SOLUTION: Prove these identity (1-SinA + CosA)^2 = 2(1-SinA)(1+ CosA)

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Question 985310: Prove these identity
(1-SinA + CosA)^2 = 2(1-SinA)(1+ CosA)
Found 3 solutions by Edwin McCravy, ThePhysicsMathGuru, josgarithmetic:
Answer by Edwin McCravy(20062) About Me  (Show Source):
You can put this solution on YOUR website!
%281%5E%22%22-sin%28A%29%5E%22%22%2Bcos%28A%29%29%5E2  %22%3F=%3F%22  2%281-sin%28A%29%5E%22%22%29%5E%22%22%281%2Bcos%28A%29%5E%22%22%29

Square out the left side:

1%2Bsin%5E2%28A%29%2Bcos%5E2%28A%29-2sin%28A%29%2B2cos%28A%29-2sin%28A%29cos%28A%29

Replace sinē(A) by 1-cosē(A):



Remove parentheses:

1%2B1-cos%5E2%28A%29%2Bcos%5E2%28A%29-2sin%28A%29%2B2cos%28A%29-2sin%28A%29cos%28A%29

Combine like terms:

2-2sin%28A%29%2B2cos%28A%29-2sin%28A%29cos%28A%29

Factor out 2:

2%281-sin%28A%29%5E%22%22%2Bcos%28A%29-sin%28A%29cos%28A%29%29

Factor by grouping:  I.e., In the parentheses factor 1 out 
of the first two terms and factor cos(A) out of the last 
two terms:

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

Factor 1-sin(A) out of both terms:

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

Edwin

Answer by ThePhysicsMathGuru(3) About Me  (Show Source):
You can put this solution on YOUR website!
Rewrite the LHS as
%281+-+Sin+A+%2B+Cos+A%29%281+-+Sin+A+%2B+Cos+A%29+
Now Multiply Thru
1+-+Sin+A+%2B+Cos+A+-+Sin+A+%2B+Sin%5E2A+-+Sin+A+Cos+A+%2B+Cos+A+-+Sin+A+Cos+A+%2B+Cos%5E2A+
Now all that is left is to collect like terms
1+-+2Sin+A+%2B+2Cos+A+-+2SinA+Cos+A+%2B+%28Sin%5E2A+%2B+Cos%5E2A%29+
We now end up with
2+-+2Sin+A+%2B+2Cos+A+-+2SinACosA+
Now you can either do Factoring By grouping on the previous expression or you can simply expand the RHS of the equation and the two parts are EQUAL with the 2 Factored out of the LHS
QED

Answer by josgarithmetic(39627) About Me  (Show Source):
You can put this solution on YOUR website!
Using lower case instead of A.
ID to prove: %281-sin%28a%29%2Bcos%28a%29%29%5E2=2%281-sin%28a%29%29%281%2Bcos%28a%29%29

LEFT SIDE:
1-2sin%28a%29%2B2cos%28a%29%2Bsin%5E2%28a%29-2sin%28a%29cos%28a%29%2Bcos%5E2%28a%29
1%2B%28sin%5E2%28a%29%2Bcos%5E2%28a%29%29-2sin%28a%29%2B2cos%28a%29-2sin%28a%29cos%28a%29
1%2B1-2sin%28a%29%2B2cos%28a%29-2sin%28a%29cos%28a%29
2%2B2cos%28a%29-2sin%28a%29-2sin%28a%29cos%28a%29
highlight_green%282%281%2Bcos%28a%29-sin%28a%29-sin%28a%29cos%28a%29%29%29
-------You might or might not realize how the longer expression IS factorable.
Notice the terms 1 and -sin(a)cos(a), which may help suggest the possible factorization.

If you now take the right hand side, and keep the factor 2 separate, and multiply
the two binomials according to First, Outer, Inner, Last, you will obtain:
2%281%2Bcos%28a%29-sin%28a%29-sin%28a%29cos%28a%29%29, essentially showing that its factorization will be
the RIGHT SIDE of the identity to be proved; and that this is identical to the
expression found when starting from the LEFT SIDE.