0

I need to eliminate $\theta$ from the equations $x=\sin\theta+\cos\theta$ and $y=\tan\theta+\cot\theta$. I am actually provide with a hint: consider $x^2y$ , which worked nicely for me. However I am wondering if it was possible to "see" this solution without a hint. I was very nice when $\tan\theta+\cot\theta$ popped back out and I was able to let that simply equal to $y$, but What if the hint wasn't given?

Unless this is simply one of those "know the famous first step" questions

user71207
  • 1,543

2 Answers2

1

You have $\tan \theta$ and $\cot\theta$ and they're in addition. So first express them in terms of $\sin$ and $\cos$.

$y = \tan \theta + \cot\theta = \dfrac{\sin^2\theta + \cos^2\theta}{\sin\theta \cos\theta} = \dfrac{1}{\sin\theta \cos\theta} $

Now we need, $\sin\theta \cos\theta$ to be eliminated.

Remember $(a+b)^2 = a^2+2\color{red}{ab} + b^2$ and luckily here, $a^2 + b^2 = \sin^2\theta + \cos^2\theta$ would yield $1$.

So, $\boxed{x^2 = 1+ 2\sin\theta \cos\theta = 1+ \dfrac{2}{y}} \Rightarrow \boxed{x^2y = y+2}$

19aksh
  • 12,768
1

You could say $$y=\tan \theta+\cot \theta = \frac{\sin \theta}{\cos \theta} + \frac{\cos \theta}{\sin \theta} = \frac{\sin^2 \theta+\cos^2 \theta}{\sin \theta \cos \theta} =\frac{1}{\sin \theta \cos \theta}$$

So we want the other equation to give something related to $\sin \theta \cos \theta$, and we know $\sin^2 \theta+\cos^2 \theta=1$ so squaring looks useful. We get $$x^2 = (\sin \theta +\cos \theta)^2 = \sin^2 \theta + 2\sin \theta \cos \theta +\cos^2 \theta =1 + 2\sin \theta \cos \theta$$

and thus $$x^2 = 1+\frac{2}{y}$$ which you can tidy up to $x^2y = y +2$

Henry
  • 157,058