math question destroyer weird !

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spoog74

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question asks what is equal to [sin(x)+cos (x)]^2 . I thought the answer was 1 since sin ^2x + Cos^2 x=1 but i was wrong.

Anyway the questions gets weirder.... The answer claims the answer is :
(sin(x)+cost(x)) * Sin (x) + Cos (x) = sin (x)^2 + 2Sin(x)cos(x) + cos (x)^2. <<--- why isnt there a 2 behind the cos? Isnt it 2sinx+2cosx? WTF?


Basically the final answer c omes down to 1+sin2x which i have no clue how that becomes the answer since cosx+cosx = 2cosx.

Someone chime in?
 
(a+b)^2=a^2+2*a*b+b^2.

Let a=sin(x), b=cos(x).
Then (a+b)^2=sin (x)^2 + 2Sin(x)cos(x) + cos (x)^2.
As you have said, sin ^2x + Cos^2 x=1,
so the answer becomes 1 + 2Sin(x)cos(x).

Note: 2Sin(x)cos(x)=sin(2x) ------double angle identity rule.
 
(a+b)^2=a^2+2*a*b+b^2.

Let a=sin(x), b=cos(x).
Then (a+b)^2=sin (x)^2 + 2Sin(x)cos(x) + cos (x)^2.
As you have said, sin ^2x + Cos^2 x=1,
so the answer becomes 1 + 2Sin(x)cos(x).

Note: 2Sin(x)cos(x)=sin(2x) ------double angle identity rule.

i guess im missing something in the first step. Wouldnt it be a^2 + ab+ab+b^2 and NOT a^2+2ab+b^2? Arent we using the foil method for multiplication? I feel so stupid not knowing how you derived 2ab instead of 2a+2b....
 
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