The angle $\theta$ lies in Quadrant IV with point $P$ on the terminal arm and $\tan\theta=-\dfrac{3}{5}$? My friend explained that.

I'm not sure if he is correct. In Quadrant IV the $\sin\theta$ and $\tan\theta$ are going to be negative numbers, from your tangent ratio of $-\dfrac{3}{5}$, the hypotenuse will be the $\sqrt{34}$. Therefore your $\sin\theta$ will be $-\dfrac{3}{\sqrt{34}}$, or more correctly written as $-\dfrac{3\sqrt{34}}{34}$, since you don't want a radical in the denominator.

I need to show full work for this type of question I don't think this cuts it...


1 Answer 1


Your friend was "half-way" right:

$(1)$ First finishing on your work, for $\,\tan\theta = \left(-\dfrac{3}{5}\right),\;$ if $\,\sin\theta < 0,\,$ we'd also need

$$\cos\theta = \dfrac{5}{\sqrt{34}} = \dfrac{5 \sqrt{34}}{34} > 0.\;$$ This makes sense since

$$\;\tan\theta = \dfrac{\sin\theta}{\cos\theta} = \dfrac{\frac{-3\sqrt{34}}{34}}{\frac{5\sqrt{34}}{34}} = \left(-\dfrac 35\right). $$

This, together with your work, gives that $P$ would then be in the fourth quadrant.

$(2)$ But... $P$ could also be located in Quadrant II: Since $\;\tan\theta = \dfrac{\sin\theta}{\cos\theta} = \left(-\dfrac 35\right)\lt 0,\;$ then exactly one of $\cos\theta, \sin \theta\,$ must be negative. $(1)$ gives one possible way this can happen.

But we might also have that $\cos \theta \lt 0, \sin\theta > 0$, putting $P$ in Quadrant II.

$$\;\tan\theta = \dfrac{\sin\theta}{\cos\theta} = \dfrac{\frac{3\sqrt{34}}{34}}{\frac{-5\sqrt{34}}{34}} = \left(-\dfrac 35\right).$$

Recall that $\,\sin\theta\,$ corresponds to the $\,y$-coordinate of $\,P\,$ in a unit circle, and $\,\cos \theta\,$ with its $\,x$-coordinate. When $\,x > 0\,$ and $\,y< 0,\,$ $\,(x, y)\,$ is in the Fourth Quadrant; when $\,x \lt 0,\;\text{and}\; y\gt 0,\;$ $(x, y)\,$ is in the Second Quadrant.

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    – amWhy
    Jun 20, 2013 at 18:03

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