tan (A + B)

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May 5th 2009, 09:18 PM
magentarita

tan (A + B)

If A = arctan(2/3) and B = arctan(1/2), find tan (A + B).
May 5th 2009, 09:35 PM
Prove It

Quote:

Originally Posted by magentarita

If A = arctan(2/3) and B = arctan(1/2), find tan (A + B).

Hint:

$\tan{(\alpha + \beta)} = \frac{\tan{\alpha} + \tan{\beta}}{1 - \tan{\alpha}\tan{\beta}}$ .
May 6th 2009, 05:23 AM
magentarita

I know but...

Quote:

Originally Posted by Prove It

Hint:

$\tan{(\alpha + \beta)} = \frac{\tan{\alpha} + \tan{\beta}}{1 - \tan{\alpha}\tan{\beta}}$ .

I know this is the correct formula to use. My problem concerns the arctan values given. I also know that arctan means tangent inverse but how do I use the given arctan values in this formula?
May 6th 2009, 02:43 PM
Prove It

$\tan{(\arctan{\theta})} = \theta$ .
May 6th 2009, 06:16 PM
magentarita

are you

Quote:

Originally Posted by Prove It

$\tan{(\arctan{\theta})} = \theta$ .

Are you saying that arctan(2/3) = 2/3 and that
arctan(1/2) = 1/2?

If that's the case, then A = 2/3 and B = 1/2. Is this what needs to plugged into the formula for tan (A + B)?
May 6th 2009, 06:55 PM
Chris L T521

Quote:

Originally Posted by magentarita

Are you saying that arctan(2/3) = 2/3 and that
arctan(1/2) = 1/2?

If that's the case, then A = 2/3 and B = 1/2. Is this what needs to plugged into the formula for tan (A + B)?

No. What he's saying is that $\tan\!\left(\arctan\!\left(\tfrac{2}{3}\right)\rig ht)=\tfrac{2}{3}$ and $\tan\!\left(\arctan\!\left(\tfrac{1}{2}\right)\rig ht)=\tfrac{1}{2}$ .

Your $A$ and $B$ values are still those arctangent values, but when you substitute them into your identity, you take into consideration what ProveIt told you (and what I elaborated above).
May 7th 2009, 06:05 AM
magentarita

can you

Quote:

Originally Posted by Chris L T521

No. What he's saying is that $\tan\!\left(\arctan\!\left(\tfrac{2}{3}\right)\rig ht)=\tfrac{2}{3}$ and $\tan\!\left(\arctan\!\left(\tfrac{1}{2}\right)\rig ht)=\tfrac{1}{2}$ .

Your $A$ and $B$ values are still those arctangent values, but when you substitute them into your identity, you take into consideration what ProveIt told you (and what I elaborated above).

Can you show me how this is done?
May 7th 2009, 06:12 AM
mr fantastic

Quote:

Originally Posted by magentarita

Can you show me how this is done?

$A = \arctan \left( \frac{2}{3} \right) \Rightarrow \tan A = \frac{2}{3}$ .

$B = \arctan \left( \frac{1}{2} \right) \Rightarrow \tan B = \frac{1}{2}$ .
May 7th 2009, 06:13 AM
magentarita

ok

Quote:

Originally Posted by mr fantastic

$A = \arctan \left( \frac{2}{3} \right) \Rightarrow \tan A = \frac{2}{3}$ .

$B = \arctan \left( \frac{1}{2} \right) \Rightarrow \tan B = \frac{1}{2}$ .

This is exactly what I thought.