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anonymous

  • 5 years ago

Prove that (sinA+sin3A)/(cosA+cos3A)=tan2A

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  1. anonymous
    • 5 years ago
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    We split sin(3A) and cos(3A) into sin(2A+A) , cos (2A+A), and use trig identities to rewrite them. sin(2A+A)=sin(A)cos(2A)+cos(A)sin(2A) cos(2A+A)=cos(A)cos(2A)-sin(A)sin(2A) sin(2A) is 2sin(A)cos(A) so in the numerator we have sin(A)*(1+cos(2A)+2cos^2(A)) denominator: cos(A)*(1+cos(2A)-2sin^2(A)) 1+cos(2A)=2cos^2(A) 2cos^2(A)-2sin^2(A)=2cos(2u) so now our denominator is 2cos(A)cos(2A) numerator is sin(A)*4cos^2(A) using the same identities. cancel cos(A) and we have 4sin(A)cos(A)/(2cos(2A)) 4sin(A)cos(A)=2sin(2A) cancel a 2 and we have sin(2A)/cos(2A)=tan(2A). Maybe there's a nicer way to do this but this works.

  2. anonymous
    • 5 years ago
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    Thanks rsvitale. its clear for me

  3. anonymous
    • 5 years ago
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    you're welcome

  4. watchmath
    • 5 years ago
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    Is this nicer rsvitale? http://openstudy.com/groups/mathematics#/groups/mathematics/updates/4dd3dc1bd95c8b0ba9e754c4

  5. anonymous
    • 5 years ago
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    yes much nicer. The method watchmath used for the linked problem works for this problem as well, so I would recommend doing that instead.

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