# Solving trigonometrics functions/equations

My niece ask me to help her with a school assignment, but I can't identify what type of equation are we solving.

For example: $\DeclareMathOperator{\tg}{tg}$ $$\sin\alpha=\frac{8\sqrt{11}}{9}$$

or:

$$\tan\alpha=\frac{2\sqrt{5}}{12}$$

The notes she have are these, for example:

$$\sin\alpha=\frac{1}{\csc\alpha}$$

and:

$$\tan\alpha=\frac{\sin\alpha}{\cos\alpha}$$

I'm trying to browse for help, but I don't know wath to search for. Where to begin.

I was able to talk with a classmate of hers, she explain me this:

In each problem, we need to use all the needed functions (Right angled triangle definitions).

In the first problem:

$$\sin\alpha=\frac{8\sqrt{11}}{9}$$

We need to solve the $\cos$ then $\sec$ then $\csc$ then $\tan$ then $\cot$. Does this makes any sense?

• Here is a hint for one of them. $-1 \le \sin(x) \le 1$. Is $\frac{ 8 \sqrt{11}}{9}$ in that interval ? Apr 10 '15 at 0:36
• For the first equation, $\frac{8\sqrt{11}}{9}$ is greater than $1$, so there is no real answer. Apr 10 '15 at 0:36
• By the way $\sin(x)=\frac{1}{\csc(x)} \neq \frac{1}{\cos(x)}$ Apr 10 '15 at 0:40
• These seem kinda like calculator questions to me (well the second one does). tutorial.math.lamar.edu/Classes/CalcI/TrigEquations_CalcI.aspx I really like paul's notes. Apr 10 '15 at 0:50
• And I mean you can really use a calculator for both, but the calculator will tell you something like error or whatever it says for the $\sin^{-1}(\frac{8 \sqrt{11}}{9})$ . I forgot exactly what those handheld thingys say. (Haven't owned one in ages. :p) Apr 10 '15 at 1:01

The first two are a slightly different kind of question than the bottom two. The first are basically asking "is (value on the right) anywhere in the range of (trig function on the left)?"

For $$\sin{\alpha} = \frac{8\sqrt{11}}{9}$$ there is no $\alpha$ that satisfies the equation since the range of $\sin$ is $[-1, 1]$, and the right-hand value is bigger than that.

For $$\tan{\alpha} = \frac{2\sqrt{5}}{12}$$ there is an answer, since $\tan$'s range is (almost) everything. The answer will equal $\tan^{-1}(\frac{2\sqrt{5}}{12})$.

The latter two you can view as being about whether the graphs of certain trig functions cross each other or not. Actually, though $$\sin{\alpha} = \frac{1}{\cos{\alpha}}$$ is amenable to a range argument as well. The range of $\cos{\alpha}$ is $[-1, 1]$; what range do we get if we take the reciprocal of every number in there? How does it compare to the range of $\sin{\alpha}$? (They do both attain $1$ and $-1$ -- you have to show that they don't intersect there, which shouldn't be too hard).

As for the last problem, remember that $\frac{\sin{\alpha}}{\cos{\alpha}} = \tan{\alpha}$. So

\begin{aligned} \tan{\alpha} &= \sqrt{\frac{(\sin{\alpha})^2}{(\cos{\alpha})^2}}\\ &= \sqrt{\left(\frac{\sin{\alpha}}{\cos{\alpha}}\right)^2}\\ &= \sqrt{(\tan{\alpha})^2}\\ &= |\tan{\alpha}|\\ \end{aligned}

(remember the absolute value). So the problem is just asking for what $\alpha$ it's true that $\tan{\alpha} = |\tan{\alpha}|$.

• Thanks! The two first were the actual problems, the other two were trigonometric functions for the sin and the tan of an angle. I'm going to update my question Apr 10 '15 at 1:43
• Okay. In that case, the notes are a bit wrong (if they weren't, the solution for the last two would be $\alpha = \text{anything}$.) $\sin{\alpha}$ just does not equal $\frac{1}{\cos{\alpha}}$, I think $\frac{1}{\csc{\alpha}}$ was probably meant. The last one is right except for that absolute value (because when you square and take the square root, the result can only be positive). Apr 10 '15 at 1:46
• You're totally right. They were typos. =( I've fixed them now Apr 10 '15 at 2:09