Hi Guys,
I have another question:calculate the exact value of the following expression:
$\displaystyle
tan(arcsin(\frac{2}{3}))
$
I don't know how to calculate the exact value of $\displaystyle arcsin(\frac{2}{3})$
Can anyone help?
Hi Guys,
I have another question:calculate the exact value of the following expression:
$\displaystyle
tan(arcsin(\frac{2}{3}))
$
I don't know how to calculate the exact value of $\displaystyle arcsin(\frac{2}{3})$
Can anyone help?
1. Draw a sketch. (see attachment)
2. Use proportions:
$\displaystyle \dfrac{\frac23}1 = \dfrac{y}{\sqrt{1+y^2}}$
3. Since $\displaystyle y = \tan\left(\arcsin\left(\frac23}\right) \right)$ solve the equation at 2. for y.
4. I've got $\displaystyle y = \tan\left(\arcsin\left(\frac23}\right) \right) = \dfrac25 \sqrt{5}$
Hello, spycrab!
Calculate the exact value of: .$\displaystyle \tan\left[\arcsin (\frac{2}{3}\right] $
Let $\displaystyle \theta = \arcsin(\frac{2}{3})\right$
Then: .$\displaystyle \sin\theta \:=\:\frac{2}{3} \:=\:\frac{opp}{hyp}$
That is, $\displaystyle \theta$ is in a right triangle with $\displaystyle opp = 2,\;hyp = 3$
Pythagorus tells us that: .$\displaystyle adj \,=\,\sqrt{5}$
Hence: .$\displaystyle \tan\theta \:=\: \frac{opp}{adj} \:=\:\frac{2}{\sqrt{5}}$
Therefore: .$\displaystyle \tan\left[\arcsin(\frac{2}{3})\right] \;=\;\dfrac{2}{\sqrt{5}}$