Math Doubts

Derivative of sinx Proof

The derivative of sin function with respect to a variable is equal to cosine. If $x$ represents a variable, then the sine function is written as $\sin{x}$. Therefore, the differentiation of the $\sin{x}$ with respect to $x$ is equal to $\cos{x}$ and it can be proved mathematically from first principle.

Beginner’s Method

In this method, the differentiation of the sine function is derived mathematically by using the trigonometric identities mostly.

Write Derivative of function in Limit form

As per definition of the derivative, the derivative of the function in terms of $x$ is written in the following limiting operation form.

$\dfrac{d}{dx}{\, f(x)}$ $\,=\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{f(x+h)-f(x)}{h}}$

Take $f{(x)} = \sin{x}$, then $f{(x+h)} = \sin{(x+h)}$. On the basis of this information, the proof of differentiation of $\sin{x}$ with respect to $x$ can be started to derive from the first principle.

$\implies \dfrac{d}{dx}{\, (\sin{x})}$ $\,=\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\sin{(x+h)}-\sin{x}}{h}}$

Simplify the trigonometric expression

The first term of the trigonometric expression in the numerator is sin of sum two angles and it can be expanded by the angle sum identity of sin function. Now, concentrate on simplifying the trigonometric expression.

$=\,\,\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\sin{x}\cos{h}+\cos{x}\sin{h}-\sin{x}}{h}}$

$=\,\,\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\sin{x}\cos{h}-\sin{x}+\cos{x}\sin{h}}{h}}$

$=\,\,\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\sin{x}{(\cos{h}-1)}+\cos{x}\sin{h}}{h}}$

$=\,\,\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\sin{x}{(-(1-\cos{h}))}+\cos{x}\sin{h}}{h}}$

$=\,\,\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{-\sin{x}{(1-\cos{h})}+\cos{x}\sin{h}}{h}}$

$=\,\,\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \Bigg[ \dfrac{-\sin{x}{(1-\cos{h})}}{h}+\dfrac{\cos{x}\sin{h}}{h}\Bigg]}$

$=\,\,\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{{(-\sin{x})}{(1-\cos{h})}}{h}}$ $+$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\cos{x}\sin{h}}{h}}$

The input of the limiting operation is in terms of $h$. So, $-\sin{x}$ and $\cos{x}$ become constants and they can be written separately by the multiple constant rule of limits.

$=\,\,\,$ ${(-\sin{x})} \displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{1-\cos{h}}{h}}$ $+$ $\cos{x} \times \displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\sin{h}}{h}}$

Evaluating Limits of trigonometric functions

According to limits of trigonometric functions formulas, the limit of ratio of $\sin{h}$ to $h$ as $h$ tends to zero is equal to one.

$=\,\,\,$ ${(-\sin{x})} \displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{1-\cos{h}}{h}}$ $+$ $\cos{x} \times 1$

$=\,\,\,$ ${(-\sin{x})} \displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{1-\cos{h}}{h}}$ $+$ $\cos{x}$

Now, evaluate the limit of the trigonometric function as $h$ approaches $0$ by the direct substitution method.

$=\,\,\,$ ${(-\sin{x})} \displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{1-\cos{(0)}}{0}}$ $+$ $\cos{x}$

$=\,\,\,$ ${(-\sin{x})} \displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{1-1}{0}}$ $+$ $\cos{x}$

$=\,\,\,$ ${(-\sin{x})} \displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{0}{0}}$ $+$ $\cos{x}$

In this case, it is not possible to determine the limit of the function. So, let us try to evaluate this trigonometric function in another method.

$=\,\,\,$ ${(-\sin{x})} \displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{1-\cos{h}}{h}}$ $+$ $\cos{x}$

Use the power reducing trigonometric identity to express the numerator in square form.

$=\,\,\,$ ${(-\sin{x})} \displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{2\sin^2{\Big(\dfrac{h}{2}\Big)}}{h}}$ $+$ $\cos{x}$

$=\,\,\,$ ${(-\sin{x})} \displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\sin^2{\Big(\dfrac{h}{2}\Big)}}{\dfrac{h}{2}}}$ $+$ $\cos{x}$

$=\,\,\,$ ${(-\sin{x})} \displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\sin{\Big(\dfrac{h}{2}\Big)} \times \sin{\Big(\dfrac{h}{2}\Big)}}{\dfrac{h}{2}}}$ $+$ $\cos{x}$

$=\,\,\,$ ${(-\sin{x})} \displaystyle \large \lim_{h \,\to\, 0}{\normalsize \Bigg[\sin{\Big(\dfrac{h}{2}\Big)} \times \dfrac{\sin{\Big(\dfrac{h}{2}\Big)}}{\dfrac{h}{2}}\Bigg]}$ $+$ $\cos{x}$

Now, use the power rule of limits for evaluating limit of product of two functions as product of their limits.

