Derivatives/Integrals, properties of sin/cos

**crabchef** · July 19th 2008, 11:29 PM

Q: ʃʃ cos (x +2y) dy dx where 0 < x < pi; 0 < y < pi/2

When I integrate cos(x+2y) wrt y, do I get sin(x+2y)/2? How do I evaluate sin(x + pi)?

Does sin (x+c) = sin(x) + sin(c)?

**Mathstud28** · July 19th 2008, 11:38 PM

Originally Posted by crabchef

Does sin (x+c) = sin(x) + sin(c)?

Not, not at all. But $\sin(x+\pi)=-\sin(x)$

**crabchef** · July 19th 2008, 11:45 PM

Originally Posted by Mathstud28

Not, not at all. But $\sin(x+\pi)=-\sin(x)$

thanks...but why does sin(x+pi)=-sin(x)?

also, what does sin(x+c)= ? or are there no general properties associated with that?

**Marine** · July 20th 2008, 12:01 AM

Originally Posted by crabchef

thanks...but why does sin(x+pi)=-sin(x)?

also, what does sin(x+c)= ? or are there no general properties associated with that?

1. check with the unit circle

...or

2. plot the graph of sin(x+pi) , i.e. translation by pi to the left. It comes -sinx out

3. use trig identities (I don't recommend the last method)

**Mathstud28** · July 20th 2008, 12:14 AM

Originally Posted by crabchef

thanks...but why does sin(x+pi)=-sin(x)?

also, what does sin(x+c)= ? or are there no general properties associated with that?

How are you doing double integrals but you don't know the addition formulas for sin?

$\sin\left(A+B\right)=\sin\left(A\right)\cos\left(B \right)+\cos\left(A\right)\sin\left(B\right)$

**crabchef** · July 20th 2008, 12:18 AM

Originally Posted by Mathstud28

How are you doing double integrals but you don't know the addition formulas for sin?

$\sin\left(A+B\right)=\sin\left(A\right)\cos\left(B \right)+\cos\left(A\right)\sin\left(B\right)$

The last math course I took was 6 years ago...and then I went straight to multi-variate calc without any review.

Anyway, the formula you gave is if A and B are both variables right? What about in the case that one is a variable and the other is a contant? Do I just treat the constant as a variable and plug it through?

**Mathstud28** · July 20th 2008, 12:19 AM

Originally Posted by crabchef

The last math course I took was 6 years ago...and then I went straight to multi-variate calc without any review.

Anyway, the formula you gave is if A and B are both variables right? What about in the case that one is a variable and the other is a contant? Do I just treat the constant as a variable and plug it through?

Yes.

**wingless** · July 20th 2008, 12:43 AM

$\int_0^{\pi} \int_0^{\frac{\pi}{2}}\cos (x+2y)~dy~dx$

Ignore the outer integral and do the inner one first.

$\int_0^{\frac{\pi}{2}}\cos (x+2y)~dy$

You can treat x as a constant here. Let $u=x+2y$ . Don't forget to change the integration limits. Then you can plug this in the outer integral and the rest is easy..

Also you can use the trigonometric addition formula and Fubini's theorem.

**Mathstud28** · July 20th 2008, 12:44 AM

Originally Posted by wingless

$\int_0^{\pi} \int_0^{\frac{\pi}{2}}\cos (x+2y)~dy~dx$

Ignore the outer integral and do the inner one first.

$\int_0^{\frac{\pi}{2}}\cos (x+2y)~dy$

You can treat x as a constant here. Let $u=x+2y$ . Don't forget to change the integration limits. Then you can plug this in the outer integral and the rest is easy..

Also you can use the trigonometric addition formula and Fubini's theorem.

Would there really be a need for Fubini's Theorem here?

**wingless** · July 20th 2008, 12:54 AM

Originally Posted by Mathstud28

Would there really be a need for Fubini's Theorem here?

I would use it to split $\int\int \sin (A) \cos (B)~dy~dx$ to $\int \sin(A)~dx ~\int \cos(B)~dy$ , A being a function of x and B being a function of y. Oh, this is not the theorem itself, but a use of it.

**crabchef** · July 20th 2008, 01:41 AM

Originally Posted by Mathstud28

How are you doing double integrals but you don't know the addition formulas for sin?

$\sin\left(A+B\right)=\sin\left(A\right)\cos\left(B \right)+\cos\left(A\right)\sin\left(B\right)$

thank you both for your help. I was wondering if I hate integrated the first part right. I know you recommended substitution to solve it, but it seems like it wouldn't be completely necessary. Thoughts?

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