Was hoping to get some help on this. Not sure how to approach.
Sin[2Arcsin(-4/5)]
Thank you!
For some reason I didn't think $\displaystyle \sin(2 \alpha) = 2 \sin (\alpha)$
Not sure why.
So it's in quadrant 4 because the domain of the arcsine function is
$\displaystyle (\pi /2) -> -(\pi /2)$ right?
Now if I plug -4/5 into the equation I have
$\displaystyle 2 \sin(-4/5) \cos(-4/5)$ ?
or would it be
$\displaystyle 2 \sin(-4/5) \cos(3/5)$ ?
Thanks!
All right, I think I've got it.
Because :
$\displaystyle 2 \sin x = 2 \sin x \cos x $
and if then after using the pythagorean theorem
we get 4^2 + 3^2 = 5^2
so
$\displaystyle 2( \sin -4/5 \cos 3/5) $
Luckily, in this case, I know the answer is $\displaystyle -24/25$
So I can see how the $\displaystyle (-4/5)(3/5) $ multiply to make
$\displaystyle 2(-12/25) = -24/25$
and then we can make it back into $\displaystyle \arcsin (-24/25)$
then
$\displaystyle \sin (\arcsin(-24/25)$
And because of the identity property the answer is -24/25.
Does that seem right?
Thank you
No, no, no. Watch your notation! -4/5 and 3/5 are not angles! They are the ratios.
Look at what Mr. fantastic wrote:
When plugging into the double angle formula
$\displaystyle \sin (2 \alpha) = 2 \sin \alpha \cos \alpha$
you get this:
$\displaystyle \sin (2 \alpha) = 2 \sin \left(\arcsin \left(-\frac{4}{5} \right)\right) \cos \left(\arcsin \left(-\frac{4}{5} \right)\right)$
This is not the same as this:
$\displaystyle 2( \sin -4/5 \cos 3/5) $
We are saying that $\displaystyle \alpha$ is in Quadrant IV, so
$\displaystyle \sin \left(\arcsin \left(-\frac{4}{5} \right)\right) = -\frac{4}{5}$.
You figured out correctly that
$\displaystyle \cos \left(\arcsin \left(-\frac{4}{5} \right)\right) = \frac{3}{5}$.
So we substitute and get
$\displaystyle \sin (2 \alpha) = 2\left(-\frac{4}{5}\right)\left(\frac{3}{5}\right)$
I'll say it again: this is not the same as this:
$\displaystyle 2( \sin -4/5 \cos 3/5) $
I'm sorry if I sound irritated -- I can't tell you how many times I see errors in notation like this. If this was an assignment and I was grading it I would only give partial credit.
01
Hello ImpalordYour answer is more or less correct, but rather confused. Mr F has given you the formula $\displaystyle \sin2\alpha = 2\sin\alpha\cos\alpha$, and the fact that $\displaystyle \alpha$ is in the fourth quadrant (your reason for this is correct, by the way) and you've correctly worked out that $\displaystyle \cos\alpha = \frac35$.
So you can simply write:
$\displaystyle \sin\left(2\arcsin(-\frac45)\right)=\sin2\alpha = 2\sin\alpha\cos\alpha = 2\times(-\tfrac45)\times\tfrac35 = -\tfrac{24}{25}$
Grandad
I really appreciate your help!
For some reason I wasn't seeing the identity because I wasn't seeing
$\displaystyle \sin[2 \arcsin (-4/9)] = \sin(2 \alpha)$
Replacing the $\displaystyle \arcsin -4/9 $
with $\displaystyle \alpha$ the way Mr. Fantastic pointed out. Now that I see that, the problem is pretty straight forward. Thank you all again!