Simplifying expressions

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January 12th 2007, 12:29 PM
mike1

Simplifying expressions

Could someone help me with this problem?

Question: Simplify the expression cos(2x) + 2sin(squared)x.
January 12th 2007, 12:43 PM
CaptainBlack

Quote:

Originally Posted by mike1

Could someone help me with this problem?

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

Simplify

$\cos(2x) + 2\sin^2(x)$

Now we have a trig identity (which may be derived from $\cos(A+B)=\cos(A)\cos(B)-\sin(A)\sin(B)$ ):

$\cos(2A)=1-2\sin^2(A)$

which may be used to give:

$\cos(2x) + 2\sin^2(x)=1-2\sin^2(x)+1\sin^2(x)=1$

RonL
January 12th 2007, 02:18 PM
Soroban

Hello, Mike!

Another approach . . .

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

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

Quote:

Simplify: . $\cos(2x) + 2\sin^2(x)$

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

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

. . . $= \;\cos(2x) + 1 - \cos(2x) \;= \;\boxed{1}$
January 12th 2007, 10:28 PM
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.

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.
January 12th 2007, 11:20 PM
CaptainBlack

Quote:

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:

$\sin(A+B)=\sin(A)\cos(B)+\cos(A)\sin(B)$

and:

$\sin(A-B)=\sin(A)\cos(B)-\cos(A)\sin(B)$

adding these gives another identity you are supposed to know:

$\sin(A+B)+\sin(A-B)=2\sin(A)\cos(B)$

So put $A=6u$ , and $B=4u$ , gives:

$2\sin(6u)\cos(4u)=\sin(10u)+\sin(2u)$

RonL
January 12th 2007, 11:24 PM
CaptainBlack

Quote:

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.

Thanks for any advice.

For this work the same identity as we used in the last problem backwards:

$\sin(A+B)+\sin(A-B)=2\sin(A)\cos(B)$ .

Put $A+B=6t$ , and $A-B=2t$ , so $A=4t$
and $B=2t$ and so:

$\sin(6t)+\sin(2t)=2\sin(4t)\cos(2t)$ .

RonL