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exact values of trig., inverse-trig expressions
- Thread starter lyssaboo
- Start date
Re: exact values of trig. expressions
Hello, lyssaboo!
Edit: made a stupid blunder . . .
Recall that an inverse trig expression is an angle.
Let \(\displaystyle \L\theta\,=\,\sin^{-1}\left(\frac{\sqrt{2}}{3}\right)\)
Then: \(\displaystyle \L\:\sin\theta \:=\:\frac{\sqrt{2}}{3}\)
\(\displaystyle \theta\) is an angle in a right triangle with: \(\displaystyle \,opp\,=\,\sqrt{2},\;hyp\,=\,3\)
Using Pythagorus, we find that: \(\displaystyle adj\,=\,\pm\sqrt{7}\)
Therefore: \(\displaystyle \L\:\fbox{\cos\theta \:=\:\pm\frac{3}{\sqrt{7}}}\)
Thanks for pointing it out, G!
Let \(\displaystyle \L\theta \:=\:\csc^{-1}\left(\frac{2\sqrt{3}}{3}\right)\)
Then: \(\displaystyle \L\:\csc\theta\:=\:\frac{2\sqrt{3}}{3}\)
Hence: \(\displaystyle \L\:\sin\theta\:=\:\frac{3}{2\sqrt{3}} \:=\:\frac{\sqrt{3}}{2}\)
Therefore: \(\displaystyle \L\:\theta \:=\:\begin{Bmatrix}\frac{\pi}{3}\,+\,2\pi n \\ \frac{2\pi}{3}\,+\,2\pi n\end{Bmatrix}\)
Hello, lyssaboo!
Edit: made a stupid blunder . . .
Recall that an inverse trig expression is an angle.
\(\displaystyle \cos \left[\sin^{-1}\left(\frac{sqrt{2}}{3}\right)\right]\)
Let \(\displaystyle \L\theta\,=\,\sin^{-1}\left(\frac{\sqrt{2}}{3}\right)\)
Then: \(\displaystyle \L\:\sin\theta \:=\:\frac{\sqrt{2}}{3}\)
\(\displaystyle \theta\) is an angle in a right triangle with: \(\displaystyle \,opp\,=\,\sqrt{2},\;hyp\,=\,3\)
Using Pythagorus, we find that: \(\displaystyle adj\,=\,\pm\sqrt{7}\)
Therefore: \(\displaystyle \L\:\fbox{\cos\theta \:=\:\pm\frac{3}{\sqrt{7}}}\)
Thanks for pointing it out, G!
\(\displaystyle \csc^{-1}\left(\frac{2\sqrt{3}}{3}\right)\)
Let \(\displaystyle \L\theta \:=\:\csc^{-1}\left(\frac{2\sqrt{3}}{3}\right)\)
Then: \(\displaystyle \L\:\csc\theta\:=\:\frac{2\sqrt{3}}{3}\)
Hence: \(\displaystyle \L\:\sin\theta\:=\:\frac{3}{2\sqrt{3}} \:=\:\frac{\sqrt{3}}{2}\)
Therefore: \(\displaystyle \L\:\theta \:=\:\begin{Bmatrix}\frac{\pi}{3}\,+\,2\pi n \\ \frac{2\pi}{3}\,+\,2\pi n\end{Bmatrix}\)
Re: exact values of trig. expressions
There're few problems with your solution:
adj catet seems to be longer than hypotenuse
cos is resiprocal to what it supposed to be compensating for the first mistake
only positive solution is valid since arcsin range is -pi/2 to pi/2 where cos is non- negative.
how about this:
cos(arcsin(sqrt(2)/3)) = sqrt(1-sin^2(arcsin(sqrt(2)/3)) = sqrt(1-2/9) = sqrt(7)/3
soroban said:Hello, lyssaboo!
Recall that an inverse trig expression is an angle.
\(\displaystyle \cos \left[\sin^{-1}\left(\frac{sqrt{2}}{3}\right)\right]\)
Let \(\displaystyle \L\theta\,=\,\sin^{-1}\left(\frac{\sqrt{2}}{3}\right)\)
Then: \(\displaystyle \L\:\sin\theta \:=\:\frac{\sqrt{2}}{3}\)
\(\displaystyle \theta\) is an angle in a right triangle with: \(\displaystyle \,opp\,=\,\sqrt{2},\;hyp\,=\,3\)
Using Pythagorus, we find that: \(\displaystyle adj\,=\,\pm\sqrt{11}\)
Therefore: \(\displaystyle \L\:\fbox{\cos\theta \:=\:\pm\frac{3}{\sqrt{11}}}\)
There're few problems with your solution:
adj catet seems to be longer than hypotenuse
cos is resiprocal to what it supposed to be compensating for the first mistake
only positive solution is valid since arcsin range is -pi/2 to pi/2 where cos is non- negative.
how about this:
cos(arcsin(sqrt(2)/3)) = sqrt(1-sin^2(arcsin(sqrt(2)/3)) = sqrt(1-2/9) = sqrt(7)/3