Could someone help me with this problem?
Question: Simplify the expression cos(2x) + 2sin(squared)x.
Simplify
$\displaystyle \cos(2x) + 2\sin^2(x)$
Now we have a trig identity (which may be derived from $\displaystyle \cos(A+B)=\cos(A)\cos(B)-\sin(A)\sin(B)$):
$\displaystyle \cos(2A)=1-2\sin^2(A)$
which may be used to give:
$\displaystyle \cos(2x) + 2\sin^2(x)=1-2\sin^2(x)+1\sin^2(x)=1$
RonL
Hello, Mike!
Another approach . . .
I assume you're familiar with the Double-angle Identities.
In particular: .$\displaystyle \sin^2x\:=\:\frac{1 - \cos2x}{2}$
We have: .$\displaystyle \cos(2x) + 2\underbrace{\sin^2(x)}$Simplify: .$\displaystyle \cos(2x) + 2\sin^2(x)$
. . . $\displaystyle = \;\cos(2x) + 2\left(\frac{1 - \cos(2x)}{2}\right)$
. . . $\displaystyle = \;\cos(2x) + 1 - \cos(2x) \;= \;\boxed{1}$
Thanks, I appreciate the help. I guess I was doubting myself that it was that simple.
I do have another two that I am not sure about if you could help:
Question 1: Express as a sum or difference 2sin(6u)cos(4u).
I think it's either sin(10u)-sin(2u) or sin(10u)+sin(2u), not too sure.
Question 2: Express as a product sin(6t) + sin(2t).
I think it's either 2sin(4t)cos(2t) or 2cos(4t)sin(2t), again not sure.
Thanks for any advice.
As:
$\displaystyle \sin(A+B)=\sin(A)\cos(B)+\cos(A)\sin(B)$
and:
$\displaystyle \sin(A-B)=\sin(A)\cos(B)-\cos(A)\sin(B)$
adding these gives another identity you are supposed to know:
$\displaystyle \sin(A+B)+\sin(A-B)=2\sin(A)\cos(B)$
So put $\displaystyle A=6u$, and $\displaystyle B=4u$, gives:
$\displaystyle 2\sin(6u)\cos(4u)=\sin(10u)+\sin(2u)$
RonL
For this work the same identity as we used in the last problem backwards:
$\displaystyle \sin(A+B)+\sin(A-B)=2\sin(A)\cos(B)$.
Put $\displaystyle A+B=6t$, and $\displaystyle A-B=2t$, so $\displaystyle A=4t$
and $\displaystyle B=2t$ and so:
$\displaystyle \sin(6t)+\sin(2t)=2\sin(4t)\cos(2t)$.
RonL