Question

using sin(alpha-beta)=(sin alpha)(cos beta)-(cos alpha)(sin beta)

use the identity to derive the result
proof for : sin(alpha-beta)
PLEASE PLEASE PLEASE HELP

Answer 1

so you want us to use the identity:
sin(A - B) = sinAcosB - cosAsinB to prove

sin(A - B) = sinAcosB - cosAsinB ?

mmmhhh?

did you mean sin(A+B) ??

Answer 2

Nope, assuming I want to prove cos2theta is cos 2theta =1-2sin^2theta
cos2theta=(1-sin^2theta)-sin^2theta
cos2theta=1-2sin^2theta

Answer 3

That's not what you said originally, look at your post.

So which identity are we deriving ?

Answer 4

there are 2

sin(alpha+beta)=(sin alpha)(cos beta)+(cos alpha)(sin beta)

cos(alpha+beta)=(sis alpha)(cos beta)+(sin alpha)(sin beta)
these identities to derive
i)sin(alpha-theta)
ii)cos(alpha-theta)
iii)sin2theta
I just want one of them solved in steps to be able to implement the steps to rest of the following.

Answer 5

Ok, starting to make sense now

So I will assume you are given:
sin(alpha+beta)=(sin alpha)(cos beta)+(cos alpha)(sin beta)
and
cos(alpha+beta)=(sis alpha)(cos beta)+(sin alpha)(sin beta)

I will use A and B for easier typing

You must also know that
1. sin(-A) = -sinA
2. cos(-A) = cosA

so lets rewrite
sin(A - B) as sin(A + (-B) )
which now is
sinAcos(-B) + cosAsin(-B)
= sinAcosB - cosAsinB
DONE!

Do the same for cos(A - B)

for sin 2A, write it as sin(A + A) and use our first identity
sin 2A
= sin(A+A)
= sinAcosA + cosAsinA
= 2sinAcosA

for cos 2A, do the same thing
cos2A = cos(A+A)
= cosAcosA - sinAsinA
= cos^2 A - sin^2 A

but we also know cos^2 A = 1 - sin^2 A
and sin^2 A = 1 - cos^2 A

replace those in your cos2A result to get

cos^2 A - sin^2 A
= cos^2 A - (1-cos^2 A(
= 2cos^2 A - 1

do the same thing to get the third version of cos 2A

Answer 6

thank you so much !!