Word problems

[rachelmaddie]

Can someone please check this?
Sin 3pi/2

1. Draw the triangle in the 3rd quadrant.

Cos 3pi / 2 = 0 because the adjacent = 0 and the hypotenuse = -1 at theta = 3pi/2

Therefore cos 3pi/2 = 0 / -1

2. Sin theta in the third quadrant = opp / hyp
= negative / 1

Therefore your answer should be negative.

3. Therefore sin 3pi/2 = 0 / -1 = 0

 

[pka]

Actually $\left(\dfrac{3\pi}{2}\right)$ is equivalent $\left(\dfrac{-\pi}{2}\right)$ so that it is not in any quadrant but is a special angle.
$\sin(\left(\dfrac{-\pi}{2}\right)=-1$ & $\cos\left(\dfrac{-\pi}{2}\right)=0$

 

[rachelmaddie]

Actually $\left(\dfrac{3\pi}{2}\right)$ is equivalent $\left(\dfrac{-\pi}{2}\right)$ so that it is not in any quadrant but is a special angle.
$\sin(\left(\dfrac{-\pi}{2}\right)=-1$ & $\cos\left(\dfrac{-\pi}{2}\right)=0$

Do I need to make any corrections?

 

[JeffM]

Do I need to make any corrections?

Yes, You came up with [MATH]sin \left ( \dfrac{3 \pi}{2} \right ) = 0.[/MATH]
That is wrong. You were shown one way to think about this by pka. Here is another way.

[MATH]sin(\theta \pm \pi) = - sin(\theta) \text { and } cos(\theta \pm \pi) = - cos ( \theta).[/MATH]
[MATH]\dfrac{3 \pi}{2} = \dfrac{\pi}{2} + \pi.[/MATH]
What are [MATH]sin \left ( \dfrac{\pi}{2}\right ) \text { and } cos \left ( \dfrac{\pi}{2}\right )[/MATH] respectively?

Therefore WHAT?

 

[rachelmaddie]

Yes, You came up with [MATH]sin \left ( \dfrac{3 \pi}{2} \right ) = 0.[/MATH]
That is wrong. You were shown one way to think about this by pka. Here is another way.

[MATH]sin(\theta \pm \pi) = - sin(\theta) \text { and } cos(\theta \pm \pi) = - cos ( \theta).[/MATH]
[MATH]\dfrac{3 \pi}{2} = \dfrac{\pi}{2} + \pi.[/MATH]
What are [MATH]sin \left ( \dfrac{\pi}{2}\right ) \text { and } cos \left ( \dfrac{\pi}{2}\right )[/MATH] respectively?

Therefore WHAT?

sin(pi/2) = 1 and cos(pi/2) = 0

 

[JeffM]

sin(pi/2) = 1 and cos(pi/2) = 0

Indeed. So given what was said about [MATH]sin( \theta + \pi)[/MATH] and [MATH]cos(\theta + \pi)[/MATH]
what do you conclude about [MATH]sin \left ( \dfrac{3\pi}{2} \right )[/MATH] and [MATH]cos \left ( \dfrac{3\pi}{2} \right )[/MATH]?

 

[Deleted member 4993]

sin(pi/2) = 1 and cos(pi/2) = 0

But what about sin(3π/2) and cos(3π/2) = ?

 

[pka]

But what about sin(3π/2) and cos(3π/2) = ?

But $\left(\dfrac{3\pi}{2}\right)$ is $\left(\dfrac{-\pi}{2}\right)$

 

[Deleted member 4993]

But $\left(\dfrac{3\pi}{2}\right)$ is $\left(\dfrac{-\pi}{2}\right)$

Correct - but there was no negative sign in response #5 !!

 

[rachelmaddie]

But what about sin(3π/2) and cos(3π/2) = ?

sin(3pi/2) = -1 and cos(3pi/2) = 0

 

[JeffM]

sin(3pi/2) = -1 and cos(3pi/2) = 0

Yes. Do you see why now?

 

[rachelmaddie]

Yes. Do you see why now?

how do I write this out in steps to show my work?

 

[JeffM]

In your first post, you said you were to use the unit circle to solve the problem. In the context of the unit circle y = sin(theta). So what is y at an angle of 3pi/2? Not knowing what your teacher expects as a proof, my best guess is that something as simple as that will suffice.