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Math Help - Solving a*cos(x) + b*sin(x) = 1

  1. #1
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    Solving a*cos(x) + b*sin(x) = 1

    Hi! The solution to this question may be completely trivial, but I still managed to get stuck on this equation. So I wish to solve:

    a \cos(x) + b \sin(x) = 1

    for x. Hence, I want to express x as a function of a and b. And for which values of a and b does there exist a solution?
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  2. #2
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    I found that on the unit circle, i.e. a^2 + b^2 = 1, one finds the solution:

    a = \cos(x)
    b = \sin(x)
    \Rightarrow~x = \tan^{-1}\left(\frac{b}{a}\right)

    Is this the only solution to the problem?
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by sitho View Post
    Hi! The solution to this question may be completely trivial, but I still managed to get stuck on this equation. So I wish to solve:

    a \cos(x) + b \sin(x) = 1

    for x. Hence, I want to express x as a function of a and b. And for which values of a and b does there exist a solution?
    Put \tan(\theta)=a/b then:

    a \cos(x)+b \sin(x)=\frac{1}{\sqrt{a^2+b^2}}[\sin(\theta)\cos(x)+\cos(\theta)\sin(x)]=\frac{1}{\sqrt{a^2+b^2}}\sin(x+\theta)=1

    For this to have any real solutions requires that a^2+b^2 \le 1.

    CB
    Last edited by CaptainBlack; April 19th 2010 at 04:08 AM. Reason: reality check
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