Answer:
B
Step-by-step explanation:
(To save time, I'm going to use x instead of θ)
So we have the expression:
[tex](\sin(x)-\cos(x))^2+(\sin(x)+\cos(x))^2[/tex]
First, expand these binomials:
[tex]=(\sin^2(x)-2\sin(x)\cos(x)+\cos^2(x))+(\sin^2(x)+2\sin(x)\cos(x)+\cos^2(x))[/tex]
Combine like terms:
[tex]=\sin^2(x)+\sin^2(x)+\cos^2(x)+\cos^2(x)-2\sin(x)\cos(x)+2\sin(x)\cos(x)\\=2\sin^2(x)+2\cos^2(x)[/tex]
Factor out a 2:
[tex]=2(\sin^2(x)+\cos^2(x))[/tex]
The expression inside the parentheses is the Pythagorean Identity:
[tex]\sin^2(x)+\cos^2(x)=1[/tex]
Substitute:
[tex]=2(1)\\=2[/tex]
The answer is B.
