Question

sinx+sin3x+...+sin(2n-1)x=sin^2.nx/sinx solve this question by using principle of Mathematical induction

Answer 1

Heya....hope it helps u....

@skb

Answer 2

Answer:

Testing n=1:

$sinx = \dfrac {sin^2x} {sinx} \\\\\text {P_{n} is true for n = 1}$

Now let's assume $P_n$ is true for n = k

$sinx+ sin3x +.......... + sin(2k - 1)x = \dfrac {sin^2kx}{sinx}$

Now let's assume n = k + 1

$sinx + sin3x +........... + sin(2(k + 1) - 1)x = \dfrac {sin^2(k + 1)x}{sinx}..(1)$

Now add sin(2k + 1) in equation (1)

$sinx + sin3x + ........... + sin(2(k + 1) - 1)x + sin(2k + 1)= \dfrac {sin^2(k + 1)x}{sinx} + sin(2k + 1)$

$\dfrac {sin^2(k + 1)x}{sinx} + sin(2k + 1)$

$= \dfrac {sin^2 kx + sin(2kx + x)sinx}{sinx}\\= \dfrac {sin^2 kx + \dfrac {cos(2kx + x - x)- cos(2kx + x + x)}{2}}{sinx}\\\\= \dfrac {sin^2 kx + \dfrac {cos2kx- cos2(k + 1)x}{2}}{sinx}\\\\= \dfrac {sin^2 kx + sin^2(k + 1)x - sin^2kx}{sinx}\\= \dfrac {sin^(k + 1)x}{sinx}$