sin A + sin 3A/ cos A + cos 3A = tan 2A

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Solution

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).

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