Using the formula for \sin(A+B) and \cos(A+B) show that: \frac{\cos(A-B)-\cos(A+B)}{\sin(A-B)-\sin(A+B)} =

zamenjenot7k

zamenjenot7k

Answered question

2022-04-21

Using the formula for sin(A+B) and cos(A+B) show that:
cos(AB)cos(A+B)sin(AB)sin(A+B)=tan(A)
anybody knows how to get the answer? I keep getting stuck at 2sin(A)sin(B)2cos(A)cos(B)

Answer & Explanation

Volsa280

Volsa280

Beginner2022-04-22Added 16 answers

cos(ab)cos(a+b)sin(ab)sin(a+b)
cosacosb+sinasinbcosacosb+sinasinbsinacosbcosasinbsinacosbcosasinb
2sinasinb2cosasinb
sinacosa
tana
Note that you have the incorrect sign from the original proposition, suggesting something was transposed in the post.
i.e. the identity was supposed to be.
cos(ab)cos(a+b)sin(a+b)sin(ab)
And in your work, it appears that you have applied the incorrect addition identity for
sin(a+b),sin(ab)

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