Helpjebal
  • Helpjebal
tan (arcsin (2/5)) find the exact value. Not sure how to go about this... can anyone point me in the right direction?
Mathematics
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SOLVED
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chestercat
  • chestercat
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zepdrix
  • zepdrix
\[\rm \tan\left[\color{orangered}{\arcsin\left(\frac{2}{5}\right)}\right]\]Let's examine the inside first.\[\rm \color{orangered}{\arcsin\left(\frac{2}{5}\right)}\]
zepdrix
  • zepdrix
Recall that when you take the `sine` of an `angle`, you end up with the `ratio` of a side to another side, namely, opposite to hypotenuse. So when we take the `inverse sine` of a `ratio`, you end up with an `angle`.\[\rm \color{orangered}{\arcsin\left(\frac{2}{5}\right)=\theta}\]
zepdrix
  • zepdrix
We can rewrite this in terms of the sine function, instead of the inverse. Do you remember how to do that?

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Helpjebal
  • Helpjebal
no... is that when I'd need he special calculator?
Helpjebal
  • Helpjebal
*the
zepdrix
  • zepdrix
Here is one approach: Applying sine function to each side gives us:\[\large\rm \sin\left(\arcsin\left(\frac{2}{5}\right)\right)=\sin(\theta)\]Since the left side is the composition of a function and it's inverse, we simply get the argument as a result,\[\large\rm \frac{2}{5}=\sin(\theta)\] If that process is confusing, a simpler way to think about it is... When switching from inverse sine to sine, the two things switch places.\[\rm \color{orangered}{\arcsin\left(\frac{2}{5}\right)=\theta}\qquad\to\qquad \sin (\theta)=\frac{2}{5}\]
zepdrix
  • zepdrix
\[\rm \sin(\theta)=\frac{2}{5}=\frac{opposite}{hypotenuse}\]We want to draw a triangle to show this relationship.
zepdrix
  • zepdrix
|dw:1447217473657:dw|We'll arbitrarily place theta in the bottom left corner here. Do you understand where the 2 and 5 would go in this diagram?
Helpjebal
  • Helpjebal
the 2 would be on the right, vertical straight part of the triangle and the 5 would be on the diagonal part, right?
zepdrix
  • zepdrix
Yes, good. |dw:1447217721081:dw|From there, we'll apply our Pythagorean Theorem to find the missing side length. So what do you get for that length? :) No decimal, leave it as an ugly number if that's what you end up with.
Helpjebal
  • Helpjebal
\[b=\sqrt{21}\]
zepdrix
  • zepdrix
|dw:1447217924455:dw|Cool
zepdrix
  • zepdrix
Ok we've gone pretty deep here. Let's go back and try to remember what we're working on.\[\rm \tan\left[\color{orangered}{\arcsin\left(\frac{2}{5}\right)}\right]\]We determined that the inverse sine of some ratio has to be an angle. So we can actually rewrite this expression as,\[\large\rm \tan\left[\color{orangered}{~\theta~}\right]\]
zepdrix
  • zepdrix
We did that lengthy work to establish a triangle. And now the problem simply boils down to this, using the triangle to find tangent of the angle.
Helpjebal
  • Helpjebal
how do we use the triangle to find the tangent of the angle>
Helpjebal
  • Helpjebal
wait!
Helpjebal
  • Helpjebal
tan = opp/adjacent so that means the exact value is \[2/\sqrt{21}\]
zepdrix
  • zepdrix
Ah very good! There is one more little clean up step. \(\rm \sqrt{21}\) is an irrational number. We don't like to leave irrationals in the denominator of a fraction. Remember how to fix that?
Helpjebal
  • Helpjebal
we multiply \[2/\sqrt{21}\] by \[\sqrt{21}/\sqrt{21}\]
Helpjebal
  • Helpjebal
which is \[2\sqrt{21}/21\]
zepdrix
  • zepdrix
\[\large\rm \tan\left[\arcsin\left(\frac{2}{5}\right)\right]\quad=\quad\frac{2\sqrt{21}}{21}\]Yessss! Good job! :)
Helpjebal
  • Helpjebal
thank you so much! this feels great to know I can use this in the future and be more comfortable with it. Thank you!!!!!
zepdrix
  • zepdrix
np :)

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