Question

calculate the amount sinx + sin 3x+.....+sin(2n +1)

Answer 1

Answer:

sin²(n+1)x/cosx

Step-by-step explanation:

Hi,

Let the sum S = sinx + sin 3x+.....+sin(2n +1),

Now multiplying  by sinx on both sides we get,

sinx.S = sinx(sinx + sin 3x+.....+sin(2n +1)x)

= sin²x + sinx.sin3x + ........+sinx.sin(2n+1)x

= 1/2[ 1 - cos2x + cos2x - cos4x +.........+ cos(2n)x - cos(2n+2)x]

=1/2[1 - cos2(n+1)x]

=1/2*2sin²(n+1)x

= sin²(n+1)x

⇒cosx.S = sin²(n+1)x

⇒S = sin²(n+1)x/cosx

Hope, it helped !