Trigonometry: Prove:

Question

Sin3A Cos4A - SinA Cos2A/Sin4A sinA + Cos6A CosA = Tan 2A?

Answer 1

Given that,

                    (Sin3A Cos4A - SinA Cos2A)/(Sin4A sinA + Cos6A CosA)

Diving numerator and denominator dy 2 we get,

                 2X(Sin3A Cos4A - SinA Cos2A)/2X(Sin4A sinA + Cos6A CosA)

Let us take the numerator  2(sin3A Cos4A - SinA Cos2A)

                             = 2sin3A Cos4A - 2SinA Cos2A

                             = (sin 7A + sin A) - (sin 3A + sin A)  [Since 2 sin A cos B = sin (A + B) + sin (A-B)]

                             = sin 7A + sin A - sin 3A - sin A

                             = sin 7A - sin 3 A                            [Since sin C - sin D = 2 cos (C+D)/2 sin (C-D)/2]

                             = 2 cos 5A sin 2 A                  -> (1)

Let us take the denominator 2(Sin4A sinA + Cos6A CosA)

                             = 2Sin4A sinA + 2Cos6A CosA    [Since 2 sin A sin B = cos (A-B) - cos (A+ B)]

                             = cos 3A - cos 5 A + cos 5A + cos 7A     [and 2cosAcosB = cos(A-B) + cos(A+ B)]

                             = cos 3A + cos 7A

                             = 2 cos 5A cos 2A              -> (2)

Dividing (1) by (2) we get,

  (Sin3A Cos4A - SinA Cos2A)/(Sin4A sinA + Cos6A CosA)  = 2 cos 5A sin 2 A / 2 cos 5A cos 2A       

                                                                                              = sin 2A / cos 2A

                                                                                              = tan 2A

Hence proved.