Hi, I was working on the proof for the tangent rule (a-b/a+b = tan(A-B/2)/tan(A+B/2) and the website I was studying it on said that sin(A)=sin(A+B2)cos(A−B2)+cos(A+B2)sin(A−B2) . I do not get how is it possible for sin(A) to equal to that. Can someone provide me with the proof for this please?
First you need to use grouping symbols. For example did the website have
\(\displaystyle sin(A) = sin(\frac{A+B}{2})\, cos(\frac{A-B}{2})\, +\, sin(\frac{A-B}{2})\, cos(\frac{A+B}{2})\), i.e
sin(A) = sin((A+B)/2) cos((A−B)/2) + cos((A+B)/2) sin((A−B)/2)
which would be more reasonable.
What are your thoughts? What have you done so far? Please show us your work even if you feel that it is wrong so we may try to help you. You might also read
http://www.freemathhelp.com/forum/threads/78006-Read-Before-Posting
Hint: Just use the rules for sum and differences of trig functions and plow through it. It might easier to use a=A/2 and b=B/2 [assuming that grouping above is correct] so that you have less writing. Actually, I just use sa, ca, for sine and cosine of a and sb, cb for sine and cosine of b when working out something like this. BTW, I generally write other trig functions in terms of the sine and cosine, i.e. tan(a) as sa/ca but not always.
EDIT: The tangent rule for the tangent of the sum and difference of quantities x and y is
\(\displaystyle tan(x\, \pm\, y) = \frac{tan(x)\, \pm\, tan(y)}{1\, \mp\, tan(x)\, tan(y)}\)
I'm not sure what you mean by what you have.
