Question

Prove : cos4θ + cos2θ / sin2θ - sin4θ = -cotθ

Answer 1

Do you mean\[\frac{ \cos 4 \theta+\cos 2 \theta }{ \sin 2 \theta-\sin 4\theta }\]or\[\cos 4\theta+\frac{ \cos2\theta }{ \sin 2\theta }-\sin 4\theta\]I think you meant the first, but you wrote the 2nd ;)

Answer 2

yeah it's the first..
help please?

Answer 3

OK, I'll give it a try!

Answer 4

Using\[\cos 2x=2\cos^2x-1\]and \[\sin 2x=2 \sin x \cos x\]with \[x=2\theta\]i get:\[\frac{ 2\cos^22\theta-1+\cos 2 \theta }{ \sin 2\theta - 2 \sin 2\theta \cos 2\theta }=\]\[\frac{ 2\cos^2 \theta+\cos 2\theta -1 }{ \sin 2 \theta(1-\cos 2\theta) }=\]Now we have to factor the numerator:\[\frac{ (2\cos 2\theta-1)(\cos 2\theta +1) }{-\sin 2\theta(2\cos 2\theta-1) }=\]

Answer 5

As you can see, there is a common factor, so that cancels:\[-\frac{ \cos 2 \theta+1 }{ \sin 2 \theta }=-\frac{ 2\cos^2 \theta-1+1 }{ 2\sin \theta \cos \theta }\]Do you see where this is getting to?

Answer 6

TYPO: in the first part of my explanation, the 2nd fraction should have \[2\cos^22\theta\]in the numerator!!

Answer 7

thank you very much sir :D

Answer 8

Welcome!