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Question: If \(\sin \theta +\cos \theta =0\) and \(\theta \) lies in the fourth quadrant, then find the values...

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

Explanation

Solution

Hint: Use the fact that in the fourth quadrant sin⁡x\sin x is negative while cos⁡x\cos x is positive. Divide the equation by cos⁡θ\cos \theta and hence find the value of tan⁡θ\tan \theta . Use the fact that if tan⁡θ=a\tan \theta =a, then sin⁡θ=±a1+a2\sin \theta =\pm \dfrac{a}{\sqrt{1+{{a}^{2}}}} and cos⁡θ=±11+a2\cos \theta =\pm \dfrac{1}{\sqrt{1+{{a}^{2}}}}. Hence find the values of sin⁡θ\sin \theta and cos⁡θ\cos \theta .

Complete step-by-step answer:
We have sin⁡θ+cos⁡θ=0\sin \theta +\cos \theta =0
Dividing both sides of the equation by cos⁡θ\cos \theta , we get
sin⁡θcos⁡θ+cos⁡θcos⁡θ=0\dfrac{\sin \theta }{\cos \theta }+\dfrac{\cos \theta }{\cos \theta }=0
We know that sin⁡θcos⁡θ=tan⁡θ\dfrac{\sin \theta }{\cos \theta }=\tan \theta .
Hence, we have tan⁡θ+1=0\tan \theta +1=0
Subtracting 1 from both sides of the equation, we get
tan⁡θ=−1\tan \theta =-1
Now, we know that if tan⁡θ=a\tan \theta =a, then sin⁡θ=±a1+a2\sin \theta =\pm \dfrac{a}{\sqrt{1+{{a}^{2}}}} and cos⁡θ=±11+a2\cos \theta =\pm \dfrac{1}{\sqrt{1+{{a}^{2}}}}. Hence find the values of sin⁡θ\sin \theta and cos⁡θ\cos \theta .
Hence, we have sin⁡θ=±−11+(−1)2=±12\sin \theta =\pm \dfrac{-1}{\sqrt{1+{{\left( -1 \right)}^{2}}}}=\pm \dfrac{1}{\sqrt{2}} and cos⁡θ=±11+(−1)2=±12\cos \theta =\pm \dfrac{1}{\sqrt{1+{{\left( -1 \right)}^{2}}}}=\pm \dfrac{1}{\sqrt{2}}.
Now since θ\theta is in the fourth quadrant, we have cos⁡θ\cos \theta is positive and sin⁡θ\sin \theta is negative.
Hence, we have cos⁡θ=12\cos \theta =\dfrac{1}{\sqrt{2}} and sin⁡θ=−12\sin \theta =\dfrac{-1}{\sqrt{2}}.

Note: Alternative solution:
We have sin⁡θ+cos⁡θ=0 (i)\sin \theta +\cos \theta =0\text{ (i)}
Subtracting cos⁡θ\cos \theta from both sides, we get
sin⁡θ=−cos⁡θ\sin \theta =-\cos \theta
Squaring both sides of the equation, we get
sin⁡2θ=cos⁡2θ{{\sin }^{2}}\theta ={{\cos }^{2}}\theta
We know that cos⁡2θ=1−sin⁡2θ{{\cos }^{2}}\theta =1-{{\sin }^{2}}\theta
Hence, we have
sin⁡2θ=1−sin⁡2θ{{\sin }^{2}}\theta =1-{{\sin }^{2}}\theta
Adding sin⁡2θ{{\sin }^{2}}\theta on both sides of the equation, we get
2sin⁡2θ=12{{\sin }^{2}}\theta =1
Dividing both sides of the equation by 2, we get
sin⁡2θ=12{{\sin }^{2}}\theta =\dfrac{1}{2}
Hence, we have
sin⁡θ=±12\sin \theta =\pm \dfrac{1}{\sqrt{2}}
Since θ\theta lies in the fourth quadrant, we have sin⁡θ\sin \theta is negative. Hence, we have sin⁡θ=−12\sin \theta =\dfrac{-1}{\sqrt{2}}.
Substituting the value of sin⁡θ\sin \theta in equation (i), we get
−12+cos⁡θ=0\dfrac{-1}{\sqrt{2}}+\cos \theta =0
Adding 12\dfrac{1}{\sqrt{2}} on both sides of the equation, we get
cos⁡θ=12\cos \theta =\dfrac{1}{\sqrt{2}}
Hence, we have sin⁡θ=−12\sin \theta =\dfrac{-1}{\sqrt{2}} and cos⁡θ=12\cos \theta =\dfrac{1}{\sqrt{2}}