Welcome to NRICH.

 
Expansion of tan(A+B)
By Margaret Brunt on Monday, June 21, 1999 - 01:11 pm:

Hi,

Does anyone know of a way to prove the expansion of tan(A+B) other than dividing sin(A+B) by cos(A+B).

Thanks
Margaret Brunt
The British School of Costa Rica.

By Alex Barnard (Agb21) on Monday, June 21, 1999 - 05:23 pm:

What about this...

Think of tan(x) as the gradient of a line drawn at x degrees to the horizontal.

Consider the following matrix, which I call R(x):

æ
ç
ç
ç
è
1
m
-m
1
ö
÷
÷
÷
ø		 where m=tan(x)
It sends a horizontal line (1 0) to (1 m) which is a line of gradient m. And it does something similar to the vertical line (0 1). In otherwords it is a rotation [and an enlargement, but you don't notice that on lines through the origin] taking the horizontal to gradient m.

Hence R(x)R(y) takes the horizontal line to one at an angle of (x+y) to the horizontal - ie. one with gradient tan(x+y).

But R(x).R(y) is (m=tan(x), n=tan(y)):
æ
ç
ç
ç
è
1
m
-m
1
ö
÷
÷
÷
ø		 æ
ç
ç
ç
è
1
n
-n
1
ö
÷
÷
÷
ø		 = æ
ç
ç
ç
è
1-mn
n+m
-n-m
1-mn
ö
÷
÷
÷
ø
So (1 0) goes to ( 1-mn n+m ) which has gradient:

(n+m)/(1-mn).

Hence tan(x+y) = (tan(x) + tan(y))/(1-tan(x)tan(y))

Hope this is okay,
AlexB.