geeze this stuff never ends! trig
- Thread starter she18
- Start date
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
Using the identities
sin<sup>-1</sup>(y/x)=tan<sup>-1</sup>(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<sup>-1</sup>(2/\(\displaystyle sqrt[5]\))/(1-tan(pi)*tan(tan<sup>-1</sup>(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.
sin<sup>-1</sup>(y/x)=tan<sup>-1</sup>(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<sup>-1</sup>(2/\(\displaystyle sqrt[5]\))/(1-tan(pi)*tan(tan<sup>-1</sup>(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.
Hello, she18!
. . 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: .\(\displaystyle \tan\left(\arctan\frac{2}{\sqrt{5}}\right) \:= \:\frac{2}{\sqrt{5}}\)
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: .\(\displaystyle \tan\left(\arctan\frac{2}{\sqrt{5}}\right) \:= \:\frac{2}{\sqrt{5}}\)