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Math Help - Addition Formula question expressing function in terms of Rsin(a+b)

  1. #1
    dix
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    Addition Formula question expressing function in terms of Rsin(a+b)

    I don't know where I'm going wrong. The answer at the end is supposed to be "tan a= 3/4" but going through it step by step I've gone wrong but I don't know how.

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    MHF Contributor
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    Draw a right triangle.
    Designate an acute angle, Angle a.
    Label the legs consistent with the given value of the tangent of 'a'.
    Contemplate the hypotenuse.
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  3. #3
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    Quote Originally Posted by dix View Post
    I don't know where I'm going wrong. The answer at the end is supposed to be "tan a= 3/4" but going through it step by step I've gone wrong but I don't know how.

    You are blindly using a formula.

    Surely if you compare R(\sin (\theta) \cos (a) + \cos (\theta) \sin (a)) with the expression in the question, it is obvious that R sin(a) = 3 and R cos(a) = 4 and so tan(a) = 3/4.
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  4. #4
    dix
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    Quote Originally Posted by TKHunny View Post
    Draw a right triangle.
    Designate an acute angle, Angle a.
    Label the legs consistent with the given value of the tangent of 'a'.
    Contemplate the hypotenuse.
    Thanks TKHunny, I only got this far with your explanation but it sounds intersting if if could help with these types of questions, I can never get them right.
    Quote Originally Posted by mr fantastic View Post
    You are blindly using a formula.

    Surely if you compare R(\sin (\theta) \cos (a) + \cos (\theta) \sin (a)) with the expression in the question, it is obvious that R sin(a) = 3 and R cos(a) = 4 and so tan(a) = 3/4.
    I know that I should surely be able to understand but I don't, I wouldn't be asking if I did. I don't get it
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    Quote Originally Posted by dix View Post
    [snip]
    I know that I should surely be able to understand but I don't, I wouldn't be asking if I did. I don't get it
    What part of what I said do you not understand? Do you understand how to compare the two expressions?
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  6. #6
    dix
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    Quote Originally Posted by mr fantastic View Post
    What part of what I said do you not understand? Do you understand how to compare the two expressions?
    I don't understand where I've went wrong in my step by step thing. I don't understand how R sin(a) = 3 and not 4. It says in the function '...4sin \theta "
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    Yes, that's exactly why it is 3 and not 4. It is sin(a) you want, not sin(\theta).

    Compare Rcos(\theta)sin(a)+ Rsin(\theta)cos(a)
    and 3cos(\theta)+ 4sin(\theta). You should see immediately that you want to have Rsin(a)= 3 and Rcos(a)= 4. Set the coefficients of sin(\theta) and cos(\theta) in the two equations equal.
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  8. #8
    dix
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    Quote Originally Posted by HallsofIvy View Post
    Yes, that's exactly why it is 3 and not 4. It is sin(a) you want, not sin(\theta).

    Compare Rcos(\theta)sin(a)+ Rsin(\theta)cos(a)
    and 3cos(\theta)+ 4sin(\theta). You should see immediately that you want to have Rsin(a)= 3 and Rcos(a)= 4. Set the coefficients of sin(\theta) and cos(\theta) in the two equations equal.
    So in this one: Rcos(\theta)sin(a)+ Rsin(\theta)cos(a) you put a/b (3/4) not b/a (4/3) because sin (a) comes before cos (a)? So the method for doing these questions is to take the general formula of R(\sin (\theta) \cos (a) + \cos (\theta) \sin (a) and swap it around like you have and take it that way?
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  9. #9
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    Quote Originally Posted by dix View Post
    So in this one: Rcos(\theta)sin(a)+ Rsin(\theta)cos(a) you put a/b (3/4) not b/a (4/3) because sin (a) comes before cos (a)? So the method for doing these questions is to take the general formula of R(\sin (\theta) \cos (a) + \cos (\theta) \sin (a) and swap it around like you have and take it that way?
    Don't overthink this.

    You have essentially said at the beginning of the question (correctly):

    Let 3\cos\theta + 4\sin\theta = R\sin\alpha\cos\theta + R\cos\alpha\sin\theta

    In order for both sides of the equation to be equal, both sides have to have 3\cos\theta

    This means that, looking at the right hand side, R\sin\alpha =3.

    Both sides have to have 4\sin\theta.

    So, looking at the right hand side, R\cos\alpha=4

    And then it's just an application of \tan\alpha =\frac{\sin\alpha}{\cos\alpha}}
    Last edited by Quacky; June 4th 2011 at 05:55 AM.
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  10. #10
    dix
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    Thanks Quacky. So you just replace whatever the opposites are? I've done another using you method, is it right?

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  11. #11
    Super Member Quacky's Avatar
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    Quote Originally Posted by dix View Post
    Thanks Quacky. So you just replace whatever the opposites are? I've done another using you method, is it right?

    Not quite. Firstly, you forgot the 'R'

    12\cos\theta+5\sin\theta=R\cos\alpha\cos\theta+R \sin\alpha\sin\theta

    Now look at the left hand side. You have 12\cos\theta

    Now look at the right hand side. You have R\cos\alpha\cos\theta

    This means that R\cos\alpha=12 because you need to have the same amount of \cos\theta on both sides. It doesn't have anything to do with opposites.
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  12. #12
    dix
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    Thanks for the explanation Quacky, I think I finally get it
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