If x = sin t, y = sin kt, show that

Given: x = sin t

and y = sin kt

Proof: We have, x = sin t

Differentiate with respect to t, we get

…(i)

Now, y = sin kt

Differentiate with respect to t, we get

…(ii)

[from (i) & (ii)]

⇒ y’ cos t = k cos kt …(iii)

Now, we have to find y’’

By Quotient Rule

⇒ y’’ (cos² t)(cos t) = -k² cos t sin kt + k cos kt sin t

We know that,

cos²x + sin²x = 1

⇒ y’’ cos t (1 – sin² t) = -k² cos t sin kt + k cos kt sin t

⇒ y’’ cos t(1 – x²) = -k² cos t y + k cos kt x

[given: x = sin t & y = sin kt]

⇒ (1 – x²)cos t y’’ = -k²ycos t + kx cos kt

[from (iii)

Hence proved