How do I deal with sines or cosines greater than $1$?

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

• $\tan \varphi = \frac{a}{b}$ does not simplify to $\sin\varphi = a$ and $\cos \varphi = b$ in general. Jan 18, 2014 at 23:23
• 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 $b=d$. Jan 18, 2014 at 23:23
• 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! Jan 18, 2014 at 23:25
• 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. Jan 18, 2014 at 23:29
• sin(x) and cos(x) are between -1 and +1 but tan(x) is between minus infinity and plus infinity. Jan 19, 2014 at 6:11

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.

• ...except that you should not need a calculator for this particular value of the tangent function. Jan 18, 2014 at 23:30
• Yup, just edited that in, before I saw your comment. Jan 18, 2014 at 23:30
• Awesome! Thanks a lot... Jan 18, 2014 at 23:33
• No problem, my pleasure! Jan 18, 2014 at 23:34
• @evamvid That was a typo--he meant $\frac{\sqrt{3}}{3} = \frac{1}{\sqrt{3}}$ not $\frac{3}{\sqrt{3}}$. Jan 19, 2014 at 0:00

$$\tan(3x) = \frac{\sqrt{3}}{3} = \frac{1}{\sqrt{3}}$$ $$\frac{\sin(3x)}{\cos(3x)} = \frac{1}{\sqrt{3}}$$ $$\sqrt{3}\sin(3x) = \cos(3x)$$ $$1 = (\sin (3x))^2 + (\cos (3x))^2 = (\sin (3x))^2 + (\sqrt{3}\sin(3x))^2$$ $$1 = (\sin (3x))^2 + 3(\sin (3x))^2 = 4(\sin(3x))^2$$ $$(\sin(3x))^2 = \frac 1 4$$ $$\sin(3x) = \pm\frac 1 2$$ $$\cos(3x) = \pm\frac{\sqrt{3}}{2}$$ (If one of them is "$+$" then so is the other; otherwise they're both "$-$".)

• Thanks for another way to solve (and for helping me out!) Jan 18, 2014 at 23:44

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.