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x = sin θ + cos θ and y = sin θ − cos θ. Prove that 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 ?

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    $\begingroup$ It would truly help to know what "this expression" is! Cheers! $\endgroup$ – Robert Lewis Jan 22 at 0:59
  • $\begingroup$ x = sin θ + cos θ and y = sin θ − cos θ $\endgroup$ – sophia Jan 22 at 1:00
  • $\begingroup$ What do you mean by "independent of $\theta$"? $\endgroup$ – Brian Jan 22 at 1:03
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    $\begingroup$ Normally I would view the expression as defining a point $(x,y)$, which is clearly not independent of $\theta$. An expression independent of $\theta$ would be $\sin^2 \theta + \cos^2 \theta$ for which the value is the same whatever $\theta$ you plug in. $\endgroup$ – Ross Millikan Jan 22 at 1:17
  • $\begingroup$ Where did the question come from? $\endgroup$ – Jam Jan 24 at 13:10
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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 $(0,0)$.

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    $\begingroup$ The question is very badly stated, but this is a reasonable guess for what is wanted. $\endgroup$ – Ross Millikan Jan 22 at 1:14
  • $\begingroup$ Yes, thank you. i understand what the question is trying to ask now ! $\endgroup$ – sophia Jan 22 at 1:32
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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?

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$x = sin(\theta) + cos(\theta)$

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 $\theta$.

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