Question 1151636
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{{{(A+B)^2 = A^2 + 2*A*B + B^2}}}


{{{(sin(x)+cos(x))^2 = sin^2(x) + 2*sin(x)*cos(x) + cos^2(x)}}} Plug in A = sin(x) and B = cos(x)


{{{(sin(x)+cos(x))^2 = sin^2(x) + cos^2(x) + 2*sin(x)*cos(x)}}}


{{{(sin(x)+cos(x))^2 = highlight(sin^2(x) + cos^2(x)) + 2*sin(x)*cos(x)}}} The portion in red


{{{(sin(x)+cos(x))^2 = highlight(1) + 2*sin(x)*cos(x)}}} is equivalent to 1 due to the pythagorean trig identity


{{{(sin(x)+cos(x))^2 = 1 + 2*sin(x)*cos(x)}}}


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{{{(sin(x)+cos(x))^2 = 1 + 2*sin(x)*cos(x)}}}


{{{(sin(x)+cos(x))^2 + C = 1 + 2*sin(x)*cos(x) + C}}} Add C to both sides


We'll let {{{C = - 3(cos(x)+sin(x))+2}}}


{{{(sin(x)+cos(x))^2 - 3(cos(x)+sin(x))+2 = 1 + 2*sin(x)*cos(x) - 3(cos(x)+sin(x))+2}}} Plug in {{{C = - 3(cos(x)+sin(x))+2}}}


{{{(sin(x)+cos(x))^2 - 3(cos(x)+sin(x))+2 = 1 + 2*sin(x)*cos(x) - 3cos(x)-3sin(x)+2}}} Distribute


{{{(sin(x)+cos(x))^2 - 3(cos(x)+sin(x))+2 = 2*sin(x)*cos(x)-3sin(x) - 3cos(x)+3}}} Combine like terms and do a bit of rearranging.


The RHS (right hand side) matches the original expression
So,{{{(sin(x)+cos(x))^2 - 3(cos(x)+sin(x))+2}}} is equivalent to {{{2*sin(x)*cos(x)-3sin(x) - 3cos(x)+3}}} for all real numbers x.


<font size=4 color=red>Therefore, the final answer is choice A</font>


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Through graphing technology, we can use programs like <a href = "https://www.geogebra.org/">GeoGebra</a> to visually verify we have the proper answer.


Let
{{{f(x) = 2*sin(x)*cos(x)-3sin(x) - 3cos(x)+3}}}
{{{g(x) = (sin(x)+cos(x))^2 - 3(cos(x)+sin(x))+2}}}
The graph below shows a red and blue striped curve. This is really a solid red curve underneath with a blue dashed curve directly on top of it. 
f(x) = red solid curve, g(x) = blue dashed curve
The table of values show that f(x) = g(x) for any real x. While this isn't an exhaustive proof by any means, it's a quick way to verify the answer. 
<img width="60%" src = "https://i.imgur.com/aUdjlxg.png">
I recommend you play around with GeoGebra (or any equivalent graphing technology) to turn off/on the second graph to see how it is directly over top the first graph.
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