Question

If $\sin \theta +\cos \theta =0$ and $\theta $ lies in the fourth quadrant, then find the values of $\sin \theta $ and $\cos \theta $.

Answer 1

Hint: Use the fact that in the fourth quadrant $\sin x$ is negative while $\cos x$ is positive. Divide the equation by $\cos \theta $ and hence find the value of $\tan \theta $. Use the fact that if $\tan \theta =a$, then $\sin \theta =\pm \dfrac{a}{\sqrt{1+{{a}^{2}}}}$ and $\cos \theta =\pm \dfrac{1}{\sqrt{1+{{a}^{2}}}}$. Hence find the values of $\sin \theta $ and $\cos \theta $.

Complete step-by-step answer:
We have $\sin \theta +\cos \theta =0$
Dividing both sides of the equation by $\cos \theta $, we get
$\dfrac{\sin \theta }{\cos \theta }+\dfrac{\cos \theta }{\cos \theta }=0$
We know that $\dfrac{\sin \theta }{\cos \theta }=\tan \theta $.
Hence, we have $\tan \theta +1=0$
Subtracting 1 from both sides of the equation, we get
$\tan \theta =-1$
Now, we know that if $\tan \theta =a$, then $\sin \theta =\pm \dfrac{a}{\sqrt{1+{{a}^{2}}}}$ and $\cos \theta =\pm \dfrac{1}{\sqrt{1+{{a}^{2}}}}$. Hence find the values of $\sin \theta $ and $\cos \theta $.
Hence, we have $\sin \theta =\pm \dfrac{-1}{\sqrt{1+{{\left( -1 \right)}^{2}}}}=\pm \dfrac{1}{\sqrt{2}}$ and $\cos \theta =\pm \dfrac{1}{\sqrt{1+{{\left( -1 \right)}^{2}}}}=\pm \dfrac{1}{\sqrt{2}}$.
Now since $\theta $ is in the fourth quadrant, we have $\cos \theta $ is positive and $\sin \theta $ is negative.
Hence, we have $\cos \theta =\dfrac{1}{\sqrt{2}}$ and $\sin \theta =\dfrac{-1}{\sqrt{2}}$.

Note: Alternative solution:
We have $\sin \theta +\cos \theta =0\text{ (i)}$
Subtracting $\cos \theta $ from both sides, we get
$\sin \theta =-\cos \theta $
Squaring both sides of the equation, we get
${{\sin }^{2}}\theta ={{\cos }^{2}}\theta $
We know that ${{\cos }^{2}}\theta =1-{{\sin }^{2}}\theta $
Hence, we have
${{\sin }^{2}}\theta =1-{{\sin }^{2}}\theta $
Adding ${{\sin }^{2}}\theta $ on both sides of the equation, we get
$2{{\sin }^{2}}\theta =1$
Dividing both sides of the equation by 2, we get
${{\sin }^{2}}\theta =\dfrac{1}{2}$
Hence, we have
$\sin \theta =\pm \dfrac{1}{\sqrt{2}}$
Since $\theta $ lies in the fourth quadrant, we have $\sin \theta $ is negative. Hence, we have $\sin \theta =\dfrac{-1}{\sqrt{2}}$.
Substituting the value of $\sin \theta $ in equation (i), we get
$\dfrac{-1}{\sqrt{2}}+\cos \theta =0$
Adding $\dfrac{1}{\sqrt{2}}$ on both sides of the equation, we get
$\cos \theta =\dfrac{1}{\sqrt{2}}$
Hence, we have $\sin \theta =\dfrac{-1}{\sqrt{2}}$ and $\cos \theta =\dfrac{1}{\sqrt{2}}$