Question

Prove the following:

Answer 1

\[ \frac{ sinA-sinB }{ cosA+cosB }=\tan(\frac{ A-B }{ 2 })\] please help someone.. ive managed to get it simplified to sin(A-B)/cos(A+B) but im stuck.. thanks in advance

Answer 2

how did u write sinA-sin B ??

Answer 3

which standard formula u used ?

Answer 4

im pretty sure i did it wrong.. but what i did was use sinA-sinB = sin(A-B)
and cosA+cosB = cos(A+B)

Answer 5

but i think i have to do it with tangent half angle formula tan ((A+B)/2) = sinA+sinB/cosA+cosB ...

Answer 6

that is so incorrect!
use this :
SinC - SinD = 2Cos[(C+D)/2]Sin[(C-D)/2]
CosC + CosD = 2Cos[(C+D)/2]Cos[(C-D)/2]

Answer 7

and in 2 steps u get the answer

Answer 8

thankyou! ill try that now :)

Answer 9

are they identities?

Answer 10

yes!
they can be used directly...unless your teacher doesn't allow

Answer 11

yea i think ill have to now where they are derived from and i keep thinking it looks like sin2x=sinxcosx. Do you know which identity they come from?

Answer 12

they can be derived by many ways, one way is to use cos(A+B) and cos(A-B) formulas, and evaluate right side to get left side

Answer 13

i got the answer doing your way really easily btw, thanks heaps. what do you mean evaluate right side to get left side?

Answer 14

take this : 2Cos[(C+D)/2]Sin[(C-D)/2]
put the formulas and get it = sin C -sinD

Answer 15

u want list of formulas ?

Answer 16

nah ive got the formulas thanks, im fair sure i get what you mean now. thanks so much

Answer 17

welcome ^_^