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Math Help - Inverse trig questions =/

  1. #1
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    Inverse trig questions =/

    Alright. While doing homework i stumbled on these questions. If you can assist me on some, it would be greatly appreciated:

    1) Prove that Cos(arcsinx) = (1-x^2)^(1/2)

    Find the derivative and simplify:

    2) y = (x)(arccosx) - (1-x^2)^(1/2)

    3) y = arctan(x-(1+x^2)^(1/2))

    4) y = arctan (x/9) + ln9 ( (x-9)/(x+9) ) ^ (1/2) )

    and find the limit of

    5) Lim as x goes to 0 from the right, arctan(lnx)


    Any help would be very much-ly received and appreciated!!
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by 3deltat View Post
    Alright. While doing homework i stumbled on these questions. If you can assist me on some, it would be greatly appreciated:

    1) Prove that Cos(arcsinx) = (1-x^2)^(1/2)
    Draw a triangle that represents the angle \vartheta=\sin^{-1}x. We see that the side opposite the angle has a value of x, and the hypotenuse of the triangle has a value of 1. This implies that the adjacent has the value of a^2+x^2=1^2\implies a=\sqrt{1-x^2}

    Thus, we see that the \cos\vartheta=\frac{adj}{hyp}=\frac{\sqrt{1-x^2}}{1}=\color{red}\boxed{\sqrt{1-x^2}}

    Find the derivative and simplify:

    2) y = (x)(arccosx) - (1-x^2)^(1/2)
    Keep in mind that \frac{d}{dx}\cos^{-1}x=-\frac{d}{dx}\sin^{-1}x=-\frac{1}{\sqrt{1-x^2}}. You will also need to apply the product rule to the x\cos^{-1}x term.

    3) y = arctan(x-(1+x^2)^(1/2))
    Keep in mind that \frac{d}{dx}\tan^{-1}u=\frac{1}{1+u^2}\cdot\frac{\,du}{\,dx}

    4) y = arctan (x/9) + ln( ( (x-9)/(x+9) ) ^ (1/2) )
    Again, keep in mind that \frac{d}{dx}\tan^{-1}u=\frac{1}{1+u^2}\cdot\frac{\,du}{\,dx}, but recall that \frac{d}{dx}\ln(u)=\frac{1}{u}\cdot\frac{\,du}{\,d  x}

    and find the limit of

    5) Lim as x goes to 0 from the right, arctan(lnx)


    Any help would be very much-ly received and appreciated!!
    \lim_{x\to{0^+}}\tan^{-1}\left(\ln x\right)

    Note that as x\to{0^+}, \ln x\to{-\infty}

    In essence, we are finding \tan^{-1}(-\infty)

    This is the same thing as asking what value of theta causes \tan(\vartheta)=-\infty?

    I'll let you think about that.

    I hope this helps!

    --Chris
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