document.write( "Question 985310: Prove these identity
\n" ); document.write( "(1-SinA + CosA)^2 = 2(1-SinA)(1+ CosA) \n" ); document.write( "

Algebra.Com's Answer #606442 by josgarithmetic(39627) $\"\"$ $\"About$

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Using lower case instead of A.\r
\n" ); document.write( "\n" ); document.write( "ID to prove: $\"%281-sin%28a%29%2Bcos%28a%29%29%5E2=2%281-sin%28a%29%29%281%2Bcos%28a%29%29\"$ \r
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\n" ); document.write( "\n" ); document.write( "LEFT SIDE:
\n" ); document.write( " $\"1-2sin%28a%29%2B2cos%28a%29%2Bsin%5E2%28a%29-2sin%28a%29cos%28a%29%2Bcos%5E2%28a%29\"$
\n" ); document.write( " $\"1%2B%28sin%5E2%28a%29%2Bcos%5E2%28a%29%29-2sin%28a%29%2B2cos%28a%29-2sin%28a%29cos%28a%29\"$
\n" ); document.write( " $\"1%2B1-2sin%28a%29%2B2cos%28a%29-2sin%28a%29cos%28a%29\"$
\n" ); document.write( " $\"2%2B2cos%28a%29-2sin%28a%29-2sin%28a%29cos%28a%29\"$
\n" ); document.write( " $\"highlight_green%282%281%2Bcos%28a%29-sin%28a%29-sin%28a%29cos%28a%29%29%29\"$
\n" ); document.write( "-------You might or might not realize how the longer expression IS factorable.
\n" ); document.write( "Notice the terms 1 and -sin(a)cos(a), which may help suggest the possible factorization.\r
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\n" ); document.write( "\n" ); document.write( "If you now take the right hand side, and keep the factor 2 separate, and multiply
\n" ); document.write( "the two binomials according to First, Outer, Inner, Last, you will obtain:
\n" ); document.write( " $\"2%281%2Bcos%28a%29-sin%28a%29-sin%28a%29cos%28a%29%29\"$ , essentially showing that its factorization will be
\n" ); document.write( "the RIGHT SIDE of the identity to be proved; and that this is identical to the
\n" ); document.write( "expression found when starting from the LEFT SIDE.
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