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Math Help - Show that B = √5 and ѳ = tan ⁻

  1. #1
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    Show that B = √5 and ѳ = tan ⁻

    Hi All,
    Been given this one, and honestly I am looking forward to seeing you guys tackle it.
    Consider the equation, 2 cosx + sinx = 1
    Let 2cosx + sinx = B(cos(x ѳ), where B>0 and 0 < ѳ < 360.
    Show that B = √5 and ѳ = tan ⁻


    Thanks Heaps
    Joel
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  2. #2
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    Quote Originally Posted by Joel View Post
    Hi All,
    Been given this one, and honestly I am looking forward to seeing you guys tackle it.
    Consider the equation, 2 cosx + sinx = 1
    Let 2cosx + sinx = B(cos(x – ѳ), where B>0 and 0 < ѳ < 360.
    Show that B = √5 and ѳ = tan ⁻


    Thanks Heaps
    Joel
    Use the "difference formula": cos(a-b)= cos(a)cos(b)+ sin(a)sin(b). cos(x-\theta)= cos(\theta)cos(x)+ sin(\theta)sin(x) so we would like to have cos(\theta)= 2, sin(\theta)= 1.

    But that's impossible because that would mean that cos^2(\theta)+ sin^2(\theta)= 2^2+ 1= 5 when we know it must be 1. So factor out a \sqrt{5}:

    \sqrt{5}(\frac{2}{\sqrt{5}}cos(x)+ \frac{1}{\sqrt{5}}sin(x))= \sqrt{5}(cos(\theta)cos(x)+ sin(\theta)sin(x)) if and only if cos(\theta)= \frac{2}{\sqrt{5}} and sin(\theta)= \frac{1}{\sqrt{5}}. Now we have cos^2(\theta)+ sin^2(\theta)= \frac{4}{5}+ \frac{1}{5}= 1 so that is possible.

    And, of course, we must have tan(\theta)= \frac{sin(\theta)}{cos(\theta)} = \frac{ \frac{1}{\sqrt{5}} }{ \frac{2}{\sqrt{5}} }= \frac{1}{2}.
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  3. #3
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    Recall that cos(x-A) = cosx cos A + sin x sinA
    Then B cos(x-A) = B cosA cosx + B sinA sinx
    This equals 2 cos x + sin x, so B cosA = 2 and B sin A =1 (equating the terms)
    Then you have 2 simul equations to solve for B and A. (to give B= sqr 5 and A = tan^-1(1/2)
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