geeze this stuff never ends! trig

[she18]

evaluate tan(pi+ arcsin 2/3)

 

[pka]

Once again because of the period, $\displaystyle \tan (\pi + \arcsin (x)) = \tan (\arcsin (x))$.
Now draw the right triangle so that one acute angle has sine equal 2/3.
The tangent of that angle will be 2/[√5].

 

[wjm11]

evaluate tan(pi+ arcsin 2/3)

First evaluate arcsine 2/3 by drawing yourself a picture: a triangle, in the first quadrant (where sine is positive), “opposite side” of length 2 and hypotenuse of length 3. Find the third side by the Pythagorean theorem: sqrt5. Now we add pi to the angle in the picture: extend the hypotenuse line through the origin, into the third quadrant. Draw another, identical triangle, this time in the third quadrant. The legs of the triangle will now be negative: -2 and –sqrt5. The tangent of the angle will, therefore, be 2/sqrt5, or 2*sqrt5/5.

tan(pi+ arcsin 2/3) = 2*sqrt5/5

 

[Gene]

Using the identities
sin^-1(y/x)=tan^-1(y/$\displaystyle sqrt[x^2-y^2]$)
and
tan(u+v) = (tan(u)+tan(v))/(1-tan(a)*tan(v))

tan(pi+ arcsin 2/3) =
tan(pi+ arctan(2/$\displaystyle sqrt[5]$) =
(tan(pi)+tan(tan^-1(2/$\displaystyle sqrt[5]$)/(1-tan(pi)*tan(tan^-1(2/$\displaystyle sqrt[5])$
Since tan(pi) = 0 you can take it from there.

Oh what the heck. I've typed too long getting this right, and it is a slightly different approach. I'll post it anyway.

 

[soroban]

Hello, she18!

EvaluateL $\displaystyle \tan\left(\pi\,+\,\arcsin\frac{2}{3}\right)$

First, we note note that: .$\displaystyle \tan(x\,+\,\pi)\,=\,\tan(x)$

. . Hence: .$\displaystyle \tan\left(\pi + \arcsin\frac{2}{3}\right)\,=\,\tan\left(\arcsin\frac{2}{3}\right)$

$\displaystyle \arcsin\frac{2}{3}$ is just some angle, $\displaystyle \theta$, such that: $\displaystyle \sin\theta\,=\,\frac{2}{3}$

Since $\displaystyle \sin\theta\,=\,\frac{2}{3}\,=\,\frac{opp}{hyp}$, we find that: $\displaystyle adj = \sqrt{5}\;\;\Rightarrow\;\;\tan\theta\,=\,\frac{2}{\sqrt{5}}\;\;\Rightarrow\;\;\theta\,=\,\arctan\frac{2}{\sqrt{5}}$

. . And we have: .$\displaystyle \tan\left(\arctan\frac{2}{\sqrt{5}}\right) \:= \:\frac{2}{\sqrt{5}}$