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Thread: Trigonometric identities

  1. #1
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    Trigonometric identities

    √3 sinx+cosx=2, solve for x, please explain how I have to do this
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  2. #2
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    Quote Originally Posted by shane99 View Post
    √3 sinx+cosx=2, solve for x, please explain how I have to do this
    $\displaystyle \sqrt{3} \cdot \sin{x} = 2 - \cos{x}$

    square both sides ...

    $\displaystyle 3\sin^2{x} = 4 - 4\cos{x} + \cos^2{x}$

    use the Pythagorean identity ...

    $\displaystyle 3(1 - \cos^2{x}) = 4 - 4\cos{x} + \cos^2{x}$

    $\displaystyle 3 - 3\cos^2{x} = 4 - 4\cos{x} + \cos^2{x}$

    $\displaystyle 0 = 4\cos^2{x} - 4\cos{x} + 1
    $

    $\displaystyle 0 = (2\cos{x} - 1)^2$

    $\displaystyle \cos{x} = \frac{1}{2}$

    $\displaystyle x = \frac{\pi}{3} \, , \, \frac{5\pi}{3}$

    check for extraneous solution(s) (see if they work in the original equation)
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  3. #3
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    Hello, shane99!

    Another approach . . . clever, but obscure.


    Solve for $\displaystyle x\!:\;\;\sqrt{3}\sin x + \cos x \:=\:2$

    Divide by 2: .$\displaystyle \tfrac{\sqrt{3}}{2}\sin x + \tfrac{1}{2}\cos x \;=\;1$

    Note that: .$\displaystyle \begin{array}{ccc}\sin\frac{\pi}{6} &=& \frac{1}{2} \\ \\[-4mm] \cos\frac{\pi}{6} &=& \frac{\sqrt{3}}{2} \end{array}$


    So we have: .$\displaystyle \underbrace{\cos\tfrac{\pi}{6}\sin x + \sin\tfrac{\pi}{6}\cos x} \:=\:1$

    . . . . . . . . . . . . . $\displaystyle \sin(x + \tfrac{\pi}{6}) \:=\:1$

    . . . . . . . . . . . . . . . .$\displaystyle x + \tfrac{\pi}{6} \:=\:\tfrac{\pi}{2}$

    . . . . . . . . . . . . . . . . . . $\displaystyle x \:=\:\frac{\pi}{3}$

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  4. #4
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    Quote Originally Posted by Soroban View Post


    So we have: .$\displaystyle \underbrace{\cos\tfrac{\pi}{6}\sin x + \sin\tfrac{\pi}{6}\cos x} \:=\:1$

    . . . . . . . . . . . . . $\displaystyle \sin(x + \tfrac{\pi}{6}) \:=\:1$
    What I am confused about is that after you plugged in those identities into the equation how did you end up with sin(x+π/6) = 1. Can you explain what you did to get that equation, because where did cos go.
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  5. #5
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    Quote Originally Posted by shane99 View Post
    What I am confused about is that after you plugged in those identities into the equation how did you end up with sin(x+π/6) = 1. Can you explain what you did to get that equation, because where did cos go.
    Take the sine inverse of both sides and you get

    $\displaystyle x + \frac{\pi}{6} = \sin^{-1}{(1)}$

    which equals

    $\displaystyle x + \frac{\pi}{6} = \frac{\pi}{4}$
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