$=\,\,\,$ ${(-\sin{x})}\Bigg[ \displaystyle \large \lim_{h \,\to\, 0}{\normalsize \sin{\Big(\dfrac{h}{2}\Big)}}$ $\times$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\sin{\Big(\dfrac{h}{2}\Big)}}{\dfrac{h}{2}}\Bigg]}$ $+$ $\cos{x}$

If $h \to 0$, then $\dfrac{h}{2} \to \dfrac{0}{2}$. Therefore, $\dfrac{h}{2} \to 0$. It is proved that if $h$ approaches zero, then $\dfrac{h}{2}$ also approaches zero.

$=\,\,\,$ ${(-\sin{x})}\Bigg[ \displaystyle \large \lim_{h \,\to\, 0}{\normalsize \sin{\Big(\dfrac{h}{2}\Big)}}$ $\times$ $\displaystyle \large \lim_{\frac{h}{2} \,\to\, 0}{\normalsize \dfrac{\sin{\Big(\dfrac{h}{2}\Big)}}{\dfrac{h}{2}}\Bigg]}$ $+$ $\cos{x}$

Take $w = \dfrac{h}{2}$, and convert this second liming function in terms of $w$.

$=\,\,\,$ ${(-\sin{x})}\Bigg[ \displaystyle \large \lim_{h \,\to\, 0}{\normalsize \sin{\Big(\dfrac{h}{2}\Big)}}$ $\times$ $\displaystyle \large \lim_{w \,\to\, 0}{\normalsize \dfrac{\sin{w}}{w}\Bigg]}$ $+$ $\cos{x}$

As per limit of $\dfrac{\sin{x}{x}$ as $x$ approaches zero rule, the limit of the trigonometric function is one.

$=\,\,\,$ ${(-\sin{x})}\Bigg[\sin{\Big(\dfrac{0}{2}\Big)} \times 1 \Bigg]$ $+$ $\cos{x}$

$=\,\,\,$ ${(-\sin{x})}\Bigg[\sin{\Big(\dfrac{0}{2}\Big)}\Bigg]$ $+$ $\cos{x}$

$=\,\,\,$ ${(-\sin{x})}\Big[\sin{(0)}\Big]$ $+$ $\cos{x}$

According to trigonometry, the exact value of sine of zero degrees is zero.

$=\,\,\,$ ${(-\sin{x})}{(0)}$ $+$ $\cos{x}$

$=\,\,\,$ ${(-\sin{x})} \times 0$ $+$ $\cos{x}$

$=\,\,\, 0 + \cos{x}$

$\therefore \,\,\,\,\,\, \dfrac{d}{dx}{\, (\sin{x})} \,=\, \cos{x}$

Advanced Method

It is another method in which only one trigonometric identity is used for simplifying the trigonometric expression in terms of sin functions.

Express Differentiation of function in Limit form

According to definition of the derivative, the differentiation of the function in terms of $x$ is written in the following limiting operation form.

$\dfrac{d}{dx}{\, f(x)}$ $\,=\,$ $\displaystyle \large \lim_{\Delta x \,\to\, 0}{\normalsize \dfrac{f(x+\Delta x)-f(x)}{\Delta x}}$

Take $f{(x)} = \sin{x}$, then $f{(x+\Delta x)} = \sin{(x+\Delta x)}$. Now, the proof of derivative of $\sin{x}$ with respect to $x$ can be started by the first principle.

$\implies \dfrac{d}{dx}{\, (\sin{x})}$ $\,=\,$ $\displaystyle \large \lim_{\Delta x \,\to\, 0}{\normalsize \dfrac{\sin{(x+\Delta x)}-\sin{x}}{\Delta x}}$

Use difference to product transformation rule

Use difference to product identity of sin functions to combine the difference of two sine functions in the numerator of the function.

$= \,\,\,$ $\displaystyle \large \lim_{\Delta x \,\to\, 0}{\normalsize \dfrac{2\cos{\Bigg[\dfrac{x+\Delta x+x}{2}\Bigg]}\sin{\Bigg[\dfrac{x+\Delta x-x}{2}\Bigg]}}{\Delta x}}$

$= \,\,\,$ $\require{cancel} \displaystyle \large \lim_{\Delta x \,\to\, 0}{\normalsize \dfrac{2\cos{\Bigg[\dfrac{2x+\Delta x}{2}\Bigg]}\sin{\Bigg[\dfrac{\cancel{x}+\Delta x-\cancel{x}}{2}\Bigg]}}{\Delta x}}$

$= \,\,\,$ $\displaystyle \large \lim_{\Delta x \,\to\, 0}{\normalsize \dfrac{2\cos{\Bigg[\dfrac{2x+\Delta x}{2}\Bigg]}\sin{\Bigg[\dfrac{\Delta x}{2}\Bigg]}}{\Delta x}}$

Simplify the entire function

The number $2$ multiplies the numerator and it divides the denominator. So, move the number $2$ to denominator.

