How is ## \tan ( \arcsin x) = \frac{x}{\sqrt{1-x^2}} ## ?

[Rectifier]

How can I write $\tan ( \arcsin x)$ as $\frac{x}{\sqrt{1-x^2}}$? This is not a problem in itself but a part of a solution to a problem.

 

[jedishrfu]

Start by using the sin / cos definition of tan and you should begin to see the derivation.

 

[Rectifier]

$\tan ( \arcsin x) = \frac{\sin ( \arcsin x)}{\cos ( \arcsin x)} = \frac{x}{\cos ( \arcsin x)}$

I draw a triangle with sides x and 1. Therefore the hypotenuse is $\sqrt{1+x^2}$.

$\cos \left( v \right) = \frac{1}{\sqrt{1+x^2}}$

This is where I get stuck. Please help :,(

 

[Chestermiller]

If $\theta=arcsin x$, what is $\sin \theta?$

 

[Mastermind01]

$\tan ( \arcsin x) = \frac{\sin ( \arcsin x)}{\cos ( \arcsin x)} = \frac{x}{\cos ( \arcsin x)}$

I draw a triangle with sides x and 1. Therefore the hypotenuse is $\sqrt{1+x^2}$.

$\cos \left( v \right) = \frac{1}{\sqrt{1+x^2}}$

This is where I get stuck. Please help :,(

You drew the wrong triangle. You need a triangle with an angle whose sine is x. Then which of the sides needs to be x and 1?

 

[Rectifier]

If $\theta = \arcsin (\sin x)$

$x$ is the opposite side of the angle $\theta$ and 1 is the hypotenuse?.

 

[Mastermind01]

If $\theta = \arcsin (\sin x)$

$x$ is the opposite side of the angle $\theta$ and 1 is the hypotenuse?.

Does that get you the answer?

 

[Chestermiller]

Does that get you the answer?

If $\theta = \arcsin (\sin x)$

$x$ is the opposite side of the angle $\theta$ and 1 is the hypotenuse?.

No. What is the sine of the angle whose sine is x?

 

[Mastermind01]

No. What is the sine of the angle whose sine is x?

I think he got it correct this time.

 

[Rectifier]

Hmm...
$\theta = \arcsin (\sin x)$

Then $\sin \theta = \sin x$

But this looks wrong

Sorry, this is hard for me :S

 

[Mastermind01]

Hmm...
$\theta = \arcsin (\sin x)$

Then $\sin \theta = \sin x$

But this looks wrong

Sorry, this is hard for me :S

Well you drew the triangle correct, so what's $cos(\theta)$ ?

 

[Rectifier]

Wait, so one side of the triangle is equal to the angle?

$\cos x = \sqrt{1-x^2}$

 

[Mastermind01]

Wait, so one side of the triangle is equal to the angle?

$\cos x = \sqrt{1-x^2}$

So $tan(\theta)$ is?

 

[Rectifier]

$\frac{x}{\sqrt{1-x^2}}$ :D

 

[Chestermiller]

Hmm...
$\theta = \arcsin (\sin x)$

Then $\sin \theta = \sin x$

But this looks wrong

Sorry, this is hard for me :S

No. $sin \theta = x$

 

[Chestermiller]

In post #8, I asked you "What is the sine of the angle whose sine is x?" You answered sin x. What would your answer have been if I asked "What is the name of the person whose name is John?"

 

[Rectifier]

In post #8, I asked you "What is the sine of the angle whose sine is x?" You answered sin x. What would your answer have been if I asked "What is the name of the person whose name is John?"

It is x :)

 

[Chestermiller]

It is x :)

Good. So, in terms of x, what is cos theta? Then , in terms of x, what is tan theta?