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Thread: simple trig equation

  1. #1
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    simple trig equation

    I am stuck on this trig equation.

    Solve the Equation for one function of $\displaystyle \theta$ and find the angle.

    $\displaystyle sin\theta + 3 cos\theta = 2 \sqrt{2}$

    ok, step one is write in terms of one function. so looking at it I can't see any easy identity to replace. so I raise power on both sides.

    $\displaystyle
    \begin{aligned}
    sin\theta + 3 cos\theta = 2 \sqrt{2}&=(sin\theta + 3 cos\theta)^2 = (2\sqrt{2})^2 \\
    &=sin^2\theta + 6cos\theta sin\theta+9cos^2\theta = 8 \\
    &=sin^2\theta + 6cos\theta sin\theta+9(1-sin^2\theta) = 8 \\
    &=8sin^2\theta - 6cos\theta sin\theta = 1
    \end{aligned}
    $

    if I raise power again it will not square the other cos. and I can't see any elementary identity to use on it. in the book it says the answer should be $\displaystyle \frac{\pi}{4}, 45^{\circ}$
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  2. #2
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  3. #3
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    Quote Originally Posted by skeeter View Post
    If you wish to avoid an error click this link: http://www.mathhelpforum.com/math-he...-a-172498.html
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  4. #4
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    hmm, I took a look but they say to do this problem with no calculator and just the elementary identities.
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  5. #5
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    so I found a way to do this one and get correct solution with just the pencil. but I introduce extraneous solutions so I have to limit the domain.

    $\displaystyle
    \begin{aligned}
    sin\theta +3cos \theta=2\sqrt{2}
    &\equiv (3\sqrt{1-sin^2\theta})^2 = (2\sqrt{2}-sin\theta)^2 \\
    &\equiv 10sin^2\theta-4\sqrt{2}sin\theta-1= 0 \\
    &\equiv (10sin^2\theta-1)^2= (4\sqrt{2}sin\theta)^2 \\
    &\equiv 100sin^4\theta-52sin^2\theta+1= 0 \\
    &\equiv (50sin^2\theta-1)(2sin^2\theta-1) \\
    &\equiv \bigg( sin\theta=\frac{1}{\sqrt{50}} \bigg) \bigg( sin\theta=\frac{1}{\sqrt{2}} \bigg) \\
    & \therefore \theta=\frac{\pi}{4},\frac{3\pi}{4}, \lbrace 0<\theta<2\pi \rbrace \\
    \end{aligned}.
    $
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  6. #6
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    Since the right hand side is positive and greater than 1, sinθ and cosθ must be in the firs quadrant.
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