How to simplify $(\sin\theta-\cos\theta)^2+(\sin\theta+\cos\theta)^2$? Simplify: $(\sin \theta − \cos \theta)^2 + (\sin \theta + \cos \theta)^2$
Answer choices:


*

*1

*2

*$ \sin^2 \theta$

*$ \cos^2 \theta$


I am lost on how to do this. Help would be much appreciated.
 A: Hint:
1) Expand : $(a+b)^2$ and $(a-b)^2$ for all real numbers $a$ and $b$.
2) What is the value of $\sin^2(x) + \cos^2(x)$ for all real number $x$ ?
A: All you have to do is multiply it out.
$(\sin{\theta} - \cos{\theta})^{2} + (\sin{\theta} + \cos{\theta})^{2}$
$= (\sin{\theta} - \cos{\theta})(\sin{\theta} - \cos{\theta}) + (\sin{\theta} + \cos{\theta})(\sin{\theta} + \cos{\theta})$
$= \sin^{2}{\theta} - 2\sin{\theta}\cos{\theta} + \cos^{2}{\theta} + \sin^{2}{\theta} + 2\sin{\theta}\cos{\theta} + cos^{2}{\theta}$
$=\sin^{2}{\theta} + \cos^{2}{\theta} + \sin^{2}{\theta}+ cos^{2}{\theta}$
$= 1 + 1$
$= 2$
A: $$\begin{align}
&\phantom{=}\left(\sin x-\cos x\right)^2+\left(\sin x+\cos x\right)^2\\
&=\sin^2x-2\sin x\cos x+\cos^2x+\sin^2x+2\sin x\cos x+\cos^2x\\
&=2\sin^2x+2\cos^2x\\
&=2\left(\sin^2x+\cos^2x\right)\\
&=2
\end{align}$$
A: Since this is a multiple-choice question, you can very quickly narrow down the answer by plugging in values of $\theta$. For example, when $\theta = 0$, the expression becomes $(0-1)^2+(0+1)^2 = 2$.
Now, looking at the four options and evaluating at $\theta = 0$, we have:


*

*$1$

*$2$

*$\sin^2 \theta = 0^2 = 0$

*$\cos^2 \theta = 1^2 = 1$


The answer is clearly the second option.
