The derivationof sine to cosine

Question

I am new to mathematics somehow, and I know that:

$\frac{d(\sin x)}{dx}= \cos x$

But I can't understand this:

$\frac{d (a \sin(\omega.t))}{dt} = a.\omega\cos(\omega.t)$

where $a$ is the peak of the wave and $\omega$ is $2\pi f$ (where $f$ is frequency)

Could anyone please explain it for me?

Answer 1

$\frac{d (a \sin(\omega.t))}{dt} = \frac{d(a.\sin(\omega t))}{d(\omega t)}.\frac{d(\omega t)}{dt} = a . \frac{d(\sin(\omega t))}{d(\omega t)}.\frac{\omega.d(t)}{dt}=a.\cos(\omega t).\omega$.

Here this type of above method is known as chain rule .Explore it more! Hope it helps!

Answer 2

Here chain rule is used.

$\frac {d}{dt} (\sin (\omega.t)) = \cos (\omega.t). \frac {d}{dt} (\omega.t)$

= $ \cos (\omega.t). \omega. \frac{d}{dt} t$

= $ \cos (\omega.t). \omega . 1$

= $\omega. \cos (\omega.t)$