Derivative of \( sinθ \) using the First Principle

Let \(y\) = \( sinθ \)

∴ \(y + δy\) = \( sin(θ + δθ) \)

∴ \(δy\) = \( sin(θ + δθ) \) – \( sinθ \)
From Trigonometry , we have \( sin(A-B) \) = 2.\( sin \dfrac {(A-B)}{2} \).\( cos \dfrac {(A+B)}{2} \)
Using the above rule, we get
\(δy\) = 2.\( sin \dfrac {(θ+δθ – θ)}{2} \).\( cos \dfrac {(θ+δθ + θ)}{2} \)

or \(δy\) = 2\( cos (θ+ \dfrac{δθ}{2}) \) . \( sin( \dfrac {δθ}{2}) \)
∴ \(\dfrac {δy}{δθ}\) = 2 \( \dfrac { cos (θ+ \dfrac{δθ}{2}) sin( \dfrac {δθ}{2}) } {δθ}\)
or \(\dfrac {δy}{δθ}\) = 2 \(cos (θ+ \dfrac{δθ}{2}) \) \( \dfrac { sin( \dfrac {δθ}{2}) } {δθ}\)

∴ \(\dfrac {dy}{dx} = \) \( \lim_{δθ \to 0} \) 2 \(cos (θ+ \dfrac{δθ}{2}) \) \( \dfrac { sin( \dfrac {δθ}{2}) } {δθ} \)

= 2 . \( \lim_{δθ \to 0} \) \(cos (θ+ \dfrac{δθ}{2}) \) . \( \lim_{δθ \to 0} \) \( \dfrac { sin( \dfrac {δθ}{2}) } {δθ} \)

= \( \lim_{δθ \to 0} \) \(cos (θ+ \dfrac{δθ}{2}) \) . \( \lim_{δθ \to 0} \) \( \dfrac { sin( \dfrac {δθ}{2}) } {\dfrac {δθ}{2}} \)

Also, from theorems on limits, we know that \( \lim_{x \to 0} \) \( \dfrac {sin(x)} {x} = 1\)

Thus we have
\(\dfrac {dy}{dx} = \) \( \lim_{δθ \to 0} \) \(cos (θ+ \dfrac{δθ}{2}) \) . \( \lim_{\dfrac {δθ}{2} \to 0} \) \( \dfrac { sin( \dfrac {δθ}{2}) } {\dfrac {δθ}{2}} \)

\(\dfrac {dy}{dx} = \) \(cos (θ+ 0) \) . 1 = \(cosθ \)

Hence, \(\dfrac {d}{dx} \)\( sinθ \) = \( cosθ \)