Question

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

Answer 1

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.

Answer 2

Thanks rsvitale. its clear for me

Answer 3

you're welcome

Answer 4

Is this nicer rsvitale? http://openstudy.com/groups/mathematics#/groups/mathematics/updates/4dd3dc1bd95c8b0ba9e754c4

Answer 5

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