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It follows that
Thank you so much
Can you give me some clues about how to work with this kind of problems. I get really confused
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Not about the problem you just solved. But in general
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.
Can you help me with this one?
It is better to post it as a new question. Maybe others can help you. I'll try if nobody answer it :).
Ok thank you very much
In the third line, how do you get "cos(3A)sin(4A)"