Question

Prove sin(A±B)=sinAcosB ± cosAsinB
using the following formulae:

sinθ=cos(90º-θ)
cos(A-B)=cosAcosB + sinAsinB
cos(A+B)=cosAcosB - sinAsinB

Hint: sin(A+B)=cos(90º-(A+B))

Answer 1

in cos(a-b) formulla,,substitute b as (90-b) ..[which means on both sides,ofcorse]
also note that cos(-x)=cos x
so you get the formulla for sin(a+b)
now for sin(a-b),,make a similar subsn in cos(a+b) formulla..
are you getting it buddy ?

Answer 2

Ummmm not quite sure. Would you be able to show me the first bit please?

Answer 3

hmm..
i'll show you 1st part..you show me 2nd
after making subsn,,we have :
cos(a-(90-b))=cosa cos(90-b) + sina sin(90-b)
=>cos(a+b-90) = cos(90-(a+b))=cosa sinb + sina cosb = sin(a+b)
hope that was fine..
so your turn now! ;)

Answer 4

Thank you :)