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

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- Oct 23rd 2009, 01:06 PMalexandrabel90addition theorems
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 - Oct 23rd 2009, 01:29 PMadkinsjr
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.

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

Divide the numerator and the denominator by $\displaystyle cosAcosB$

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

$\displaystyle =\frac{tanA+tanB}{1-tan(A)tan(B)}$ - Oct 23rd 2009, 01:30 PMtonio

Of course it gives the above formula:

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

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

Tonio