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C3 Trig Functions Help Watch

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    Hi,

    There is a question (no. 7) about an inverse trig function which I really don't understand how to do. I'd be really grateful if someone could explain.

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    Recall that:
    \cos{\theta}=\frac{A}{H}
    By definition, the hypothenuse has a greater length than the adjacent side, so H>A. The ration between A and H will always be within [0,1], but with the unit circle, A can be negative, so it will always be within [-1,1], hence the domain of the function.
    Given that cos^{-1}{\theta}=\arccos{\theta} is the inverse function of \cos{\theta}, the domain of \arccos{\theta} is the range of \cos{\theta}, and the range of the first is the domain of the second.
    For ii), you could draw a triangle in a unit circle where A<0 such that \frac{A}{H}=-0.5, and find \arccos{\theta} from there.
    For iii), the inverse function is a matter of flipping y and x and isolating y.
 
 
 
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