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Math Help - tan

  1. #1
    Senior Member pacman's Avatar
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    tan

    Without the aid of a calculator, what is the value of this expresion, tan(arcsin 1/3 + arccos 1/2)?
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  2. #2
    MHF Contributor Calculus26's Avatar
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    Use formula for tan(a+b)




    Note tan(arcsin(1/3)) = 1/sqrt(8) Draw a rt triangle with an angle whose sine is 1/3 you should get a 1-sqrt(8)- 3 triangle

    tan(arcos(1/2)) = sqrt(3) Draw a rt triangle with an angle whose cosine is 1/2 you should get a 1 - sqrt(3) - 2 triangle
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  3. #3
    MHF Contributor red_dog's Avatar
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    Quote Originally Posted by Calculus26 View Post

    Note tan(arcsin(1/3)) = 1/sqrt(8) Draw a rt triangle with an angle whose sine is 1/3 you should get a 1-sqrt(8)- 3 triangle

    tan(arcos(1/2)) = sqrt(3) Draw a rt triangle with an angle whose cosine is 1/2 you should get a 1 - sqrt(3) - 2 triangle
    \arcsin x=\alpha\Rightarrow\sin\alpha=x, \alpha\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right]

    \tan\alpha=\frac{\sin\alpha}{\cos\alpha}=\frac{\si  n\alpha}{\sqrt{1-\sin^2\alpha}}

    Then \tan(\arcsin x)=\frac{x}{\sqrt{1-x^2}}

    \arccos x=\beta\Rightarrow \cos\beta=x, \ \beta\in[0,\pi]

    \tan\beta=\frac{\sqrt{1-\cos^2\beta}}{\cos\beta}

    Then \tan(\arccos x)=\frac{\sqrt{1-x^2}}{x}
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  4. #4
    Senior Member pacman's Avatar
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    Red_dog, your personal message is awesome, the solution was step by step. thanks

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