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Thread: Express in terms of x:sin(...

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

    I have such a problem:

    a) Find$\displaystyle \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 $\displaystyle x:sin(2tan^{-1}(x))$

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

    $\displaystyle 8.88:sin(2tan^{-1}(8.88))$, 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 $\displaystyle \theta = \tan^{-1}x \Longrightarrow x = \frac x1 = \tan \theta$

    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 $\displaystyle \theta$, the opposite side is $\displaystyle x$ and the adjacent side is $\displaystyle 1$. You can use Pythagoras to find the other side. With the fully populated triangle, you can find all trig functions of $\displaystyle \theta$. 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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    Senior 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 $\displaystyle \theta = \tan^{-1}x \Longrightarrow x = \frac x1 = \tan \theta$

    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 $\displaystyle \theta$, the opposite side is $\displaystyle x$ and the adjacent side is $\displaystyle 1$. You can use Pythagoras to find the other side. With the fully populated triangle, you can find all trig functions of $\displaystyle \theta$. Would you know how to proceed?
    OK, so by Pthagorian's we have the hypotenuse = $\displaystyle \sqrt{1 + x^{2}}$, but further...

    we would have $\displaystyle \sin2 \frac{x}{\sqrt{1 + x^{2}}}$ am I right... do I need to do something else?
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    Senior 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 . . .


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

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

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


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

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

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

    Multiply by $\displaystyle \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}} $

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

    . . . . . . $\displaystyle =\;\frac{8\sqrt{21} + 25\sqrt{3}}{9}$
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    Senior 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 $\displaystyle \alpha = \sin^{-1}\!\left(\tfrac{2}{5}\right) \quad\Rightarrow\quad \sin\alpha = \tfrac{2}{5} \quad\Rightarrow\quad \tan\alpha = \tfrac{2}{\sqrt{21}}$ .[1]

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


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

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

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

    Multiply by $\displaystyle \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}} $

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

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