Integral question: Product to Sum Angle Formula for Trigonometric Functions

[Kuma]

Homework Statement

Here is the question given:

Homework Equations

The Attempt at a Solution

SO i used the product to sum angle formula:
sin u sin v = 1/2[cos(u-v) - cos (u + v)]

so I get
from ∏ to -∏

1/2 ∫cos(k-n)x dx - ∫cos(k+n)x dx

= 1/2 [1/k-n (sin (k-n)∏ - sin(k-n)-∏) -1/k+n(sin(k+n)∏ - sin(k+n)-∏)]

so when k isn't equal to ±n, i think the second term becomes 0, but the first term doesn't have to be 0 as well (for the first condition).

 

[micromass]

What is $\sin(n\pi)$ if $n\in \mathbb{Z}$?


Also note that your method does not work if n=k. Indeed, you seem to have a term $\frac{1}{n-k}$ there. But if n=k, then you will divide by zero. So if n=k, then you'll need to do something different.

 

[SammyS]

Homework Statement

Here is the question given:

b]3. The Attempt at a Solution [/b]

SO i used the product to sum angle formula:
sin u sin v = 1/2[cos(u-v) - cos (u + v)]

so I get
from ∏ to -∏

1/2 ∫cos(k-n)x dx - ∫cos(k+n)x dx

= 1/2 [1/(k-n) (sin (k-n)∏ + sin(k-n)∏) -1/(k+n)(sin(k+n)∏ + sin(k+n)∏)]

so when k isn't equal to ±n, i think the second term becomes 0, but the first term doesn't have to be 0 as well (for the first condition).

sin(mπ) = 0 for all integers, m. So when k≠n, all those terms are zero.

When k=n, the last term is zero. The first term must be evaluated as a limit n → k .

 

[Kuma]

Yes you are right about the second condition when k = n i will get 1/0, but how do I eliminate that term then

 

[SammyS]

Yes you are right about the second condition when k = n i will get 1/0, but how do I eliminate that term then

Actually it's of the form 0/0 .

 

[micromass]

Yes you are right about the second condition when k = n i will get 1/0, but how do I eliminate that term then

If k=n, then

$$\int_{-\pi}^\pi \cos(n-k)x dx=\int_{-\pi}^\pi \cos(0)dx=\int_{-\pi}^\pi dx$$

 

[SammyS]

If n=k, the integrand becomes sin²(kx).

 

[Kuma]

sin(mπ) = 0 for all integers, m. So when k≠n, all those terms are zero.

When k=n, the last term is zero. The first term must be evaluated as a limit n → k .

I see what you did but k+n doesn't necessarily have to be an integer does it? What if k+n= 0.5, it doesn't make it 0.

 

[micromass]

I see what you did but k+n doesn't necessarily have to be an integer does it? What if k+n= 0.5, it doesn't make it 0.

I think the question implies that both n and k are integers. If n and k are not integers, then the result is not true.