Finding all the values of $\theta$ for which $\tan(\theta)=\sqrt3$; problem with understanding.

Question

My textbook has a section where it says a possible way that $\tan(\theta)$ can be thought of is:

For acute angles $\theta$, $\tan(\theta)$ is the $y$-coordinate of the point on the terminal side of $\theta$ which lies on the vertical line $x=1$, which is tangent to the Unit Circle.

It also provides an illustration which I particularly like as it provides me with some intuition as to where the tangent function is in relation with sine and cosine on the Unit Circle:

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Bearing this in mind I am then asked to find all the possible angles which satisfy the following equation: $\tan(\theta)=\sqrt3$.

I know from memory that $\tan(\theta)=\sqrt3$ for one of the common angles $60^\circ$ or $\frac{\pi}{3}$ radians. All other solutions in quadrant I must be coterminal with this angle so I state simply: $\theta=\frac{\pi}{3}+2\pi k$ where $k$ is an integer.

However the answer states another solution is to be found in the third quadrant that is given by $\theta=\frac{4\pi}{3}+2\pi k$ for integers $k$. And as $\tan(\theta)$ is periodic every $\pi$ radians a simplified formula for all solutions is given by: $\theta=\frac{\pi}{3}+\pi k$ for integers $k$.

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My confusion arises from the above diagram. According to this diagram:

1. It is not possible for $\theta$ to be a quadrant III angle, as the terminal side of $\theta$ will never intersect the vertical line $x=1$, as illustrated by this rudimentary picture below:

2. Furthermore it is not possible for $\tan\theta=\sqrt 3$ in the third quadrant where $\theta=\frac{4\pi}{3}$ as the terminal side of the angle will not be able to intersect $y=\sqrt 3$.

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What have I misunderstood? How is it possible to retain the intuition from this diagram and apply it to angles that are greater than $\frac{\pi}{2}$ radians?

Any answers should be written with the knowledge that I am a precalculus student. Thanks.

Answer 1

You can use the same intuition when $\theta$ lies in other quadrants, drawing the same line.

Answer 2

If your book does not give you the definition $\tan\theta$ for non-acute $\theta$ then there is nothing you can do for non-acute angles. However, according to I. J. Kennedy's answer, your book define $\tan\theta$ as $\tan\theta=\frac{\sin\theta}{\cos\theta}$. So let's take it from there.

Your rule for visualizing $\tan$ is only for acute angles, as the rule states. Therefore you can't use that for $4\pi\over 3$ which is non-acute. But you can still retain your intuition and apply similar geometric rule for other quadrants to solve your example:

First, why this rule works from definition of $\tan$?

From similar triangles $\overset{\Delta}{OAP}$ and $\overset{\Delta}{OBQ}$, we get:

$$ \frac{|OA|}{|OB|}=\frac{|AP|}{|BQ|}\\ \frac{\cos\theta}{1}=\frac{\sin\theta}{|BQ|}\\ |BQ|=\tan\theta $$

If we are do this to third quadrant;

we get:

$$ |BQ|=\frac{\sin\theta^\prime}{\cos\theta^\prime}=\frac{\sin(\theta-\pi)}{\cos(\theta-\pi)}=\frac{-\sin\theta}{-\cos\theta}=\tan\theta $$

From this, we derive:

$$ \color{blue}{\tan\theta}=\frac{\sin\theta^\prime}{\cos\theta^\prime}=\tan\theta^\prime=\color{blue}{\tan(\theta-\pi)}=\tan(\theta-\pi+2\pi)=\color{blue}{\tan(\theta+\pi)} $$

Following this process yields:

$$ \forall k\in\mathbb{Z} \space , \space \tan\theta=\tan(\theta+k\pi) $$

In particular, if $\theta=\frac{\pi}{3}$ then $\tan\left(\frac{\pi}{3}+k\pi\right)=\sqrt3$ for all $k\in\mathbb{Z}$.

Second and fourth quadrants need not checking since sine and cosine have opposite signs there, making tangent negative, therefore not $\sqrt3$.

