Question

Calc 2

Can you explain to me this step?
http://screencast.com/t/4xD83OizO2

Answer 1

I better let King George help you :D
@KingGeorge

Answer 2

@agent0smith will be glad to help:)

Answer 3

Assuming the arrows point to the problematic part:

Consider \[\cos(a+b)=\cos(a)\cos(b)-\sin(a)\sin(b)\]
Now if we let a=b=u then \[\cos(2u)=\cos^2(u)-\sin^2(u)\]
Since \[\sin^2(u)+\cos^2(u)=1\]
the double angle relation above can be rewritten as \[\cos(2u)=\cos^2(u)-(1-\cos^2(u))\]
So \[\cos(2u)=2\cos^2(u)-1\] which rearranges to \[1+\cos(2u)=2\cos^2(u)\] as required.

For the second part, simply differentiate the result to see that it gives the original integrand \[\frac{ d }{ dx }[\frac{\sin(2u)}{2}]=\frac{\cos(2u)}{2}\times 2=\cos(2u)\]

Answer 4

ty

Answer 5

You can also see that one on a table of identities... scroll down to double angle identities http://www.purplemath.com/modules/idents.htm :

\[\large \cos(2x) = \cos^2(x) – \sin^2(x) = 1 – 2\sin^2(x) = 2\cos^2(x) – 1 \]
so use the last result on the end\[\large \cos(2x) = 2 \cos^2 x -1\]then just add 1 to both sides.