• October 23rd 2009, 02:06 PM
alexandrabel90
Hi!

How do you prove that tan(∅+Φ) = (tan ∅ + tan Φ)/ (1- tanΦtan∅)?

I'm able to prove the formulas for sin and cos (∅+Φ) but when i use sin/ cos = tan, it doesnt give me the above formula..

thanks
• October 23rd 2009, 02:29 PM
Quote:

Originally Posted by alexandrabel90
Hi!

How do you prove that tan(∅+Φ) = (tan ∅ + tan Φ)/ (1- tanΦtan∅)?

I'm able to prove the formulas for sin and cos (∅+Φ) but when i use sin/ cos = tan, it doesnt give me the above formula..

thanks

This is really a topic for the trigonometry section. But all you need to know are the formulae for the sine and cosine of a sum.

$tan(A+B)=\frac{sin(A+B)}{cos(A+B)}=\frac{sinAcosB+ cosAsinB}{cosAcosB-sinAsinB}$

Divide the numerator and the denominator by $cosAcosB$

$\frac{\frac{sinA}{cosA}+\frac{sinB}{cosB}}{1-\frac{sinAsinB}{cosAcosB}}$

$=\frac{tanA+tanB}{1-tan(A)tan(B)}$
• October 23rd 2009, 02:30 PM
tonio
Quote:

Originally Posted by alexandrabel90
Hi!

How do you prove that tan(∅+Φ) = (tan ∅ + tan Φ)/ (1- tanΦtan∅)?

I'm able to prove the formulas for sin and cos (∅+Φ) but when i use sin/ cos = tan, it doesnt give me the above formula..

thanks

Of course it gives the above formula:

$\tan (x+y)=\frac{\sin (x+y)}{\cos (x+y)}=\frac{\sin x\cos y +\sin y\cos x}{\cos x\cos y-\sin x\sin y}$

Now simply multiply the right hand above by $\displaystyle{\frac{\frac{1}{\cos x\cos y}}{\frac{1}{\cos x\cos y}}=1}$ and develop.

Tonio