Question

sinA+sin3A\cosA+cos3A=tanA

Answer 1

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Answer 2

recall the identities cosA - cos B = -2sin((A+B)/2) .sin((A-B)/2)

and sinA - sinB = 2cos((A+B)/2) .sin((A-B)/2)

(cosA-cos3A)/(sin(3A-sinA)

= [ -2sin(4A/2).sin(-2A/2) ] / [ 2cos(4A/2).sin(2A/2) ]

= [2sin(2A).sin(A) ] / [ 2cos(2A).sin(A) ] as sin(-A) = -sinA

= sin2A/cos2A

= tan2A