anonymous
  • anonymous
Prove that (sinA+sin3A)/(cosA+cos3A)=tan2A
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
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.
anonymous
  • anonymous
Thanks rsvitale. its clear for me
anonymous
  • anonymous
you're welcome

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

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

Looking for something else?

Not the answer you are looking for? Search for more explanations.