12$ sin^2 heta$$ cos^2 heta$

I am lost on exactly how to execute this. Help would be lot appreciated.

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1) broaden : $(a+b)^2$ and also $(a-b)^2$ because that all actual numbers $a$ and also $b$.

2) What is the value of $sin^2(x) + cos^2(x)$ for all actual number $x$ ?


All you have to do is main point it out.

$(sin heta - cos heta)^2 + (sin heta + cos heta)^2$

$= (sin heta - cos heta)(sin heta - cos heta) + (sin heta + cos heta)(sin heta + cos heta)$

$= sin^2 heta - 2sin hetacos heta + cos^2 heta + sin^2 heta + 2sin hetacos heta + cos^2 heta$

$=sin^2 heta + cos^2 heta + sin^2 heta+ cos^2 heta$

$= 1 + 1$

$= 2$


$$eginalign&phantom=left(sin x-cos x ight)^2+left(sin x+cos x ight)^2\&=sin^2x-2sin xcos x+cos^2x+sin^2x+2sin xcos x+cos^2x\&=2sin^2x+2cos^2x\&=2left(sin^2x+cos^2x ight)\&=2endalign$$


Since this is a multiple-choice question, girlfriend can very quickly narrow under the price by plugging in values of $ heta$. For example, when $ heta = 0$, the expression i do not care $(0-1)^2+(0+1)^2 = 2$.

Now, looking in ~ the four choices and examining at $ heta = 0$, us have:

$1$$2$$sin^2 heta = 0^2 = 0$$cos^2 heta = 1^2 = 1$

The prize is plainly the second option.

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