$x = sin θ + cos θ$ and $y = sin θ − cos θ$. Prove that this expression is independent of $θ $by simplifying it 
Let $x = \sin θ + \cos θ$ and $y = \sin θ − \cos θ$. Prove that this expression is independent of $θ$ by simplifying it.

My first problem is that I don't know what 'independent of $θ$' even means. Am I supposed to solve it simultaneously or add the equations together?
 A: Hint:
In general for $x=a\cos t+b\sin t$
and $y=c\cos t+d\sin t$
Solve the two simultaneous equations for $\sin t,\cos t$
Then use $$\cos^2t+\sin^2t=1$$ to eliminate $t$
Can you recognize $a,b,c,d$ here?
A: $x = sin(\theta) + cos(\theta)$
Hence:
$$\sqrt2/2 \cdot x = \sqrt2/2 \cdot sin(\theta) + \sqrt2/2 \cdot cos(\theta)$$
$$    =cos(\frac \pi 4) \cdot cos(\theta) + sin(\frac \pi 4) \cdot sin(\theta)$$
$$    =cos(\theta) \cdot cos(\frac \pi 4) + sin(\theta) \cdot sin(\frac \pi 4)$$
$$    =cos(\theta - \frac \pi 4)$$
$y = sin(\theta) - cos(\theta)$
Hence:
$$\sqrt2/2 \cdot y = \sqrt2/2 \cdot sin(\theta) - \sqrt2/2 \cdot cos(\theta)$$
$$    =cos(\frac \pi 4) \cdot sin(\theta) - sin(\frac \pi 4) \cdot cos(\theta)$$
$$    =sin(\theta) \cdot cos(\frac \pi 4) - cos(\theta) \cdot sin(\frac \pi 4)$$
$$    =sin(\theta - \frac \pi 4)$$
From that, you can easily derive:
$$x^2/2 + y^2/2 = 1$$ or:  
$$x^2 + y^2 = 2$$
This means: you can write an equation, linking $x$ and $y$, without any $\theta$.
A: If you mean to express this curve independently from $\theta$ and in a way to find a function $f$ such that $$f(x,y)=0$$, then we can write$$x^2+y^2{=(\sin\theta+\cos\theta)^2+(\sin\theta-\cos\theta)^2\\=1+2\sin\theta\cos\theta+1-2\sin\theta\cos\theta\\=2}$$
which is a circle with radius $\sqrt 2$ centered at $(0,0)$.
