# Thread: Inverse trig questions =/

1. ## 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!!

2. Originally Posted by 3deltat
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