Integrate cos(x)-cos(x-c) from 0 to c/2

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Homework Statement


Integrate cos(x)-cos(x-c) from 0 to c/2


Homework Equations


sin(x-c)=sinxcosc-cosxsinc


The Attempt at a Solution


sin(x)-sin(x-c) from 0 to c/2
=sin(c/2)-sin(c/2)cos(c)+cos(c/2)sin(c)-sin(0)+sin(-c)
=sin(c/2)-sin(c/2)cos(c)+cos(c/2)sin(c)-0-sin(c)

Correct response:2sin(c/2)-sin(c)

What am I doing wrong?
 
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sugarxsweet said:

Homework Statement


Integrate cos(x)-cos(x-c) from 0 to c/2

Homework Equations


sin(x-c)=sinxcosc-cosxsinc

The Attempt at a Solution


sin(x)-sin(x-c) from 0 to c/2
=sin(c/2)-sin(c/2)cos(c)+cos(c/2)sin(c)-sin(0)+sin(-c)
=sin(c/2)-sin(c/2)cos(c)+cos(c/2)sin(c)-0-sin(c)

Correct response:2sin(c/2)-sin(c)

What am I doing wrong?
You're doing nothing wrong. They are equivalent.

To get the answer in a more straight forward manner, simply plug the limits of integration into the anti-derivative, sin(x)-sin(x-c) .
 
Sorry, I guess this is a question of trig then - how do I simplify my answer to the correct answer? It's probably something stupid but I'm having trouble figuring it out!
 
sugarxsweet said:

Homework Statement


Integrate cos(x)-cos(x-c) from 0 to c/2

Homework Equations


sin(x-c)=sinxcosc-cosxsinc

The Attempt at a Solution


sin(x)-sin(x-c) from 0 to c/2
=sin(c/2)-sin(c/2)cos(c)+cos(c/2)sin(c)-sin(0)+sin(-c)
=sin(c/2)-sin(c/2)cos(c)+cos(c/2)sin(c)-0-sin(c)

Correct response:2sin(c/2)-sin(c)

What am I doing wrong?

$$sin(\frac{c}{2})-sin(\frac{c}{2})cos(c)+cos(\frac{c}{2})sin(c)-sin(c)$$
$$sin(\frac{c}{2})-sin(c)+cos(\frac{c}{2})sin(c)-sin(\frac{c}{2})cos(c)$$

You have to realize (from trig addition identity) that ##cos(\frac{c}{2})sin(c)-sin(\frac{c}{2})cos(c) = sin(c-\frac{c}{2}) = sin(\frac{c}{2})##

Using the above, we have:

$$sin(\frac{c}{2})-sin(c)+sin(\frac{c}{2})$$
$$2sin(\frac{c}{2})-sin(c)$$
 
Thanks!
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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