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Thread: [SOLVED] inverse of angles

  1. #1
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    [SOLVED] inverse of angles

    Hey everyone,
    I have a test tomorrow and having some difficulty with inverses. so if anyone can help, I'd really appreciate it. Also, I am a visual learner so if it is shown I understand a lot more and can able to apply the same principles I have seen to other problems.

    So can someone show me how to do these:

    1) Find the inverse function for

    , with .

    Answer:


    2) Evaluate each of the inverse trig functions in radians rounded to 4 decimal places


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  2. #2
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    Hello lsnyder
    Quote Originally Posted by lsnyder View Post
    1) Find the inverse function for

    , with .

    Answer:
    $\displaystyle 0\le x\le\frac{\pi}{14}\Rightarrow 0\le 7x\le \frac{\pi}{2}$ and $\displaystyle f(x)$ is therefore is a one-to-one function, with values in the range $\displaystyle 0\le f(x) \le 1$. So we can find its inverse, using $\displaystyle \arccos$, taking the angle in the range $\displaystyle 0$ to $\displaystyle \frac{\pi}{2}$.

    So, let $\displaystyle y = f(x) = \cos(7x)$

    Then $\displaystyle \arccos(y) = 7x$

    $\displaystyle \Rightarrow x = \tfrac17\arccos(y)$

    So the inverse function (replacing $\displaystyle y$ by $\displaystyle x$) is $\displaystyle f^{-1}(x) = \tfrac17\arccos(x)$

    2) Evaluate each of the inverse trig functions in radians rounded to 4 decimal places


    I assume that you have a calculator that will give you inverse tangents and inverse sines. Then:

    $\displaystyle x = \text{arccot} (1.5)$

    $\displaystyle \Rightarrow\cot(x) = 1.5$

    $\displaystyle \Rightarrow \tan(x) = \frac{1}{1.5} = \frac23$

    $\displaystyle \Rightarrow x = \arctan\Big(\frac23 \Big) = 0.5880$ to 4 d.p.

    And $\displaystyle x = \csc^{-1}(8)$

    $\displaystyle \Rightarrow \csc(x) = 8$

    $\displaystyle \Rightarrow \sin(x) = \tfrac18 $

    $\displaystyle x = 0.1253$ to 4 d.p.

    Grandad
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