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Math Help - Simplifying Trigonometric Expressions

  1. #1
    Senior Member topher0805's Avatar
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    Simplifying Trigonometric Expressions

    How do I simplify this?

    tan(arcsin(x))

    I have to know how to solve these problems by tomorrow morning and I'm dying here! Any help is appreciated.
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  2. #2
    Bar0n janvdl's Avatar
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    Quote Originally Posted by topher0805 View Post
    How do I simplify this?

    tan(arcsin(x))

    I have to know how to solve these problems by tomorrow morning and I'm dying here! Any help is appreciated.
    Always write everything in terms of sin and cos.
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  3. #3
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by topher0805 View Post
    How do I simplify this?

    tan(sin-1(x))

    I have to know how to solve these problems by tomorrow morning and I'm dying here! Any help is appreciated.
    i'll start you off. Let \theta = \sin^{-1} x

    \Rightarrow \sin \theta = x

    so we can draw a right triangle with an acute angle \theta where the opposite side is x and the hypotenuse is 1. by Pythagoras, the adjacent side will be \sqrt{1 - x^2}

    so now, \tan \left( \sin^{-1} x \right) = \tan \theta = ... ?

    ok, so i almost did the entire problem...don't complain, i left the punch line for you
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  4. #4
    Senior Member topher0805's Avatar
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    Quote Originally Posted by janvdl View Post
    Always write everything in terms of sin and cos.
    I tried that and ended up with x/(cos(arcsin(x))). I feel like I have tried everything. I vaguely remember my prof saying something about using a triangle to solve these kinds of problems, but I have no clue what that even means.
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  5. #5
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by topher0805 View Post
    ... I vaguely remember my prof saying something about using a triangle to solve these kinds of problems, but I have no clue what that even means.
    now you know. do you understand how and why i set up the triangle the way i did?

    and what's the answer??
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  6. #6
    Senior Member topher0805's Avatar
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    Quote Originally Posted by Jhevon View Post
    i'll start you off. Let \theta = \sin^{-1} x

    \Rightarrow \sin \theta = x

    so we can draw a right triangle with an acute angle \theta where the opposite side is x and the hypotenuse is 1. by Pythagoras, the adjacent side will be \sqrt{1 - x^2}

    so now, \tan \left( \sin^{-1} x \right) = \tan \theta = ... ?

    ok, so i almost did the entire problem...don't complain, i left the punch line for you
    So this works for all problems like this? Wow, I've never seen it explained so simply before. Thanks for this. This site is great, I'll be posting here a lot more often methinks. (I can help too, not just looking for homework answers)
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  7. #7
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    Quote Originally Posted by topher0805 View Post
    I tried that and ended up with x/(cos(arcsin(x))). I feel like I have tried everything. I vaguely remember my prof saying something about using a triangle to solve these kinds of problems, but I have no clue what that even means.
    Did you draw and label the triangle? The opopsite side is x. The adjacent side is \sqrt{1 - x^2}. \tan \theta = \frac{\text{opposite side}}{\text{adjacent side}} = .......

    If I haven't completely stolen the punch line ....
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  8. #8
    Senior Member topher0805's Avatar
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    Quote Originally Posted by Jhevon View Post
    now you know. do you understand how and why i set up the triangle the way i did?

    and what's the answer??
    x/
    <br />
\sqrt{1 - x^2}<br />

    Oh I see.
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  9. #9
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    should be correct

    tan(arcsin(x))
    let arcsin x =y then sin y =x, cos y =+[1-(sin y)^2]^(1/2) as we give
    arcsin x a domain of [-pi/2, pi/2]
    tan (arcsinx)= [sin (arcsin x)]/ [cos (arcsin x)]= x/(1-x^2)^(1/2)
    sin (arcsinx)= x
    should be correct, but if it is wrong
    work in similiar manner
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  10. #10
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by topher0805 View Post
    So this works for all problems like this?
    pretty much, there may be cases where things have to be treated differently. but definitely for any problem you will get of this type this will work. there are also ways to do them using formulas, and then there's calculus_jy's way.

    Wow, I've never seen it explained so simply before.
    really? i wasn't trying to explain it simple. you're just smart that's what it is. i've explained this even simpler with the aids of diagrams on this forum before. and there are other methods for doing them posted as well. do a search.

    Thanks for this.
    you're welcome
    This site is great
    yes it is!
    I'll be posting here a lot more often methinks. (I can help too, not just looking for homework answers)
    looking forward to your input

    Quote Originally Posted by topher0805 View Post
    x/
    <br />
\sqrt{1 - x^2}<br />

    Oh I see.
    that's correct...i suppose you were trying to type \frac x{\sqrt{1 - x^2}}
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  11. #11
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    how do you manage to type those fractions?
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  12. #12
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    Quote Originally Posted by calculus_jy View Post
    how do you manage to type those fractions?
    Use this latex code:

    [tex]\frac{a}{b}[/tex]

    to get the fraction \frac{a}{b}.
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  13. #13
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    \frac{sqr(xsq+1}{b}
    does not work, where do i find the codes with all the sign like integral multipy ...
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  14. #14
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by calculus_jy View Post
    \frac{sqr(xsq+1}{b}
    does not work, where do i find the codes with all the sign like integral multipy ...
    see our LaTeX Forum to learn how to use LaTeX. start at the tutorial (it is a sticky thread, so it will be the first thread in the list). there is a pdf file in the first post that gives you the codes
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