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Thread: Solving trig equation with identies

  1. #1
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    Solving trig equation with identies

    If sin(2x) =0 tanx - cos 2x, then sinx - cosx =?

    I know of all the double angle identies for sin and cosine, and I know tan=sin/cos.
    I still can't manage to find sinx-cosx
    Perhaps you can get me in the right direction.. I tried subbing a few identites, but apparently not the right ones. And note that this problem isn't for class.... This is from Princeton Review math lvl 2 review.
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    Re: Solving trig equation with identies

    Quote Originally Posted by benny92000 View Post
    If sin(2x) =0 tanx - cos 2x, then sinx - cosx =?

    I know of all the double angle identies for sin and cosine, and I know tan=sin/cos.
    I still can't manage to find sinx-cosx
    Perhaps you can get me in the right direction.. I tried subbing a few identites, but apparently not the right ones. And note that this problem isn't for class.... This is from Princeton Review math lvl 2 review.
    makes no sense ... is that 0 times tanx ?
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    Re: Solving trig equation with identies

    I apologize. There is no 0.
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    Re: Solving trig equation with identies

    If sin(2x) = tanx - cos 2x, then sinx - cosx =?
    $\displaystyle \sin(2x) = \tan{x}-\cos(2x)$

    $\displaystyle \cos(2x) = \tan{x} - \sin(2x)$

    $\displaystyle \cos^2{x} - \sin^2{x} = \tan{x} - \sin(2x)$

    $\displaystyle (\cos{x}-\sin{x})(\cos{x}+\sin{x}) = \tan{x} - \sin(2x)$

    $\displaystyle \cos{x} - \sin{x} = \frac{\tan{x} - \sin(2x)}{\cos{x}+\sin{x}}$

    $\displaystyle \sin{x} - \cos{x} = \frac{\sin(2x) - \tan{x}}{\cos{x}+\sin{x}}$
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    Re: Solving trig equation with identies

    Makes sense. They are looking for a number value, however. Choices are negative square root of 2, 0, 1, 2 times the square root of 2, or not enough information.
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    Re: Solving trig equation with identies

    Hello, benny92000!

    $\displaystyle \text{If }\sin 2x \:=\: \tan x - \cos 2x,\,\text{ then }\,\sin x - \cos x \:= $

    . . $\displaystyle (a)\;\text{-}\sqrt{2} \qquad (b)\;0 \qquad (c)\;1 \qquad (d)\;2\sqrt{2} \qquad (e)\text{ not enough information}$
    Note that: .$\displaystyle x \,\ne\,\tfrac{\pi}{2},\:\tfrac{3\pi}{2}$


    We have: .$\displaystyle \sin2x + \cos2x \:=\:\tan x $

    . $\displaystyle 2\sin x\cos x + 1 - 2\sin^2x \:=\:\tan x$

    . . . . $\displaystyle 2\sin^2\!x - 2\sin x\cos x \:=\:1 - \tan x$

    . . . . $\displaystyle 2\sin x(\sin x - \cos x) \:=\: 1 - \frac{\sin x}{\cos x}$

    . . . . $\displaystyle 2\sin x(\sin x - \cos x) \;=\;\frac{\cos x - \sin x}{\cos x}$


    Multiply by $\displaystyle \cos x\!:$
    n . . . . . . . . . . . . $\displaystyle 2\sin x\cos x(\sin x - \cos x) \;=\;\cos x - \sin x$

    . . $\displaystyle 2\sin x\cos x(\sin x - \cos x) - (\sin x - \cos x) \;=\;0$

    n . . . . . . . . $\displaystyle (\sin x - \cos x)(2\sin x\cos x - 1) \;=\;0$


    And we have two equations: .$\displaystyle \begin{Bmatrix}\sin x - \cos x &=& 0 \\ 2\sin x\cos x -1 &=& 0 \end{Bmatrix}$

    Both equations have the solution: .$\displaystyle x \:=\:\tfrac{\pi}{4} + \pi n$


    Therefore: .$\displaystyle \sin x - \cos x \;=\;0\;\;\;\text{ answer (b)}$

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    Re: Solving trig equation with identies

    That's pretty detailed. Thank you. How did you immediately eliminate pi/2 and 3pi/2 for x?
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    Re: Solving trig equation with identies

    Quote Originally Posted by benny92000 View Post
    That's pretty detailed. Thank you. How did you immediately eliminate pi/2 and 3pi/2 for x?
    Because tan(x) = sin(x)/cos(x) and we can't have cos(x) = 0.
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