Answer 3

Your book (College Algebra and Trigonometry) does not actually define $\tan(\theta)$ as you describe. The part about acute angles and the line $x=1$ is an explanation of how the tangent function got its name, not the definition. The book's definition (shown below) is not restricted to acute angles.

Answer 4

Your textbook definition of $\tan$ works for $0\leq\theta<{\pi\over2}$, and with some stretch of the imagination also for $-{\pi\over2}<\theta\leq0$. That's where the "geometric definition" of $\tan$ definitively ends. But within this range we obviously have $$\tan\theta={\sin\theta\over\cos\theta}\qquad\left(-{\pi\over2}<\theta<{\pi\over2}\right)\ .$$ On the other hand it is possible, using your drawing, to extend $\cos$ and $\sin$ to all of ${\mathbb R}$, whereby the known properties of these functions are preserved. It is then natural to extend also the definition of $\tan$ to a domain as large as possible by putting $$\tan\theta:={\sin\theta\over\cos\theta}$$ whenever $\cos\theta\ne0$, and leaving $\tan$ undefined when $\cos\theta=0$, i.e., at the odd multiples of ${\pi\over2}$. Since $$\cos(\theta+\pi)=-\cos\theta,\quad \sin(\theta+\pi)=-\sin\theta\qquad\forall\theta\in{\mathbb R}$$ it immediately follows that $\tan$ has period $\pi$ on its domain of definition. Furthermore $\tan$ is odd, and is strictly monotonically increasing on the interval $\bigl[0,{\pi\over2}\bigr[\ $, because $\sin$ is increasing and $\cos$ is decreasing there. Therefore $\tan$ maps the interval $\bigl[0,{\pi\over2}\bigr[\ $ bijectively onto ${\mathbb R}_{\geq0}$ (as we all know), and by symmetry it follows that $$\tan:\quad I:=\left]{-{\pi\over2}},{\pi\over2}\right[\ \to\ {\mathbb R}$$ is bijective. As a consequence there is exactly one $\theta\in I$ with $\tan\theta=\sqrt{3}$, namely $\theta={\pi\over3}$. The $\pi$-periodicity of $\tan$ then implies that the solution set $S$ of the equation $$\tan \theta=\sqrt{3},\quad\theta\in{\mathbb R},$$ is given by $$S=\left\{{\pi\over3}+k\pi\>\biggm|\>k\in{\mathbb Z}\right\}\ .$$

Answer 5

First, I assume you mean $\sqrt{3}$, not $3$, throughout, since $\tan 60^{\circ} = \sqrt{3}$.

Second, note that your book says "For acute angles $\theta$, ...". I bet that it continues with a discussion of what $\tan\theta$ means when $\theta$ is not an acute angle. And in fact $\tan\frac{4\pi}{3} = \tan\frac{\pi}{3} = \sqrt{3}$.

Answer 6

$\tan\theta$ is basically defined as a ratio of two line segment lengths along y-axis and x-axis respy.

$$\tan\theta={ y\over\ x }$$

The line line segments can be signed. i.e., can carry a sign.

For x moving to right or left it is plus or minus by vector direction or sense.

Likewise for y moving up or down is plus or minus by vector direction or sense.

$ \theta = \pi/6$,

$$\tan\theta={BQ\over \ OP }= +\sqrt{3}$$ in first quadrant.

$$\tan\theta={- BQ\over \ - OP }= +\sqrt{3}$$ in third quadrant exactly same, due to the ratio definition.

In case of sin or cos the denominator radius vector is unsigned, and is by convention always positive. So sign is positive in 1,2 quadrants and cos in 1,4 quadrants.

Answer 7

i can't think of a better way to see $\tan(\theta),$ but you can interpret $-1/tan(\theta)$ geometrically as the slope of the tangent at the terminal point $P(\cos \theta, \sin \theta).$ this interpretation should work for $\theta$ in any quadrant; what is more it nicely reproduces symmetric properties of $\tan()$ function. for example $\pi$- periodicity and oddness.