Question

Prove that: (sin2A+sin4A)/(cos2A+cos4A)=tan3A

Answer 1

\(\sin(2A-3A)=\sin(3A-4A)\)
\(\sin(2A)\cos(3A)-\cos(2A)\sin(3A)=\sin(3A)\cos(4A)-\cos(3A)\sin(4A)\)
\(\sin(2A)\cos(3A)+\cos(3A)\sin(4A)=\sin(3A)\cos(4A)+\sin(3A)\cos(2A)\)
\(\cos(3A)(\sin(2A)+\cos(4A))=\sin(3A)(\cos(2A)+\cos(4A))\)
It follows that
\[\frac{\sin(2A)+\cos(4A)}{\cos(2A)+\cos(4A)}=\frac{\sin(3A)}{\cos{(3A)}}\]

Answer 2

Thank you so much

Answer 3

Can you give me some clues about how to work with this kind of problems. I get really confused

Answer 4

Not about the problem you just solved. But in general

Answer 5

To be honest it took me some time as well to solve this problem. Because usually the just give you a not so hard identity where you can play around with one side of the equation.
When I allow myself to work with both sides simultaneously I allow myself to see differently.
First I rewrite the \(\tan(3A)\) as \(\sin(3A)/\cos(3A)\) and then rewrite the equation. From there I can see the expression of the form \(\sin a\cos b-\cos a\sin b\) which is \(\sin(a-b)\). Then to write the solution you just need to write it backward.

Answer 6

Can you help me with this one?
(cosA+cosB)/(cosA-cosB)=cot((1/2)(A-B))cot((1/2)(A+B))

Answer 7

It is better to post it as a new question. Maybe others can help you. I'll try if nobody answer it :).

Answer 8

Ok thank you very much

Answer 9

In the third line, how do you get "cos(3A)sin(4A)"

Answer 10

I got it now. Clear