SOLUTION: sin(a+b)-sin(a-b)=2cosasinb How do I solve this problem? I am so close to a break down over this. Help!!

Algebra ->  Trigonometry-basics -> SOLUTION: sin(a+b)-sin(a-b)=2cosasinb How do I solve this problem? I am so close to a break down over this. Help!!       Log On


   

Question 408715: sin(a+b)-sin(a-b)=2cosasinb
How do I solve this problem? I am so close to a break down over this. Help!!
Found 3 solutions by robertb, jim_thompson5910, richard1234:
Answer by robertb(5830) About Me  (Show Source):
You can put this solution on YOUR website!
The left-hand side reduces to sina*cosb + cosa*sinb - sina*cosb + cosa*sinb = 2cosa*sinb. Hence the equation is an identity, and all values for a , b are solutions.

Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
sin(a+b)-sin(a-b)=2cos(a)sin(b)


sin(a)cos(b)+cos(a)sin(b)-(sin(a)cos(b)-cos(a)sin(b))=2cos(a)sin(b)


sin(a)cos(b)+cos(a)sin(b)-sin(a)cos(b)+cos(a)sin(b)=2cos(a)sin(b)


(sin(a)cos(b)-sin(a)cos(b))+(cos(a)sin(b)+cos(a)sin(b))=2cos(a)sin(b)


0+2cos(a)sin(b)=2cos(a)sin(b)


2cos(a)sin(b)=2cos(a)sin(b)


So the original equation is an identity (ie it's true for ALL values of 'a' and 'b')


If you need more help, email me at jim_thompson5910@hotmail.com

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Jim

Answer by richard1234(7193) About Me  (Show Source):
You can put this solution on YOUR website!
. The sin%28a%29cos%28b%29 terms cancel out, so our expression is equal to sin%28b%29cos%28a%29+-+%28-sin%28b%29cos%28a%29%29+=+2sin%28b%29cos%28a%29 which is the RHS, so we're done.