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anonymous

  • 5 years ago

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

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  1. watchmath
    • 5 years ago
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    \(\sin(2A-3A)=\sin(3A-4A)\) \(\sin(2A)\cos(3A)-\cos(2A)\sin(3A)=\sin(3A)\cos(4A)-\cos(3A)\sin(4A)\) \(\sin(2A)\cos(3A)+\cos(3A)\sin(4A)=\sin(3A)\cos(4A)+\sin(3A)\cos(2A)\) \(\cos(3A)(\sin(2A)+\cos(4A))=\sin(3A)(\cos(2A)+\cos(4A))\) It follows that \[\frac{\sin(2A)+\cos(4A)}{\cos(2A)+\cos(4A)}=\frac{\sin(3A)}{\cos{(3A)}}\]

  2. anonymous
    • 5 years ago
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    Thank you so much

  3. anonymous
    • 5 years ago
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    Can you give me some clues about how to work with this kind of problems. I get really confused

  4. anonymous
    • 5 years ago
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    Not about the problem you just solved. But in general

  5. watchmath
    • 5 years ago
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    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
    • 5 years ago
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    Can you help me with this one? (cosA+cosB)/(cosA-cosB)=cot((1/2)(A-B))cot((1/2)(A+B))

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

  8. anonymous
    • 5 years ago
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    Ok thank you very much

  9. anonymous
    • 5 years ago
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    In the third line, how do you get "cos(3A)sin(4A)"

  10. anonymous
    • 5 years ago
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    I got it now. Clear

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