0
$\begingroup$

Let $x = a\sec\theta$ where $a>0$, $0<\theta<\frac{\pi}{2}$ or $\pi < \theta < \frac{3\pi}{2}$. Find the value of $\tan\theta$.

My attempt:

I want to use a geometric method to solve this question since I know how to do it via trigonometric identities. I use the following definition provided by my book:

For a general angle $\theta$, we let $P(x,y)$ be any point on the terminal side of $\theta$ and we let $r$ be the distance $|OP|$, then we define:

enter image description here

So we know $\cos\theta = \frac{a}{x}$ and for $0<\theta<\frac{\pi}{2}$, $x=a\sec\theta$ is clearly positive. So, using my book's definition, the point $P(x,y) = P(a, \sqrt{x^2-a^2})$, so $\tan\theta = \frac{\sqrt{x^2-a^2}}{a}$.

Now where I have the problem is for $\pi < \theta < \frac{3\pi}{2}$. When $\pi < \theta < \frac{3\pi}{2}$, $x = a\sec\theta<0$ and $a>0$ as per the question. But, according to my book's definition, we should be in the third quadrant for $\pi < \theta < \frac{3\pi}{2}$, but $\cos\theta=\frac{a}{x}$ doesn't make sense in the third quadrant because if you draw a right angle triangle in the third quadrant, $a$ (the base of the triangle) should be negative (i.e., have a negative $x$-coordinate) and $x$ (the hypotenuse) should be positive, but we have $a>0$ and $x<0$. Could someone clear up my understanding?

$\endgroup$
  • $\begingroup$ Which software/website have you used make that figure? $\endgroup$ – sgrmshrsm7 Mar 17 '17 at 20:21
1
$\begingroup$

Things are a little confusing since $x$ is both a coordinate in the figure you're using to define the $\sec$ function and also a variable in the problem you're trying to solve. So I'll write $t$ for the horizontal coordinate in the figure and formulas for the trigonometric functions, and therefore $$ \sec \theta = \frac rt \quad\text{and}\quad \tan \theta = \frac yt. $$

Now it appears that what you have done is, you have observed that when $0<\theta<\frac\pi2,$ so that the ray at angle $\theta$ lies in the first quadrant, there is a right triangle with leg $a$ adjacent to an angle of measure $\theta$ that "fits" perfectly between the ray and the horizontal axis, and the length of the hypotenuse of that triangle is $x.$

Now you expect to find a triangle whose side lengths you can similarly identify that fits in the same way between the horizontal axis and the ray at angle $\theta$ when $\pi < \theta < \frac{3\pi}{2}.$ There is, in fact, a suitable right triangle that has a hypotenuse of length $\lvert x \rvert > 0.$ There is just not a triangle whose hypotenuse you can naïvely label $x,$ expecting that label to makes sense as a length.

The variables $t$ and $y$ that show up in the definition of the trigonometric functions for general angles are not lengths of the sides of the triangle; they are coordinates of the other end of the hypotenuse, away from the origin. Unlike the sides of a triangle, these coordinates can be negative.

When $\pi < \theta < \frac{3\pi}{2},$ both $t < 0$ and $y < 0,$ since $(t,y)$ is the coordinates of a point in the third quadrant. But we always have $r > 0,$ that is, $r$ is actually the distance between the points $(0,0)$ and $(t,y)$. Therefore $\sec\theta = \frac rt < 0,$ and therefore $x = a\sec\theta < 0$ since $a > 0.$

You can, if you like, construct a right triangle in the third quadrant with vertices at $(0,0),$ $(-a,0),$ $(-a,y),$ choosing $y$ so that the point $(-a,y)$ is on the ray at angle $\theta.$ Then the leg of that triangle along the horizontal axis will have length $a$ (the distance between $(0,0)$ and $(-a,0)$; observe that it is perfectly sensible to have a positive distance to a point that happens to have a negative coordinate). The hypotenuse of that triangle will have length $-x.$ The length is not $x$; the definition of $\sec\theta$ for a general angle $\theta$ never says that $a\sec\theta$ will be a length when $a$ is a length.

That's about as close as you can get to an accurate labeling of a triangle in the third quadrant comparable to the triangle you can construct in the first quadrant.

Yes, it can be confusing to know when the hypotenuse of a triangle should be said to have length $x$ and when it should have length $-x.$ I think that is a big part of the reason why the figure and formulas you found define the trigonometric functions of a general angle in terms of Cartesian coordinates instead of drawing a triangle.

$\endgroup$
1
$\begingroup$

Rewritng $-x = (-a)\sec \theta$ should lead you in the direction you want.

However, it is a bit confusing to write $P(x,y)$ when the letter $x$ already stands for something other than the first coordinate of $P$, so rather than write $P(x,y)$ instead I'll write $P(\tilde{a},\tilde{b})$ for the point on the terminal side of the angle $\theta$.

Fix $\pi < \theta < \frac{3\pi}{2}$. Clearly you should have that $\tilde{a},\tilde{b} < 0$.

What if we allowed $\tilde{a} := -a < 0$? Given that our angle $\theta$ is fixed, picking $\tilde{a}$ should determine both $\tilde{b},r$ uniquely. Note that $x = a \sec \theta < 0$ implies that $$ -x = (-a) \sec \theta > 0 \quad \mathrm{and} \quad \frac{\tilde{a}}{-x} = \frac{-a}{-x} = \cos \theta = \frac{\tilde{a}}{r}.$$

Therefore $r = -x > 0$, and furthermore, $\tilde{b} = -\sqrt{r^2 - \tilde{a}^2} = - \sqrt{x^2 - a^2}<0$ (you take the negative square root because $P$ is in the third quadrant).

Thus we conclude $$ \tan \theta = \frac{\tilde{b}}{\tilde{a}} = \frac{-\sqrt{x^2 - a^2}}{-a} = \frac{\sqrt{x^2 - a^2}}{a}$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.