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

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 ?

• It would truly help to know what "this expression" is! Cheers! – Robert Lewis Jan 22 at 0:59
• x = sin θ + cos θ and y = sin θ − cos θ – sophia Jan 22 at 1:00
• What do you mean by "independent of $\theta$"? – Brian Jan 22 at 1:03
• 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. – Ross Millikan Jan 22 at 1:17
• Where did the question come from? – Jam Jan 24 at 13:10

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 $$(0,0)$$.

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

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?

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

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 $$\theta$$.