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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?

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    $\begingroup$ Here is a hint for one of them. $-1 \le \sin(x) \le 1$. Is $\frac{ 8 \sqrt{11}}{9}$ in that interval ? $\endgroup$
    – randomgirl
    Apr 10 '15 at 0:36
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    $\begingroup$ For the first equation, $\frac{8\sqrt{11}}{9}$ is greater than $1$, so there is no real answer. $\endgroup$ Apr 10 '15 at 0:36
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    $\begingroup$ By the way $\sin(x)=\frac{1}{\csc(x)} \neq \frac{1}{\cos(x)}$ $\endgroup$
    – randomgirl
    Apr 10 '15 at 0:40
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    $\begingroup$ 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. $\endgroup$
    – randomgirl
    Apr 10 '15 at 0:50
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    $\begingroup$ 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) $\endgroup$
    – randomgirl
    Apr 10 '15 at 1:01
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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 $\tan{\alpha} = |\tan{\alpha}|$.

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  • $\begingroup$ 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 $\endgroup$ Apr 10 '15 at 1:43
  • $\begingroup$ 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). $\endgroup$
    – Eli Rose
    Apr 10 '15 at 1:46
  • $\begingroup$ You're totally right. They were typos. =( I've fixed them now $\endgroup$ Apr 10 '15 at 2:09

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