MHB The Calculating $\cos\left({\sin^{-1}\left({2/3}\right)}\right)$

[karush]

$\tan\left({\arcsin\left({\frac{2}{3}}\right)}\right)=\frac{2\sqrt{5}}{5}$
Don't why this is the answer?
Supposed to use
$\cos\left({\sin^{-1}\left({x}\right)}\right)=\sqrt{1-{x}^{2}}$

 

[MarkFL]

We know that $$\arcsin\left(\frac{2}{3}\right)$$ will be an angle in the first quadrant, and can be represented as an angle in a right triangle that has the ratio of the opposite to hypotenuse of 2:3. By the Pythagorean theorem, we then know the adjacent side is $\sqrt{5}$ units. And so the tangent of this angle is $$\frac{2}{\sqrt{5}}$$, and so we may state:

$$\tan\left(\arcsin\left(\frac{2}{3}\right)\right)=\frac{2}{\sqrt{5}}=\frac{2\sqrt{5}}{5}$$

 

[topsquark]

$\tan\left({\arcsin\left({\frac{2}{3}}\right)}\right)=\frac{2\sqrt{5}}{5}$
Don't why this is the answer?
Supposed to use
$\cos\left({\sin^{-1}\left({x}\right)}\right)=\sqrt{1-{x}^{2}}$

Well, you know how to calculate [math]cos(asn(2/3)) = \sqrt{1 - (2/3)^2}[/math]. Consider:
[math]tan \left ( asn \left ( \frac{2}{3} \right ) \right ) = \frac{sin \left ( asn \left ( \frac{2}{3} \right ) \right )}{cos \left ( asn \left ( \frac{2}{3} \right ) \right )}[/math]

You have just calculated [math]cos(asn(2/3))[/math]. All you need is [math]sin(asn(2/3))[/math]. How do you find this?

-Dan

 

[karush]

Won't the $\cos(x)$ cancel out if we use the suggested identity?

 

[topsquark]

Won't the $\cos(x)$ cancel out if we use the suggested identity?

No. Using the hint the tan formula becomes:
[math]tan \left ( asn \left ( \frac{2}{3} \right ) \right ) = \frac{sin \left ( asn \left ( \frac{2}{3} \right ) \right )}{\sqrt{1 - \left ( \frac{2}{3} \right )^2}}[/math]

-Dan