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When I'm solving trigonometric equations, I occasionally end up with a sine or cosine that's greater than $1$ -- and not on the unit circle. For example, today I had one that was $3 \tan 3x = \sqrt{3}$, which simplified to $\tan 3x = \frac{\sqrt{3}}{3}$, which simplified to $\sin 3x = \sqrt{3}$ and $\cos 3x = 3$. So far as I know, I can't divide this by $3$ (to isolate $x$) until I get angles with a sine of $\sqrt{3}$ and a cosine of $3$, respectively. So, how do I reduce this cosine so that I can find an angle on the unit circle with that cosine? Or, am I doing something very wrong to get this as a cosine in the first place?

Thanks!

evamvid

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    $\begingroup$ $\tan \varphi = \frac{a}{b}$ does not simplify to $\sin\varphi = a$ and $\cos \varphi = b$ in general. $\endgroup$
    – Daniel Fischer
    Jan 18, 2014 at 23:23
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    $\begingroup$ You are doing something very wrong to get that as a cosine in the first place. Note that $\dfrac a b=\dfrac c d$ doesn't imply at all that $a=c$ and $b=d$. $\endgroup$
    – Git Gud
    Jan 18, 2014 at 23:23
  • $\begingroup$ So how do I go about doing tangents? Incidentally, this was the first tangent problem on my homework -- I'm glad I asked and didn't do any more problems like this wrong! $\endgroup$
    – evamvid
    Jan 18, 2014 at 23:25
  • $\begingroup$ Presumably, you have figured out the coordinates of the "special" points on the unit circle $\tfrac{\pi}{6}$, $\tfrac{\pi}{4}$, etc. You ought to look at the tangent ratios in those particular locations. $\endgroup$
    – Sammy Black
    Jan 18, 2014 at 23:29
  • $\begingroup$ sin(x) and cos(x) are between -1 and +1 but tan(x) is between minus infinity and plus infinity. $\endgroup$
    – Claude Leibovici
    Jan 19, 2014 at 6:11

3 Answers 3

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As pointed out in the comments, you are doing something like the following: $$\frac{2}{4} = \frac{1}{2} \qquad \Rightarrow \qquad 2=1\quad \mbox{ and } \quad 4=2.$$ This is very incorrect.

When we are at $\tan 3x = \frac{\sqrt{3}}{3}$, the correct thing to do is find $y$ such that $\tan y = \frac{\sqrt{3}}{3}$ and then let $x=y/3$. On your calculator, you would compute $\arctan(\sqrt{3}/3)$ and set $y=\arctan(\sqrt{3}/3)+\pi n$ for $n=0, \pm 1, \pm 2, \dots$. Remember, there are multiple solutions because $\tan y$ has period $\pi$.

What if you don't want to use a calculator? Note that $\sqrt{3}/3$ is $1/\sqrt{3}$. Now think about a 30-60-90 triangle. Hint: Think about the 30 degree, or $\pi/6$ angle.

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    $\begingroup$ ...except that you should not need a calculator for this particular value of the tangent function. $\endgroup$
    – Sammy Black
    Jan 18, 2014 at 23:30
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    $\begingroup$ Yup, just edited that in, before I saw your comment. $\endgroup$
    – abnry
    Jan 18, 2014 at 23:30
  • $\begingroup$ Awesome! Thanks a lot... $\endgroup$
    – evamvid
    Jan 18, 2014 at 23:33
  • $\begingroup$ No problem, my pleasure! $\endgroup$
    – abnry
    Jan 18, 2014 at 23:34
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    $\begingroup$ @evamvid That was a typo--he meant $\frac{\sqrt{3}}{3} = \frac{1}{\sqrt{3}}$ not $\frac{3}{\sqrt{3}}$. $\endgroup$
    – apnorton
    Jan 19, 2014 at 0:00
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$$ \tan(3x) = \frac{\sqrt{3}}{3} = \frac{1}{\sqrt{3}} $$ $$ \frac{\sin(3x)}{\cos(3x)} = \frac{1}{\sqrt{3}} $$ $$ \sqrt{3}\sin(3x) = \cos(3x) $$ $$ 1 = (\sin (3x))^2 + (\cos (3x))^2 = (\sin (3x))^2 + (\sqrt{3}\sin(3x))^2 $$ $$ 1 = (\sin (3x))^2 + 3(\sin (3x))^2 = 4(\sin(3x))^2 $$ $$ (\sin(3x))^2 = \frac 1 4 $$ $$ \sin(3x) = \pm\frac 1 2 $$ $$ \cos(3x) = \pm\frac{\sqrt{3}}{2} $$ (If one of them is "$+$" then so is the other; otherwise they're both "$-$".)

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  • $\begingroup$ Thanks for another way to solve (and for helping me out!) $\endgroup$
    – evamvid
    Jan 18, 2014 at 23:44
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Hint: If you are solving $\tan x = a/b$, you have to realize $a/b$ can be written as $(a/c)/(b/c)$ for any nonzero $c$. You have to pick a value of $c$ before you can try to do that. Look at $c= \pm \sqrt{a^2+b^2}$ and see if either of those values gets you further.

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