General solutions

Excerpt: The first question asks students what they can say about $A$ and $B$ if $\tan A=\tan B$ . This resource offers interactive graphs to help students find general solutions of trigonometric equations and understand the relationship between solutions.

Problem

If $\tan A=\tan B$ , what can we say about $A$ and $B$ ?

What about the corresponding questions for sine and cosine?

If $\cos A=\cos B$ , what can we say about $A$ and $B$ ?
If $\sin A=\sin B$ , what can we say about $A$ and $B$ ?

Suggestion: Think about the graphs of the trigonometric functions!

Interactive graphs and solution

Here is a graph of $y=\tan x$ , showing the solutions of $\tan x=a$ .

You can slide the point on the $y$ -axis to change the value of $a$ .

[interactive embed (geogebra-iframe): //www.geogebratube.org/material/iframe/id/wpNxQgP6/width/590/height/436/border/888888/rc/false/ai/false/sdz/false/smb/false/stb/false/stbh/false/ld/false/sri/true/asb/false/at/html5] — not part of the text import; see the source page.

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How does this help you answer the question?

Solution

We can see from the graph that $y=\tan x$ has a period of $\pi$ . This means that $\tan(x+n\pi) = \tan x$ for all integers $n$ .

Therefore $\tan A=\tan B=a$ if and only if $A = B + n\pi$ for some integer $n$ . In other words, $A$ and $B$ can only differ by an integer multiple of $\pi$ .

If, for example, $\tan A =\tan B = 1$ , then $A = \dfrac{\pi}{4}+n\pi$ for some integer $n$ . We’d also have $B = \dfrac{\pi}{4}+m\pi$ for some integer $m$ , and so $A = B + k\pi\quad \text{where }k\text{ is an integer.}$

More generally, if $\tan A=\tan B = a$ , then $A = \arctan a + n\pi$ and $B = \arctan a + m\pi$ , where $n$ and $m$ are integers, so $A = B+ k\pi$ where $k$ is an integer.

Please note that $\arctan$ and $\tan^{-1}$ mean the same thing and both are widely used in books and online. Similarly, you may see $\arcsin$ or $\sin^{-1}$ and $\arccos$ or $\cos^{-1}$ .

Here is a graph of $y=\cos x$ , showing the solutions of $\cos x=a$ .

[interactive embed (geogebra-iframe): //www.geogebratube.org/material/iframe/id/KnhSdbEH/width/590/height/436/border/888888/rc/false/ai/false/sdz/false/smb/false/stb/false/stbh/false/ld/false/sri/true/asb/false/at/html5] — not part of the text import; see the source page.

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Solution

We can see from the graph that $y=\cos x$ has a period of $2\pi$ . This means that $\cos(x+2n\pi) = \cos x$ for all integers $n$ .

Looking at the graph above, solutions to $\cos x=a$ fall into two types (what we might call the ‘left-hand solutions’ and the ‘right-hand solutions’). For example, if $a = \dfrac{1}{2}$ , then we can call $\dfrac{\pi}{3}$ a right-hand solution, and $-\dfrac{\pi}{3}$ a left-hand solution.

This means that any solution to $\cos x = a$ is of the form $\text{(a right-hand solution)}+ 2n\pi \quad \text{or} \quad \text{(a left-hand solution)} + 2n\pi \quad \text{for some integer }n\text{.}$

For example, if $\cos x= \dfrac{1}{2}$ , then $x = \dfrac{\pi}{3} + 2n\pi$ or $x = -\dfrac{\pi}{3} + 2n\pi$ for some integer $n$ .

Therefore, if $\cos A = \cos B = a$ , then $A = \arccos a + 2n\pi \quad \text{or} \quad A = -\arccos a + 2n\pi,$ and $B = \arccos a + 2m\pi \quad \text{or} \quad B = -\arccos a + 2m\pi,$

where $m$ and $n$ are integers. This means that $A = B + 2k\pi$ or $A = -B + 2k\pi$ , where $k$ is an integer.

Here is a graph of $y=\sin x$ , showing the solutions of $\sin x=a$ .

[interactive embed (geogebra-iframe): //www.geogebratube.org/material/iframe/id/T3bGC4wb/width/590/height/436/border/888888/rc/false/ai/false/sdz/false/smb/false/stb/false/stbh/false/ld/false/sri/true/asb/false/at/html5] — not part of the text import; see the source page.

Download GeoGebra file

Solution

The period of $\sin x$ is also $2\pi$ and again the solutions to $\sin x=a$ fall into two types. For example, if $a = \tfrac{1}{2}$ , then we can call $\tfrac{\pi}{6}$ a left-hand solution and $\pi -\tfrac{\pi}{6}$ a right-hand solution. Any solution to $\sin x = a$ is therefore of the form $\text{(a right-hand solution)}+ 2n\pi \quad \text{or} \quad \text{(a left-hand solution)} + 2n\pi \quad\text{for some integer }n\text{.}$

If, for example, $\sin x = \dfrac{1}{2}$ , then $A = \dfrac{\pi}{6} + 2n\pi$ or $A = \pi-\dfrac{\pi}{6} + 2n\pi$ for some integer $n$ .

If $\sin A = \sin B = a$ , then we can say $A$ and $B$ have the following forms $A = \arcsin a + 2n\pi \quad \text{or} \quad A = \pi -\arcsin a + 2n\pi,\quad \text{where }n\text{ is an integer,}$ $B = \arcsin a + 2m\pi \quad \text{or} \quad B = \pi -\arcsin a + 2m\pi \quad \text{where }m\text{ is an integer.}$

This means $A = B + 2k\pi$ or $A = \pi-B + 2k\pi$ , where $k$ is an integer.

We can think of $y=\sin x$ as a translation of $y=\cos x$ , by $\tfrac{\pi}{2}$ in the positive $x$ direction. The symmetry of $y=\cos x$ about the $y$ -axis then means $y=\sin x$ is symmetric about $x = \tfrac{\pi}{2}$ , and the solutions for $\sin x = a$ are symmetrical about this line. Therefore we could write the solutions of $\sin x=a$ as $x = \left(\arcsin a - \dfrac{\pi}{2}\right) + 2n\pi + \dfrac{\pi}{2} \quad \text{or} \quad x = \left(\dfrac{\pi}{2} - \arcsin a\right) + 2n\pi + \dfrac{\pi}{2}.$