Question

prove that (cos4x sin3x - cos2x sinx) / (sin4x sinx + cos6x cosx) = tan 2x

Answer 1

Answer:

proved

Step-by-step explanation:

Given prove that (cos4x sin3x - cos2x sinx) / (sin4x sinx + cos6x cosx) = tan 2x

  Consider the given , multiply and divide by 2 we get

   2cos4x sin3x - 2cos2x sinx) / 2sin4x sinx + 2cos6x cosx

  We know that 2 cos a sin b = sin (a + b) - sin (a - b)

                           2 sin a sin b = cos(a - b) - cos(a + b)

                           2 cos a cos b = cos(a + b) + cos(a - b)

    sin (4x + 3x) - sin (4x - 3x) - sin (2x + x) - sin (2x - x)  /  cos(4x - x) - cos(4x + x) + cos(6x + x) + cos(6x - x)

    sin 7x - sin x - sin 3x + sin x / cos 3 x - cos 5 x + cos 7x + cos 5x

   sin 7 x - sin 3 x / cos 3 x + cos 7 x

  2 cos 7x + 3x / 2 sin 7x - 3 x /2  /  2 cos 3 x + 7 x/2 cos 7 x - 3 x / 2

 2 cos 5 x sin 2 x / 2 cos 5 x cos 2 x

   sin 2x / cos 2 x

tan 2 x

Answer 2

Step-by-step explanation:

Given prove that (cos4x *sin3x - cos2x* sinx) / (sin4x* sinx + cos6x* cosx) = tan 2x

Consider the given , multiply and divide by 2 we get

=  2(cos4x* sin3x - cos2x* sinx) / 2(sin4x* sinx + cos6x* cosx)

= 2cos4x* sin3x - 2cos2x* sinx) / 2sin4x *sinx + 2cos6x* cosx

We know that, 2 cos a sin b = sin (a + b) - sin (a - b)

                         2 sin a sin b = cos(a - b) - cos(a + b)

                         2 cos a cos b = cos(a + b) + cos(a - b)

=  sin (4x + 3x) - sin (4x - 3x) - sin (2x + x) - sin (2x - x)  /  cos(4x - x) - cos(4x + x) + cos(6x + x) + cos(6x - x)

=   sin 7x - sin x - sin 3x + sin x / cos 3 x - cos 5 x + cos 7x + cos 5x

=   sin 7 x - sin 3 x / cos 3 x + cos 7 x

We Know That , sin a - sin b = 2 sin () * cos ()

                         cos a + cos b = 2 cos() * cos()

=  2 cos (7x + 3x / 2)* sin (7x - 3 x /2)  /  2 cos (3x + 7x/2) *cos (7x - 3x / 2)

= 2 cos 5 x* sin 2 x / 2 cos 5 x *cos 2 x

=  sin 2x / cos 2 x

= tan 2 x

Hence Proved

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