Question

can u tell me how we get this equation ??? at below :
cos (a+b) = cos a . cos b - sin a . sin b
thanks before for u'r answer

Answer 1

http://en.wikipedia.org/wiki/Proofs_of_trigonometric_identities#Cosine

Answer 2

@lalaly thanks for the link,,,,,,,,,,

Answer 3

or u can do it like this

\[\large{e^{ix} = \cos(x) + i \sin(x)}\]\[e^{i(a+b)} = \cos(a+b) + i \sin(a+b)\]
but
\[\large{e^{i(a+b)}=e^{ia} \times e^{ib} = ( \cos(a)+isin(a) )( \cos(b)+isin(b) )}\]
=cos(a)cos(b) + icos(a)sin(b)+ isin(a)cos(b)+i^2 sin(a)sin(b)
\[\large{=\cos(a)\cos(b) + icos(a)\sin(b)+ isin(a)\cos(b)+(-1) \sin(a)\sin(b)}\]\[\large{=\cos(a)\cos(b) - \sin(a)\sin(b) + i( \cos(a)\sin(b)+ \sin(a)\cos(b) )}\]
Now compare real part with real part and imaginary with imaginary\[[ \cos(a)\cos(b) - \sin(a)\sin(b) ] + i[ \sin(a)\cos(b) + \cos(a)\sin(b) ] = \cos(a+b) + i \sin(a+b)\]

So
cos(a+b) = cos(a)cos(b) - sin(a)sin(b)
sin(a+b) = sin(a)cos(b) + cos(a)sin(b)

but this might be too advanced for u...
either way u hav both lol

Answer 4

yes,,,, this to advanced for me,,,,, teach me in other time with this topic okey ??? thanks for u'r helps

Answer 5

do u use the theory of euler ?

Answer 6

Okay, I did this one in class, using a unit circle...

Answer 7

Hang on, I'm drawing

Answer 8

To find AB, we can use the distance formula.

AB^2 = (r cos b - r cos a)^2 + (r sin b - r sin a)^2
= r^2[cos^2 b + sin^2 b + cos^2 a + sin^2 a - 2(cos a cos b + sin a sin b)]
= 2r^2[1-(cos a cos b + sin a sin b)]

and we can also use cosine rule

AB^2 = r^2 + r^2 - 2r^2 cos (a - b)
= 2r^2 - 2r^2 cos (a - b)
= 2r^2 [ 1-cos(a-b) ]

therefore,

2r^2[1-(cos a cos b + sin a sin b)] = 2r^2 [ 1-cos(a-b) ]

and so, 1 - (cos a cos b + sin a sin b) = 1 - cos(a-b)
and cos(a-b) = cos a cos b + sin a sin b

now to get cos (a + b)

let our first b = -b to get cos(a-(-b)) = cos(a+b)
= cos a cos -b + sin a sin -b

cos (-b) = cos b
sin (-b) = -sin b

therefore cos (a + b) = cos a cos b - sin a sin b