1. Simplifying expressions

Could someone help me with this problem?

Question: Simplify the expression cos(2x) + 2sin(squared)x.

2. Originally Posted by mike1
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

3. Hello, Mike!

Another approach . . .

I assume you're familiar with the Double-angle Identities.

In particular: .$\displaystyle \sin^2x\:=\:\frac{1 - \cos2x}{2}$

Simplify: .$\displaystyle \cos(2x) + 2\sin^2(x)$
We have: .$\displaystyle \cos(2x) + 2\underbrace{\sin^2(x)}$

. . . $\displaystyle = \;\cos(2x) + 2\left(\frac{1 - \cos(2x)}{2}\right)$

. . . $\displaystyle = \;\cos(2x) + 1 - \cos(2x) \;= \;\boxed{1}$

4. 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.

5. Originally Posted by mike1
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.
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

6. Originally Posted by mike1
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.

$\displaystyle \sin(A+B)+\sin(A-B)=2\sin(A)\cos(B)$.
Put $\displaystyle A+B=6t$, and $\displaystyle A-B=2t$, so $\displaystyle A=4t$
and $\displaystyle B=2t$ and so:
$\displaystyle \sin(6t)+\sin(2t)=2\sin(4t)\cos(2t)$.