# Trigonometry: $5\sin x\cos y+4\cos x\sin y=0$; $\cot x=2$; need to find $\cot y$

Given that $5\sin x\cos y+4\cos x\sin y=0$, and that $\cot x=2$, find the value of $\cot y$.

I don't understand why I got this question wrong. For this question, I drew a right-angled triangle. I then labelled angle $x$ and derived that 2 would be adjacent to $x$ and 1 would be opposite to $x$ because $\cot x=2=\frac21$. With both sides, I was able to work out the hypotenuse as $\sqrt5$.

I labelled angle $y$ as another angle in the right-angled triangle and then got $\cot y=\frac12$.

• Trignomotortery? – TMM Oct 5 '16 at 16:16
• Yes. It is a new word! – Hans Hüttel Oct 5 '16 at 16:24
• I see it as a portmanteau word made up of the prefix tri, a gnome, and a portmanteau in the portmanteau: tortery=tortuous+tottery. – Bernard Oct 5 '16 at 16:38
• The last part might also be related to "torture/tortury", i.e. the way OP feels about these exercises. – TMM Oct 5 '16 at 17:10
• @TMM You figured it out, well done! – user375494 Oct 5 '16 at 17:39

Here's what you did wrong:

You drew a right angled triangle with an angle $x$. Then you just went and guessed that the other angle was the $y$ you needed to find. Your guess was wrong.

• Why was my guess wrong though? – user375494 Oct 5 '16 at 16:23
• Give me a better explanation as to why my guess was incorrect and I'll click your answer as the 'most useful' that's what you all strive for right? To have the most useful answers? – user375494 Oct 5 '16 at 16:28
• What do you mean "why was my guess wrong"? It's just wrong. The other angle is not $y$. You can't just draw a random picture and expect everything in that picture to have something to do with your question. – Anon Oct 5 '16 at 16:29
• Yes, and using a triangle to visualize that problem was just fine. You need to keep in mind that the triangle is just a visualization of that problem, and that not every problem can be visualized using the same drawing, or even at all. – Anon Oct 5 '16 at 16:40
• @user375494 When the drawing actually matches the problem. This problem didn't say that $x$ and $y$ were angles of a right-angled triangle, so you shouldn't draw a right-angled triangle with $x$ and $y$ until you know that such a triangle exists. For this problem, it does not exist, so your drawing made a false assumption that you then used to get a false answer. – Anon Oct 5 '16 at 16:48

by dividing both to $\sin { x } \cos { y }$ and consider the fact that $\cot { x } =2$ we get $$5\sin { x\cos { y } +4\cos { x } \sin { y } =0, } \\ 5+4\frac { \cos { x } \sin { y } }{ \sin { x } \cos { y } } =0\\ 5+4\cot { x } \tan { y } =0\\ 5+\frac { 8 }{ \cot { y } } =0\\ \cot { y } =-\frac { 8 }{ 5 }$$

• @McFry,i fixed thank you – haqnatural Oct 5 '16 at 16:21
• I know how to do it, but I'd like to know why the method I originally took is incorrect? If possible... – user375494 Oct 5 '16 at 16:21