SOLUTION: (1+sinA+cosA)(1+sinA+cosA)=2(1+sinA)(1+cosA)

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Question 1039979: (1+sinA+cosA)(1+sinA+cosA)=2(1+sinA)(1+cosA)
Answer by Edwin McCravy(20056) About Me  (Show Source):
You can put this solution on YOUR website!
%281%2Bsin%28A%29%2Bcos%28A%29%5E%22%22%29%281%2Bsin%28A%29%2Bcos%28A%29%5E%22%22%29%22%3F=%3F%222%281%2Bsin%28A%29%5E%22%22%29%281%2Bcos%28A%29%5E%22%22%29

Multiply the left side out:



Rearrange like terms together:



Combine like terms and replace sin2(A)+cos2(A) by 1:

2sin%28A%29%2B2cos%28A%29%2B2sin%28A%29cos%28A%29%2B1%2B1

2sin%28A%29%2B2cos%28A%29%2B2sin%28A%29cos%28A%29%2B2

Factor out 2:

2%28sin%28A%29%5E%22%22%2Bcos%28A%29%2Bsin%28A%29cos%28A%29%2B1%29

Move the cos(A) to the end of the parentheses:

2%28sin%28A%29%5E%22%22%2Bsin%28A%29cos%28A%29%2B1%2Bcos%28A%29%29

Out of the first two term in parentheses,
factor out sin(A), and
out of the last two terms in parentheses,
factor out 1:

2%28sin%28A%29%281%2Bcos%28A%29%5E%22%22%29%2B1%281%2Bcos%28A%29%5E%22%22%29%5E%22%22%29

Factor out out common factor %281%2Bcos%28A%29%5E%22%22%29

2%28sin%28A%29%5E%22%22%2B1%29%281%2Bcos%28A%29%5E%22%22%29

Reverse terms in first parentheses:

2%281%2Bsin%28A%29%5E%22%22%29%281%2Bcos%28A%29%5E%22%22%29

Edwin