Question

Trigonometry Proof

How do you prove that
((cos(a+b))/(cosa +sinb))=((cosa - sinb)/(cos(b-a))) ?

Answer 1

\[{\cos (a+b) \over {\cos a + \sin b}} = {{\cos a - \sin b} \over \cos (b-a)}\]

Answer 2

though no so sure .. !

Answer 3

cross multiply is good
then use cos(a±b)= cos(a)cos(b)-/+ sin(a)sin(b) on the left

Answer 4

I have tried it forwards, backwards, cross multiplying to get 1, subtracting to get 0...
I know I am missing some simple identity or transformation... but I have hit the wall.

Answer 5

cross multiply
cos(a+b)cos(a-b) = (cos(a))^2-(sin(b))^2
1/2*(cos(2a)+cos(2b)) = 1/2 ( 2 cos^2a-1+1-2sin^2b) => rhs

Answer 6

I worked on this most of last night

http://openstudy.com/updates/4f78bcb7e4b0ddcbb89e662b

and it still bothers me; missing something simple somewhere

Answer 7

(cosa-sinb)(cosa+sinb) = cos^2 a -sin^2 b
= (cos a)^2 -(1- cosb)^2
=cos ^2 a -1 +2cos ^b -cos^2 b
cos(a+b) cos(b-a) = (cosa cosb-sina sinb)(cosb cos a +sinb sina)
= (cos a cos b)^2 - (sin a sin b )^2
= (cos a cos b)^2 - ((1-cos^2 a) (1-cos ^2 b))
= (cos a cos b)^2 - (1-cos^2 a - cos ^2 b + cos^2 a cos^2 b)
= -1 +cos^2 a + cos^2 b
=(cosa-sinb)(cosa+sinb)

Answer 8

Thank you!