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Thread: [SOLVED] Getting the wrong sign for cosx+sinx=rsin(x+alpha)

  1. #1
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    [SOLVED] Getting the wrong sign for cosx+sinx=rsin(x+alpha)

    The question asks me to solve $\displaystyle 7\cos x +6\sin x = 2$

    So I put it into the form $\displaystyle r\sin (x + \alpha)$:

    $\displaystyle 7\cos x +6\sin x \equiv r\sin (x + \alpha) \equiv r\sin x\cos\alpha + r\cos x\sin\alpha$

    Then I equate the coefficients:
    of cos x: $\displaystyle 7=r\sin\alpha $ [1]
    of sinx: $\displaystyle 6=r\cos\alpha$ [2]

    Divide [1] by [2] to get:
    $\displaystyle \frac{7}{6}=tan\alpha \Rightarrow \alpha = 49.4^\circ$

    Square and add [1] and [2] to get:
    $\displaystyle 7^2+6^2=r^2(\sin^2\alpha+\cos^2\alpha) \Rightarrow r=\sqrt{85}$

    $\displaystyle \therefore 7\cos x+6\sin x \equiv \sqrt{85}\sin(x+49.4^\circ)$

    Except that last statement doesn't seem to be true, according to
    the answers in my book and when I have a look at the two curves in a
    graphing application. I tried $\displaystyle \sqrt{85}\sin(x-49.4^\circ)$ in the graphing app and
    that seems to be correct...
    but why??

    What have I done wrong? Why am I getting a positive value for
    alpha when I should be getting the same number but negative?

    The only way I can see to get -49.4 would be by introducing a minus
    sign when dividing [1] by [2]. But I don't see how to justify that.
    Have I made a simple error? I can't spot it ...
    Any help would be much appreciated, thanks in advance.
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  2. #2
    MHF Contributor

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    Your result, $\displaystyle \sqrt{85}sin(x+ 49.4)$ is correct. An easy way to check that is to take x= 0. $\displaystyle \sqrt{85} sin(49.4)$= (9.219)(0.759)= 7 while obviously $\displaystyle \sqrt{85} sin(-49.4)$= -7.
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  3. #3
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    Hello, tleave2000!

    I see nothing wrong with your work.
    Why are you convinced that alpha must be negative?

    I was taught a different approach . . .


    Solve: .$\displaystyle 7\cos x +6\sin x \:=\: 2$
    Divide by: $\displaystyle \sqrt{7^2+6^2} \:=\:\sqrt{85}\!:\quad \frac{7}{\sqrt{85}}\cos x + \frac{6}{\sqrt{85}}\sin x \:=\:\frac{2}{\sqrt{85}}$


    Let $\displaystyle \alpha$ be an angle such that: .$\displaystyle \sin\alpha = \tfrac{7}{\sqrt{85}},\;\cos\alpha = \tfrac{6}{\sqrt{85}}$

    Then we have: .$\displaystyle \sin\alpha\cos x + \cos\alpha\sin x \:=\:\frac{2}{\sqrt{85}} $

    . . which is equivalent to: .$\displaystyle \sin(x + \alpha) \:=\:\frac{2}{\sqrt{85}}$


    And we have: .$\displaystyle x + \alpha \:=\:\arcsin\left(\tfrac{2}{\sqrt{85}}\right) \;\approx\;12.53^o $

    . . Then: .$\displaystyle x \;=\;12.53^o - \alpha$


    Since $\displaystyle \sin\alpha = \frac{7}{\sqrt{85}}$, then: .$\displaystyle \alpha \:=\:\arcsin\left(\tfrac{7}{\sqrt{85}}\right) \:\approx\:49.40^o$


    . . Therefore: .$\displaystyle x \;=\;12.53^o - 49.40^o \;=\;-36.87^o $

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  4. #4
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    oops

    Ah thanks. I was convinced it should be negative because when you give degree values to a graph plotting application that's expecting radian values for it's trig functions , then when you compare the graphs of $\displaystyle 7\cos x+6\sin x $ and $\displaystyle \sqrt{85}\sin(x+49.4^\circ)$ they aren't the same... but oddly if you stick a minus sign before the offset, the graphs match up. (Is that just a coincidence?)

    So that's what I was doing wrong, I needed to convert the input by multiplying it by pi/180. Thank you both again, and it was interesting to see an alternative way to get to rsin(x+alpha).
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