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.

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

æ 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 è 1 m -m 1 ö ÷ ø æ ç è 1 n -n 1 ö ÷ ø = æ ç è 1-m n n+m -n-m 1-m n ö ÷ ø

So (1 0) goes to (1-m n n+m) which has gradient:

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

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

Hope this is okay,

AlexB.