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Math Help - Composite argument property and linear combination of cosx/sinx

  1. #1
    Member wiseguy's Avatar
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    Composite argument property and linear combination of cosx/sinx

    I have three questions in my book about the composite argument property that I'm not entirely sure about


    "On your paper, draw an appropriate sketch and use it to write y=4cosx+3sinx as a cosine with a phase displacement."

    My response:

    y=5cos(x)-0.64350169<br />

    "Draw an appropriate sketch on your paper and use it to write y=-3cosx-1sinx as a cosine with a phase displacement."

    My response:

    y=-3.1622777cosθ-3.4633437

    "Use the composite argument property to write y=7cos(x-20) as a linear combination of cos and sin ."

    My response: (gave up)
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    This site might help you
    Precalculus Resources - Keymath.com
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  3. #3
    Senior Member eumyang's Avatar
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    I have not looked at the first or second problem. But regarding the third:
    "Use the composite argument property to write y=7cos(x-20) as a linear combination of cos and sin ."

    My response: (gave up)
    Use the cosine of a difference formula:
    \begin{aligned}<br />
y &= 7 cos (x - 20) \\<br />
&= 7[(cos x)(cos 20) + (sin x)(sin 20)] \\<br />
&= 7(cos x)(cos 20) + 7(sin x)(sin 20) \\<br />
&= (7cos 20)cos x + (7sin 20)sin x \\<br />
&\approx 6.578cos x + 2.394sin x \\<br />
 \end{aligned}
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    Member wiseguy's Avatar
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    Two more roadblocks...

    I have two more questions about this section that are a little shady -

    "Suppose that cos A = 0.6 and sin A = 0.8. Suppose that cos B = 0.96 and sin B = 0.28. Use the composite argument property to find cos (A + B)."

    I responded with
    cos (A − B) = cos A cos B + sin A sin B
    cos(A-B)=0.6*0.96+0.8*0.28
    cps(A-B)=0.576+0.224
    cos(A+B)=0.129024

    arccos0.129024=A+B
    A+B=1.441411637

    ...I'm pretty sure this is correct, but a double check would be greatly appreciated ... so far so good, until-

    "Show that cos -10.6 and sin-10.8 give the same degree measure for angle A."

    How am I suppose to use the "composite argument property" to answer this question? (If I'm suppose to use it, it does not say so but it is in the same section...) All I did was punch both of these in my calculator to prove that they have the same degree measure. Surely I cannot put this as an answer, my instructor will think I'm being a smart ---. Here's another just like it-

    "Show that both cos-1 0.96 and sin -1 0.28 give the same degree measure for angle B."
    Last edited by wiseguy; July 22nd 2010 at 05:51 PM.
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  5. #5
    Senior Member eumyang's Avatar
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    Quote Originally Posted by wiseguy View Post
    I have two more questions about this section that are a little shady -

    "Suppose that cos A = 0.6 and sin A = 0.8. Suppose that cos B = 0.96 and sin B = 0.28. Use the composite argument property to find cos (A + B)."

    I responded with
    cos (A − B) = cos A cos B + sin A sin B
    cos(A-B)=0.6*0.96+0.8*0.28
    cps(A-B)=0.576+0.224
    cos(A+B)=0.129024

    arccos0.129024=A+B
    A+B=1.441411637
    Be careful with your signs. Does the original problem say to find cos (A+B) or cos (A-B)? Because in your work you started with cos (A-B). Also, you multiplied in the final step instead of added.

    If the problem is to find cos (A+B), then:
    \begin{aligned}<br />
cos\,(A + B) &= cos\,A\,cos\,B - sin\,A\,sin\,B \\<br />
&= (0.6)(0.96) - (0.8)(0.28) \\<br />
&= 0.352<br />
\end{aligned}
    There's no need to take the arccosine, because the problem didn't ask for an angle.

