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Thread: Find solutions for SIN(x)=-a, where a=SIN(α) for 0<α<pi/2

  1. #1
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    Find solutions for SIN(x)=-a, where a=SIN(α) for 0<α<pi/2

    Hey guys, really struggling understanding how to go about this question :/

    The full question is:
    "Let SIN(α)=a, where α is an element of (0, pi/2). Find, in terms of α, two values of x in [0, 2pi] which satisfies the equation SIN(x)= -a"

    I can see that SIN(x)=-SIN(α), but where do I go from here?

    We know that α is an angle in the 1st Quadrant, hence SIN(α) is positive.

    The answers are: pi+α, and 2pi-α

    Thanks in advance for any guidance,
    Regards
    Smeato
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    Re: Find solutions for SIN(x)=-a, where a=SIN(α) for 0<α<pi/2

    Have you heard of the CAST rule in trigonometry? You may not have heard it called this. Google it and see.
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  3. #3
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    Re: Find solutions for SIN(x)=-a, where a=SIN(α) for 0<α<pi/2

    Quote Originally Posted by Smeato View Post
    The full question is:
    "Let SIN(α)=a, where α is an element of (0, pi/2). Find, in terms of α, two values of x in [0, 2pi] which satisfies the equation SIN(x)= -a"
    I can see that SIN(x)=-SIN([FONT=arial]α), but where do I go from here?
    Surely you know about reference angles. Find the reference angles in $III~\&~IV$ quadrants that are equivalent to the given angle $\alpha$.
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  4. #4
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    Re: Find solutions for SIN(x)=-a, where a=SIN(α) for 0<α<pi/2

    reference angles = $\alpha$ in each quadrant of the unit circle

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