cos(tan^(-1) v)

**magentarita** · November 12th 2008, 07:13 AM

Show that cos(tan^(-1) v) = 1/(sqrt{1 + v^2})

**Chris L T521** · November 12th 2008, 07:18 AM

Originally Posted by magentarita

Show that cos(tan^(-1) v) = 1/(sqrt{1 + v^2})

Let $\vartheta=\tan^{-1}v$

We can construct a right triangle representing this angle.

On the opposite side of the angle, it has a value of $\text{opp}=v$ , and the adjacent side has a value of $\text{adj}=1$ . Now, we can find the value of the hypotenuse:

$(\text{hyp})^2=(\text{opp})^2+(\text{adj})^2\impli es (\text{hyp})^2=v^2+1^2\implies \text{hyp}=\sqrt{v^2+1}$

Now, what do you think $\cos\vartheta$ equals?

--Chris

**HallsofIvy** · November 12th 2008, 07:44 AM

I prefer the way Chris did this but since he got in first (and I just have to put in my oar!) here's another way:

Use trig identities. [tex]tan^2(\theta)+1= sec^2(\theta)= 1/cos^2(\theta)[/itex] so $cos(\theta)= \sqrt{\frac{1}{1+ tan^2(\theta)}= \frac{1}{\sqrt{1+ tan^2(\theta)}$

Replacing $\theta$ by $tan^{-1}(v)$ , that gives [tex]cos(tan^{-1}(v))= \frac{1}{\sqrt{1+ tan^2(tan^{1}(v))}= \frac{1}{\sqrt{1+ v^2}}.

**magentarita** · November 12th 2008, 04:43 PM

Originally Posted by Chris L T521

Let $\vartheta=\tan^{-1}v$

We can construct a right triangle representing this angle.

On the opposite side of the angle, it has a value of $\text{opp}=v$ , and the adjacent side has a value of $\text{adj}=1$ . Now, we can find the value of the hypotenuse:

$(\text{hyp})^2=(\text{opp})^2+(\text{adj})^2\impli es (\text{hyp})^2=v^2+1^2\implies \text{hyp}=\sqrt{v^2+1}$

Now, what do you think $\cos\vartheta$ equals?

--Chris

cos(theta) = sqrt{v^2 + 1}, right?

**Chris L T521** · November 12th 2008, 06:34 PM

Originally Posted by magentarita

cos(theta) = sqrt{v^2 + 1}, right?

You almost have it!

Note that $\cos\vartheta=\frac{\text{adj}}{\text{hyp}}$

They adjacent value is $\text{adj}=1$ and the hypotenuse value is $\text{hyp}=\sqrt{1+v^2}$

So $\cos\vartheta=\frac{\text{adj}}{\text{hyp}}=\dots$

--Chris

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