$= \,\,\,$ $\displaystyle \large \lim_{\Delta x \,\to\, 0}{\normalsize \dfrac{\cos{\Bigg[\dfrac{2x+\Delta x}{2}\Bigg]}\sin{\Bigg[\dfrac{\Delta x}{2}\Bigg]}}{\dfrac{\Delta x}{2}}}$

Now, split this function as product of two functions.

$= \,\,\,$ $\displaystyle \large \lim_{\Delta x \,\to\, 0}{\normalsize \Bigg(\cos{\Bigg[\dfrac{2x+\Delta x}{2}\Bigg]} \times \dfrac{\sin{\Bigg[\dfrac{\Delta x}{2}\Bigg]}}{\dfrac{\Delta x}{2}}\Bigg)}$

Apply product rule of limits to find the limit of the function by the product of their limits.

$= \,\,\,$ $\displaystyle \large \lim_{\Delta x \,\to\, 0}{\normalsize \cos{\Bigg[\dfrac{2x+\Delta x}{2}\Bigg]}}$ $\times$ $\displaystyle \large \lim_{\Delta x \,\to\, 0}{\normalsize \dfrac{\sin{\Bigg[\dfrac{\Delta x}{2}\Bigg]}}{\dfrac{\Delta x}{2}}}$

Evaluate Limit of the function

Now, find the limit of the cos function firstly by the direct substitution method.

$= \,\,\,$ $\cos{\Bigg[\dfrac{2x+0}{2}\Bigg]}$ $\times$ $\displaystyle \large \lim_{\Delta x \,\to\, 0}{\normalsize \dfrac{\sin{\Bigg[\dfrac{\Delta x}{2}\Bigg]}}{\dfrac{\Delta x}{2}}}$

$= \,\,\,$ $\cos{\Bigg[\dfrac{2x}{2}\Bigg]}$ $\times$ $\displaystyle \large \lim_{\Delta x \,\to\, 0}{\normalsize \dfrac{\sin{\Bigg[\dfrac{\Delta x}{2}\Bigg]}}{\dfrac{\Delta x}{2}}}$

$= \,\,\,$ $\require{cancel} \cos{\Bigg[\dfrac{\cancel{2}x}{\cancel{2}}\Bigg]}$ $\times$ $\displaystyle \large \lim_{\Delta x \,\to\, 0}{\normalsize \dfrac{\sin{\Bigg[\dfrac{\Delta x}{2}\Bigg]}}{\dfrac{\Delta x}{2}}}$

$= \,\,\,$ $\cos{x}$ $\times$ $\displaystyle \large \lim_{\Delta x \,\to\, 0}{\normalsize \dfrac{\sin{\Bigg[\dfrac{\Delta x}{2}\Bigg]}}{\dfrac{\Delta x}{2}}}$

The limit of the function in which sin function is involved, is similar to the limit of sinx/x as x approaches 0 rule but the input value is slightly different. So, make this function same as this standard result.

If, $\Delta x \,\to\, 0$, then $\dfrac{\Delta x}{2} \,\to\, \dfrac{0}{2}$. Therefore $\dfrac{\Delta x}{2} \,\to\, 0$. So, it proved that if $\Delta x$ approaches $0$, then $\dfrac{\Delta x}{2}$ also approaches zero.

$= \,\,\,$ $\cos{x}$ $\times$ $\displaystyle \large \lim_{\frac{\Delta x}{2} \,\to\, 0}{\normalsize \dfrac{\sin{\Bigg[\dfrac{\Delta x}{2}\Bigg]}}{\dfrac{\Delta x}{2}}}$

Take $q = \dfrac{\Delta x}{2}$ and convert the entire function in terms of $q$.

$= \,\,\,$ $\cos{x}$ $\times$ $\displaystyle \large \lim_{q \,\to\, 0}{\normalsize \dfrac{\sin{q}}{q}}$

The limit of trigonometric function is exactly same as the limit of $\sin{x}/x$ as $x$ tends to $0$ formula. Therefore, the limit of this trigonometric function is equal to $1$.

$= \,\,\,$ $\cos{x} \times 1$

$= \,\,\,$ $\cos{x}$

$\therefore \,\,\,\,\,\, \dfrac{d}{dx}{\, (\sin{x})} \,=\, \cos{x}$

Therefore, it is proved that the differentiation of sin function with respect to a variable is equal to cosine.



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