    "Show that cos^{-1} 0.6 and sin^{-1} 0.8 give the same degree measure for angle A."
    I don't think you need to use the sum/difference identities. Maybe draw a triangle? 0.6 = 3/5 and 0.8 = 4/5. You'll get a 3-4-5 triangle, and you can see that angle A is the same.

    "Show that both cos^{-1} 0.96 and sin^{-1} 0.28 give the same degree measure for angle B."
    Same thing as above. What are the equivalent fractions for 0.96 and 0.28?
    Last edited by eumyang; July 22nd 2010 at 06:48 PM. Reason: Added next two problems from post #4
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  6. #6
    Member wiseguy's Avatar
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    Yes, I double checked it, the problem does say cos(A+B). I copied the first line from a formula in my text book and wasn't being careful :/
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    Member wiseguy's Avatar
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    I'm pretty sure the two questions about showing that arccos0.6 and arcsin0.8 give the same degree measure are simply just hitting them into the calculator. I don't see any other way to do this.

    Edit: The triangle idea is a good one. I didn't want to just put calculator answers for my teacher.
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  8. #8
    Member wiseguy's Avatar
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    Quote Originally Posted by eumyang View Post
    ...
    What are the equivalent fractions for 0.96 and 0.28?
    0.96=3/5 and 0.28=4/5, correct me if I'm wrong.
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  9. #9
    Senior Member eumyang's Avatar
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    Quote Originally Posted by wiseguy View Post
    0.96=3/5 and 0.28=4/5, correct me if I'm wrong.
    No. 0.96 = 24/25 and 0.28 = 7/25.
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  10. #10
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    Hello, wiseguy!

    I believe this is what they expected for the first one . . .


    On your paper, draw an appropriate sketch ? and use it to write
    . . y\:=\:4\cos x+3\sin x .as a cosine with a phase displacement.

    Divide by 5: . \dfrac{y}{5} \;=\;\dfrac{4}{5}\cos x + \dfrac{3}{5}\sin x .[1]


    Let \theta be in this right triangle:

    Code:
                            *
                         *  *
                 5    *     *
                   *        * 3
                *           *
             * θ            *
          * - - - - - - - - *
                    4

    Then: . \cos\theta = \dfrac{4}{5},\;\sin\theta = \dfrac{3}{5}


    Substitute into [1]: . \dfrac{y}{5} \;=\;\cos\theta\cos x + \sin\theta\sin x

    . . . . . . . . . . . . . . . \dfrac{y}{5} \;=\;\cos(x - \theta)

    . . . . . . . . . . . . . . . y \;=\;5\cos\left(x - \arcsin\frac{3}{5}\right)

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  11. #11
    Member wiseguy's Avatar
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    EDIT: thanks allot for the help on this topic. It's really great to get help from the experts over the net.


    **
    Thanks for the help!

    I ended up figuring it out by putting the original equation in my calculator, taking the first maximum value, then using the y value as my new amplitude and the x value as the phase displacement. I ended up with this equation:

    y=5cos(x-0.64350423)

    Turns out it's exactly the same as yours

    Quote Originally Posted by Soroban View Post
    Hello, wiseguy!

    I believe this is what they expected for the first one . . .



    Divide by 5: . \dfrac{y}{5} \;=\;\dfrac{4}{5}\cos x + \dfrac{3}{5}\sin x .[1]


    Let \theta be in this right triangle:

    Code:
                            *
                         *  *
                 5    *     *
                   *        * 3
                *           *
             * θ            *
          * - - - - - - - - *
                    4

    Then: . \cos\theta = \dfrac{4}{5},\;\sin\theta = \dfrac{3}{5}


    Substitute into [1]: . \dfrac{y}{5} \;=\;\cos\theta\cos x + \sin\theta\sin x

    . . . . . . . . . . . . . . . \dfrac{y}{5} \;=\;\cos(x - \theta)

    . . . . . . . . . . . . . . . y \;=\;5\cos\left(x - \arcsin\frac{3}{5}\right)

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