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Math Help - Express in terms of x:sin(...

  1. #1
    Member dokrbb's Avatar
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    Express in terms of x:sin(...

    I have such a problem:

    a) Find \tan [sin^{-1} (\frac{2}{5}) + cos^{-1} (\frac{5}{10})] =

    (Make sure your answer is an algebraic expression with square roots but without trigonometric or inverse trigonometric functions.)

    Well, I have got the answer (how could I not, with a calculator it wasn't a big deal)

    but I'm not sure about this question - I don't really understand what I need to do?

    b) Express in terms of x:sin(2tan^{-1}(x))

    do I have to plug in the value obtained in a) or what

    8.88:sin(2tan^{-1}(8.88)), doesn't make sense for me...
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Re: Express in terms of x:sin(...

    I'd consider (b) as separate problem. Here is the trick. Let \theta = \tan^{-1}x \Longrightarrow x = \frac x1 = \tan \theta

    So now you can proceed algebraically using trig identities, or geometrically and make your own right triangle. Tangent = opposite over adjacent, and so you can make a right triangle where one of the acute angles is \theta, the opposite side is x and the adjacent side is 1. You can use Pythagoras to find the other side. With the fully populated triangle, you can find all trig functions of \theta. Would you know how to proceed?
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    is up to his old tricks again! Jhevon's Avatar
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    Re: Express in terms of x:sin(...

    BTW, for part (a), your calculator gave you an algebraic answer or a decimal expansion? Because you were asked for the former.
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    Member dokrbb's Avatar
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    Re: Express in terms of x:sin(...

    Quote Originally Posted by Jhevon View Post
    I'd consider (b) as separate problem. Here is the trick. Let \theta = \tan^{-1}x \Longrightarrow x = \frac x1 = \tan \theta

    So now you can proceed algebraically using trig identities, or geometrically and make your own right triangle. Tangent = opposite over adjacent, and so you can make a right triangle where one of the acute angles is \theta, the opposite side is x and the adjacent side is 1. You can use Pythagoras to find the other side. With the fully populated triangle, you can find all trig functions of \theta. Would you know how to proceed?
    OK, so by Pthagorian's we have the hypotenuse = \sqrt{1 + x^{2}}, but further...

    we would have \sin2 \frac{x}{\sqrt{1 + x^{2}}} am I right... do I need to do something else?
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    Member dokrbb's Avatar
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    Re: Express in terms of x:sin(...

    Quote Originally Posted by Jhevon View Post
    BTW, for part (a), your calculator gave you an algebraic answer or a decimal expansion? Because you were asked for the former.
    dose that mean that I wasn't supposed to use the calculator, I'm working in webwork and, anyways, if I enter expressions in form of radicals and fractions it gives me the equivalent in decimal expansion - I obtained the decimal expansion,

    was I supposed to apply the same method that you showed me for b) ?
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    Re: Express in terms of x:sin(...

    Hello, dokrbb!

    Here's the first one . . .


    \text{(a) Find }\:\tan \left[\sin^{-1}\!\left(\tfrac{2}{5}\right) + \cos^{-1}\!\left(\tfrac{5}{10}\right)\right]

    Let \alpha = \sin^{-1}\!\left(\tfrac{2}{5}\right) \quad\Rightarrow\quad \sin\alpha = \tfrac{2}{5} \quad\Rightarrow\quad \tan\alpha = \tfrac{2}{\sqrt{21}} .[1]

    Let \beta = \cos^{-1}\!\left(\tfrac{1}{2}\right) \quad\Rightarrow\quad \cos\beta = \tfrac{1}{2} \quad\Rightarrow\quad \tan\beta = \sqrt{3} .[2]


    We have: . \tan(\alpha + \beta) \;=\;\frac{\tan\alpha + \tan\beta}{1 - \tan\alpha\tan\beta}

    Substitute [1] and [2]: . \frac{\frac{2}{\sqrt{21}} + \sqrt{3}}{1-\left(\frac{2}{\sqrt{21}}\right)\left(\sqrt{3} \right)}

    Multiply by \tfrac{\sqrt{21}}{\sqrt{21}}:\;\;\frac{2 + \sqrt{63}}{\sqrt{21} + 2\sqrt{3}} \;=\; \frac{2+3\sqrt{7}}{\sqrt{21} - 2\sqrt{3}}

    Multiply by \tfrac{\sqrt{21}+2\sqrt{3}}{\sqrt{21}+2\sqrt{3}} :\;\;\frac{2+3\sqrt{7}}{\sqrt{21} - 2\sqrt{3}} \cdot \frac{\sqrt{21} + 2\sqrt{3}}{\sqrt{21} + 2\sqrt{3}}

    . . . . . . =\;\frac{2\sqrt{21} + 4\sqrt{3} + 21\sqrt{3} + 6\sqrt{21}}{21 - 12}

    . . . . . . =\;\frac{8\sqrt{21} + 25\sqrt{3}}{9}
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    Member dokrbb's Avatar
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    Re: Express in terms of x:sin(...

    Quote Originally Posted by Soroban View Post
    Hello, dokrbb!

    Here's the first one . . .



    Let \alpha = \sin^{-1}\!\left(\tfrac{2}{5}\right) \quad\Rightarrow\quad \sin\alpha = \tfrac{2}{5} \quad\Rightarrow\quad \tan\alpha = \tfrac{2}{\sqrt{21}} .[1]

    Let \beta = \cos^{-1}\!\left(\tfrac{1}{2}\right) \quad\Rightarrow\quad \cos\beta = \tfrac{1}{2} \quad\Rightarrow\quad \tan\beta = \sqrt{3} .[2]


    We have: . \tan(\alpha + \beta) \;=\;\frac{\tan\alpha + \tan\beta}{1 - \tan\alpha\tan\beta}

    Substitute [1] and [2]: . \frac{\frac{2}{\sqrt{21}} + \sqrt{3}}{1-\left(\frac{2}{\sqrt{21}}\right)\left(\sqrt{3} \right)}

    Multiply by \tfrac{\sqrt{21}}{\sqrt{21}}:\;\;\frac{2 + \sqrt{63}}{\sqrt{21} + 2\sqrt{3}} \;=\; \frac{2+3\sqrt{7}}{\sqrt{21} - 2\sqrt{3}}

    Multiply by \tfrac{\sqrt{21}+2\sqrt{3}}{\sqrt{21}+2\sqrt{3}} :\;\;\frac{2+3\sqrt{7}}{\sqrt{21} - 2\sqrt{3}} \cdot \frac{\sqrt{21} + 2\sqrt{3}}{\sqrt{21} + 2\sqrt{3}}

    . . . . . . =\;\frac{2\sqrt{21} + 4\sqrt{3} + 21\sqrt{3} + 6\sqrt{21}}{21 - 12}

    . . . . . . =\;\frac{8\sqrt{21} + 25\sqrt{3}}{9}
    that's perfect, thanks a lot - and I already figured how to do the second one: I used sin(2x) = 2sin(x)cos(x)
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