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Math Help - Simplifying expressions

  1. #1
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    Simplifying expressions

    Could someone help me with this problem?

    Question: Simplify the expression cos(2x) + 2sin(squared)x.
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  2. #2
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    Quote Originally Posted by mike1 View Post
    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
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  3. #3
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    Hello, Mike!

    Another approach . . .

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

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


    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}

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  4. #4
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    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.
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  5. #5
    Grand Panjandrum
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    Quote Originally Posted by mike1 View Post
    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
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  6. #6
    Grand Panjandrum
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    Quote Originally Posted by mike1 View Post
    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
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