1. ## Radian question with integral

I think I am making a silly error:

They say: "we know the angle arcsin((radical3)/2) equals the angle between [-pi/2,pi/2] who sine is (radical3)/2

Apparently 60 is wrong, but I thought sin60 is radical3/2. Are they asking for the values of arcsin60? Amateur question but I'm confused as to what they want.

2. ## Re: Radian question with integral

Originally Posted by Steelers72
I think I am making a silly error:
They say: "we know the angle arcsin((radical3)/2) equals the angle between [-pi/2,pi/2] who sine is (radical3)/2
Apparently 60 is wrong, but I thought sin60 is radical3/2. Are they asking for the values of arcsin60? Amateur question but I'm confused as to what they want.
Yes that is wrong.
$\arcsin\left(\frac{\sqrt{3}}{2}\right)=\frac{\pi}{ 3}$

3. ## Re: Radian question with integral

Originally Posted by Steelers72
I think I am making a silly error:

They say: "we know the angle arcsin((radical3)/2) equals the angle between [-pi/2,pi/2] who sine is (radical3)/2

Apparently 60 is wrong, but I thought sin60 is radical3/2. Are they asking for the values of arcsin60? Amateur question but I'm confused as to what they want.

$\sin(60^\circ)=\frac{\sqrt{3}}{2}$

$\sin\left( \frac{\pi}{3}\right)=\frac{\sqrt{3}}{2}$

4. ## Re: Radian question with integral

Thank you both. Silly error. 180/3 =60 but in radians it has a pi. Bless!

5. ## Re: Radian question with integral

I got 90. Is this correct?

6. ## Re: Radian question with integral

Originally Posted by Steelers72
I got 90. Is this correct?
Give up the use of degrees. Only the still mathematically retarded use degrees.

7. ## Re: Radian question with integral

Originally Posted by Plato
Give up the use of degrees. Only the still mathematically retarded use degrees.
I dont know if I should take that as an insult or constructive criticism, but I used radians and got a decimal and it said it was wrong I guess. It doesn't specify which to use as well.

3arcsin(pi/3)-3arcin(pi/6)
and I got 90 as an answer

8. ## Re: Radian question with integral

Originally Posted by Steelers72
I dont know if I should take that as an insult or constructive criticism, but I used radians and got a decimal and it said it was wrong I guess. It doesn't specify which to use as well.
3arcsin(pi/3)-3arcin(pi/6)
and I got 90 as an answer
"as an insult or constructive criticism" that was meant as neither.
It states a fact. Most modern mathematics in fact does not use degrees.

You clearly are confused. $\arcsin\left(\frac{\pi}{3}\right)$ is not even defined. In order for $\arcsin(t)$ to be defined, you must have $-1\le t\le 1.$ and then $\frac{-\pi}{2}\le\arcsin(t)\le\frac{\pi}{2}.$