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Thread: Trigonometry Help [Q1]

  1. #1
    Member looi76's Avatar
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    Trigonometry Help [Q1]

    Question:
    Given $\displaystyle x = \arcsin \frac{1}{3}$ , find the exact value of:
    a. $\displaystyle \cos x$
    b. $\displaystyle \tan x$
    c. $\displaystyle \cot x$
    d. $\displaystyle \sec x$
    e. $\displaystyle \csc x$

    Attempt:

    a. $\displaystyle \cos x$

    $\displaystyle = \cos (x = \arcsin \frac{1}{3})$

    $\displaystyle = 0.942809041$

    $\displaystyle \color{red}{= 2 \frac{\sqrt{2}}{3}}$

    How can I get the fraction answer?
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  2. #2
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    Quote Originally Posted by looi76 View Post
    Question:
    Given $\displaystyle x = \arcsin \frac{1}{3}$ , find the exact value of:
    a. $\displaystyle \cos x$
    b. $\displaystyle \tan x$
    c. $\displaystyle \cot x$
    d. $\displaystyle \sec x$
    e. $\displaystyle \csc x$

    Attempt:

    a. $\displaystyle \cos x$

    $\displaystyle = \cos (x = \arcsin \frac{1}{3})$

    $\displaystyle = 0.942809041$

    $\displaystyle \color{red}{= 2 \frac{\sqrt{2}}{3}}$

    How can I get the fraction answer?
    $\displaystyle x= \sin^{-1}\left(\frac{1}{3}\right)$

    $\displaystyle \Rightarrow \sin x =\frac{1}{3}$

    $\displaystyle \cos x = \sqrt{1-\sin^2 x}=\sqrt{1-\left(\frac{1}{3}\right)^2 }=\sqrt{\frac{8}{9}}=\frac{2\sqrt{2}}{3}$

    $\displaystyle \tan x = \frac{\sin x}{\cos x}=\frac{.......}{.......}$ finish it.

    $\displaystyle \cot x = \frac{1}{\tan x}=\frac{1}{.......}$ finish it.

    $\displaystyle \sec x = \frac{1}{\cos x}=\frac{1}{.......}$ finish it.

    $\displaystyle \csc x = \frac{1}{\sin x}=\frac{1}{.......}$ finish it.

    got it ???
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by Shyam View Post
    $\displaystyle x= \sin^{-1}\left(\frac{1}{3}\right)$

    $\displaystyle \Rightarrow \sin x =\frac{1}{3}$

    $\displaystyle \cos x = \sqrt{1-\sin^2 x}=\sqrt{1-\left(\frac{1}{3}\right)^2 }=\sqrt{\frac{8}{9}}=\frac{2\sqrt{2}}{3}$

    $\displaystyle \tan x = \frac{\sin x}{\cos x}=\frac{.......}{.......}$ finish it.

    $\displaystyle \cot x = \frac{1}{\tan x}=\frac{1}{.......}$ finish it.

    $\displaystyle \sec x = \frac{1}{\cos x}=\frac{1}{.......}$ finish it.

    $\displaystyle \csc x = \frac{1}{\sin x}=\frac{1}{.......}$ finish it.

    got it ???
    $\displaystyle \cos x = \pm \sqrt{1-\sin^2 x}=\pm \sqrt{1-\left(\frac{1}{3}\right)^2 }=\pm \sqrt{\frac{8}{9}}=\pm \frac{2\sqrt{2}}{3}$

    CB
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