# Trig identity proof

• Feb 5th 2010, 08:58 AM
Tweety
Trig identity proof
prove that $tan50^{o} -tan40^{o} = 2tan10^{o}$
• Feb 5th 2010, 09:45 AM
red_dog
Use these formulas:

$\tan a-\tan b=\frac{\sin(a-b)}{\cos a\cos b}$

and $\cos a\cos b=\frac{1}{2}[\cos(a+b)+\cos(a-b)]$
• Feb 5th 2010, 10:35 AM
Tweety
Quote:

Originally Posted by red_dog
Use these formulas:

$\tan a-\tan b=\frac{\sin(a-b)}{\cos a\cos b}$

and $\cos a\cos b=\frac{1}{2}[\cos(a+b)+\cos(a-b)]$

thanks,

but I have never seen these formula's or been taught them. where did you get them from?

surely If I needed to use these formula's It would be in the book, that the question is from?
• Feb 5th 2010, 11:34 AM
masters
Quote:

Originally Posted by Tweety
thanks,

but I have never seen these formula's or been taught them. where did you get them from?

surely If I needed to use these formula's It would be in the book, that the question is from?

Prove that $\tan 50 - \tan 40 = 2\tan 10$

Hi Tweety,

I don't know what formulas you have been taught, but here's another proof using the Tan sum formula:

$\tan(u+v)=\frac{\tan u+\tan v}{1-\tan u \tan v}$.

$\tan (50) = \tan (40+10)= \frac{\tan 40 + \tan 10}{1- \tan 40 \tan 10}$

$\tan 50 (1- \tan 40 \tan 10) = \tan 40 + \tan 10
$

Now, distribute:

$\tan 50 - {\color{red}\tan 50 \tan 40} \tan 10 = \tan 40 + \tan 10$

You know that ${\color{red}\tan 50 \tan 40 = 1}$

$\tan 50 - \tan 10= \tan 40 + \tan 10$

$\tan 50 = \tan 40 + 2 \tan 10$

$\tan 50 - \tan 40 = 2 \tan 10$