Trigonometry/Derivative of Sine

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To find the derivative of sin(θ).

\frac{d}{dx}\bigl[\sin(x)\bigr]=\lim_{h\to 0}\frac{\sin(x+h)-\sin(x)}{h}=\lim_{h\to 0}\frac{2\cos\bigl(x+\frac{h}{2}\bigr)\sin\bigl(\frac{h}{2}\bigr)}{h}=\lim_{h\to 0}\left[\cos\bigl(x+\tfrac{h}{2}\bigr)\frac{\sin\bigl(\frac{h}{2}\bigr)}{\frac{h}{2}}\right] .

Clearly, the limit of the first term is \cos(x) since \cos(x) is a continuous function. Write k=\frac{h}{2} ; the second term is then

\frac{\sin(k)}{k} .

Which we proved earlier tends to 1 as k\to 0 .

And since

k\to 0\text{ as } h\to 0 ,

the limit of the second term is 1 too. Thus

\frac{d}{dx}\bigl[\sin(x)\bigr]=\cos(x) .