1. ## Sine of Arcsine

Was hoping to get some help on this. Not sure how to approach.

Sin[2Arcsin(-4/5)]

Thank you!

2. Originally Posted by Impalord
Was hoping to get some help on this. Not sure how to approach.

Sin[2Arcsin(-4/5)]

Thank you!
Let $\displaystyle \alpha = \arcsin \left(- \frac{4}{5} \right) \Rightarrow \sin \alpha = - \frac{4}{5}$. Note that $\displaystyle \alpha$ is in the fourth quadrant (why?).

Now use $\displaystyle \sin (2 \alpha) = 2 \sin \alpha \cos \alpha$.

3. Originally Posted by mr fantastic
Let $\displaystyle \alpha = \arcsin \left(- \frac{4}{5} \right) \Rightarrow \sin \alpha = - \frac{4}{5}$. Note that $\displaystyle \alpha$ is in the fourth quadrant (why?).

Now use $\displaystyle \sin (2 \alpha) = 2 \sin \alpha \cos \alpha$.

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!

4. 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

5. Originally Posted by Impalord
$\displaystyle 2( \sin -4/5 \cos 3/5)$
No, no, no. Watch your notation! -4/5 and 3/5 are not angles! They are the ratios.

Look at what Mr. fantastic wrote:
Originally Posted by mr fantastic
Let $\displaystyle \alpha = \arcsin \left(-\frac{4}{5} \right)$
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

6. Originally Posted by yeongil
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

Great! thanks a bunch for clearing that up. As I was working through that, It seemed wrong. Now I know why, I really appreciate it.

7. Hello Impalord
Originally Posted by Impalord
All right, I think I've got it.

Because :

$\displaystyle 2 \sin x = 2 \sin x \cos x$ You mean sin2x. Grandad.

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
Your 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:

$\displaystyle \sin\left(2\arcsin(-\frac45)\right)=\sin2\alpha = 2\sin\alpha\cos\alpha = 2\times(-\tfrac45)\times\tfrac35 = -\tfrac{24}{25}$

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:

$\displaystyle \sin\left(2\arcsin(-\frac45)\right)=\sin2\alpha = 2\sin\alpha\cos\alpha = 2\times(-\tfrac45)\times\tfrac35 = -\tfrac{24}{25}$

$\displaystyle \sin[2 \arcsin (-4/9)] = \sin(2 \alpha)$
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!