## anonymous 5 years ago Prove that: (sin2A+sin4A)/(cos2A+cos4A)=tan3A

1. watchmath

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

2. anonymous

Thank you so much

3. anonymous

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

4. anonymous

Not about the problem you just solved. But in general

5. watchmath

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.

6. anonymous

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

7. watchmath

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

8. anonymous

Ok thank you very much

9. anonymous

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

10. anonymous

I got it now. Clear