jonnburton
Junior Member
- Joined
- Dec 16, 2012
- Messages
- 155
Hi,
I have been following a worked example in my textbook and can't understand part of the working.
The example is integrating:
\(\displaystyle \int 1-3cos\theta + 3cos^2\theta -cos^3\theta d\theta\)
the next step the book outlines is this:
\(\displaystyle \int 1 -3cos\theta +\frac{3}{2} -\frac{3cos2\theta}{2}-cos^3\theta d\theta\)
While I understand they are using the identity \(\displaystyle cos2x= 2cos^2x-1\), I can not see how \(\displaystyle 3cos^2\theta\) becomes \(\displaystyle \frac {3}{2} - \frac{3cos2\theta}{2}\). In particular, where does the minus sign come from?
This is how I would have used the identity in this case:
\(\displaystyle cos2\theta = 2 cos^2\theta - 1\)
Add one to both sides:
\(\displaystyle cos2\theta + 1 = 2cos^2\theta\)
so,
\(\displaystyle \frac{3cos2\theta +3}{2} = 3 cos^2\theta\)
But I must have failed to understand something here: can anyone tell me what it is?
I have been following a worked example in my textbook and can't understand part of the working.
The example is integrating:
\(\displaystyle \int 1-3cos\theta + 3cos^2\theta -cos^3\theta d\theta\)
the next step the book outlines is this:
\(\displaystyle \int 1 -3cos\theta +\frac{3}{2} -\frac{3cos2\theta}{2}-cos^3\theta d\theta\)
While I understand they are using the identity \(\displaystyle cos2x= 2cos^2x-1\), I can not see how \(\displaystyle 3cos^2\theta\) becomes \(\displaystyle \frac {3}{2} - \frac{3cos2\theta}{2}\). In particular, where does the minus sign come from?
This is how I would have used the identity in this case:
\(\displaystyle cos2\theta = 2 cos^2\theta - 1\)
Add one to both sides:
\(\displaystyle cos2\theta + 1 = 2cos^2\theta\)
so,
\(\displaystyle \frac{3cos2\theta +3}{2} = 3 cos^2\theta\)
But I must have failed to understand something here: can anyone tell me what it is?