}=\dfrac{-1}{\sqrt{2}}............(1.2) \\
\end{align}$
As the trigonometric ratio tan is defined as
$\tan \left( \theta \right)=\dfrac{\sin \left( \theta \right)}{\cos \left( \theta \right)}$
Using this in equation (1.2), we obtain
$\dfrac{\sin \left( \theta \right)}{\cos \left( \theta \right)}=\dfrac{-1}{\sqrt{2}}\Rightarrow \sin \left( \theta \right)=\dfrac{-1}{\sqrt{2}}\cos \left( \theta \right).......(1.3)$
Also, we have the trigonometric identity between sin and cos of any angle as
${{\sin }^{2}}\left( \theta \right)+{{\cos }^{2}}\left( \theta \right)=1................(1.4)$
Using the relation from (1.3) in (1.4), we obtain
\[\begin{align}
  & {{\left( \dfrac{-1}{\sqrt{2}}\cos \left( \theta \right) \right)}^{2}}+{{\cos }^{2}}\left( \theta \right)=1 \\
 & \Rightarrow {{\cos }^{2}}\left( \theta \right)\left( \dfrac{1}{2}+1 \right)=1\Rightarrow {{\cos }^{2}}\left( \theta \right)=\dfrac{1}{\dfrac{3}{2}}=\dfrac{2}{3} \\
 & \Rightarrow \cos \left( \theta \right)=\pm \sqrt{\dfrac{2}{3}}...................(1.5) \\
\end{align}\]
Again using the relation (1.3) in (1.5), we obtain
$\sin \left( \theta \right)=\dfrac{-1}{\sqrt{2}}\cos \left( \theta \right)=\dfrac{-1}{\sqrt{2}}\times \left( \pm \sqrt{\dfrac{2}{3}} \right)=\mp \sqrt{\dfrac{1}{3}}.......(1.6)$
Thus, from equations (1.5) and (1.6), we obtain the possible values of sine and cosine of the given angles to be $+\sqrt{\dfrac{2}{3}}\text{ and }-\sqrt{\dfrac{1}{3}}\text{ }$ or $-\sqrt{\dfrac{2}{3}}\text{ and +}\sqrt{\dfrac{1}{3}}\text{ }$ respectively which are the values we wanted to find out in the question.

Note: We should note that in the question, even though the value of $\tan \left( \theta \right)$ was given, we did not have to find out theta as we were only required to find out the values of $\sin \left( \theta \right)$ and $\cos \left( \theta \right)$ which could be obtained from $\tan \left( \theta \right)$ without explicitly solving for $\theta $.