Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Given that $\sin{x} = \frac{1}{\sqrt{5}}$ where $x$ is acute, and that $\cos(x-y) = \sin{y}$, show that $\tan{y} = \frac{\sqrt{5} + 1}{2}$

I can derive the required equation but why does $x$ have to be acute?

share|cite|improve this question
$sin x$ takes on the same value at $x+\pi$, however at that value $cos(x)$ takes on the negative of the value in question. – Foo Barrigno Mar 11 '14 at 13:30

As $\sin x=\sin(\pi-x)$

and $\sin x=\frac1{\sqrt5}>0,$ one value of $x$ lies in the First Quadrant ( where $\cos x>0$ ), and the other in the Second (where $\cos x<0$)

Now as $x$ is acute, we can safely discard the value in the Second Quadrant

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.