Derivative of Trigonometric Functions(Sin Cos Tan)

Derivatives of Trig Functions

Since from the old concept of trigonometry, we have learned that there are three main functions of trigonometry on which other trigonometric function are based on:

These trigonometric functions are:

Sin x
Cos x
Tan x

The derivatives of trig functions i.e. sin cos tan are given below:

$$ \frac{d}{dx}Sin x = Cosx $$ $$ \frac{d}{dx}Cosx = -Sinx $$ $$ \frac{d}{dx}Tan x = sec^2x $$

Now we will try to prove the derivative of trigonometric functions i.e. sin cos tan by using the definition of derivatives.

So, let’s start from the prove of derivative of sin x

Derivative of Sin x:

As we discussed we will use the definition of derivative for the proof of these derivative of trigonometric functions. So the derivative in terms of its definition may be written as:

$$ \text{Let:} \lim\limits_{h \to 0} \frac{f(x+h) - f(x)}{h} $$

Then of course the function that we are talking about is sin x. For this right here we just have to plug in the f(x)= sin x .

Also,

f(x+h) = sin(x+h)

Now put it in definition of derivative,

$$ \implies \lim\limits_{h \to 0} \frac{sin(x+h) - sin(x)}{h} $$

We can use here a trigonometry identity which is given as:

sin(x+h) = sin(x)cox(h)+cos(x)sin(h)

Now, put that identity in our derivative

$$ \implies \lim\limits_{h \to 0} \frac{sin(x)cos(h)+cos(x)sin(h) - sin(x)}{h} $$

We can also write it in such a way,

$$ \lim\limits_{h \to 0} \left( \frac{sin(x)(cos(h)-1)}{h} + \frac {cos(x)sin(h)}{h} \right) $$

Now, if we follow the sum rule of limit, we can distribute the limit to both terms which are added here.

$$ sin(x) \lim\limits_{h \to 0} \left( \frac{cos(h)-1}{h} \right) + cos(x) \lim\limits_{h \to 0} \left( \frac {sin(h)}{h} \right) $$

Now, we have to work on both of these limits which looks strange,

$$ \lim\limits_{h \to 0} \left( \frac{cos(h)-1}{h} \right) \;\;\;\;\; \text{And} \;\;\;\;\; \lim\limits_{h \to 0} \left( \frac {sin(h)}{h} \right) $$

Limit of Sin(h)/h:

Start with the limit of sin(h)

We have a limit

$$ \lim\limits_{h \to 0} \frac {sin(h)}{h} $$

By using the concept of gemotery, we will get following :

We can observe here that:

$$ \text{Area of triangle AOB} \;\;\lt\;\; \text{Area of Sector AOB} \;\;\lt\;\; \text{Area of triangle AOC} $$

Also,

$$ \frac{1}{2} r^2 sin(h) \;\;\lt\;\; \frac{1}{2} r^2 h \;\;\lt\;\; \frac{1}{2} r^2 tan(h) $$

By dividing all of these terms by ½r²sin(h), we get

$$ 1 \;\;\lt\;\; \frac{h}{sin (h)} \;\;\lt\;\; \frac{1}{cos (h)} $$ $$ 1 \;\;\gt\;\; \frac{sin (h)}{h} \;\;\lt\;\; \cos (h) $$

As, h --> 0 then cos(h) --> 1

So, we can see Sin(h)/h lies between 1 & some number which is very close to 1.

So, the derivative of sin x is:

$$ \lim\limits_{h \to 0} \frac {sin(h)}{h} = 1 $$

Limit of (cos(h)-1)/h:

Now, we will find the limit in term of cos(h)

$$ \lim\limits_{h \to 0} \left( \frac {cos(h)-1}{h} \right) $$

If we use the conjugate method of evaluating limits here, it will become

$$ \frac{(cos(h)-1)(cos(h)+1)}{h(cos(h)+1)} = \frac{(cos^2(h)-1)}{h(cos(h)+1)} $$

Now we use the trigonometric identity here:

cos²(h) + sin²(h) = 1

⇒ cos²(h) -1 = sin²(h)

Now, our fraction becomes

$$ \frac{-sin^2(h)}{h(cos(h)+1)} $$

Also, Limit as h-->0 will

$$ \lim\limits_{h \to 0} \left( \frac{-sin^2(h)}{h(cos(h)+1)} \right) $$

It will be same as:

$$ \lim\limits_{h \to 0} \left(\frac {sin(h)}{h} \right) \times \lim\limits_{h \to 0} \left( \frac{-sin (h)}{cos(h) + 1} \right) $$

We know the limit of sin(h)/h and now put the limit in second limit as h-->0.

$$ \lim\limits_{h \to 0} \left(\frac {sin(h)}{h} \right) \times \lim\limits_{h \to 0} \left( \frac{-sin (h)}{cos(h) + 1} \right) = 1 \times \frac {0}{1} = 0 $$ $$ \implies \lim\limits_{h \to 0} \frac {cos(h)-1}{h} = 0 $$

Replace back in derivative of Sin x:

$$ \text{Since,} \;\;\;\; \frac{d}{dx}sinx = sin(x) \lim\limits_{h \to 0} \left( \frac{cos(h)-1}{h} \right) + cos(x) \lim\limits_{h \to 0} \left( \frac {sin(h)}{h} \right) $$ $$ \frac{d}{dx}sinx = sin(x) \times 0 + cos(x) \times 1 $$ $$ \frac{d}{dx}sinx = cos(x) $$

We Prove it, Now we have to work for the derivative of cos(x).

Derivative of Cos x:

Now we are going to figure out the derivative of cosine x and of course one of the ways to do it is to just use the definition of derivative as we have done with sin x.

But instead of using the definition of derivative, we will do an easy way of angle sums along with the implementation of chain rule.

So, we have to take the derivative of cos x

$$ \frac{d}{dx} cos x $$

If we use the angle sum or trigonometric identity here, we can write cos in terms of sin, because sin and cos are the same but compliment in their angles.

It mean if we have cos with an angle of “x”, it will be sin with complemented angle like that:

$$ \frac{d}{dx} \left( sin(\frac{?}{2}-x) \right) $$

We can observe here, sin and cos are the same but if we complement their angles because they are perpendicular to each other.

Now, we simply take the derivative of sin but don’t forget to implement Chain rule also, Chain Rule is stated as derivative of the outer function times to the derivative of inner function.

$$ cos(\frac{?}{2}-x) \times \frac{d}{dx} \left( \frac{?}{2}-x \right) $$ $$ cos(\frac{?}{2}-x) \times (-1) $$

Now, take a look at the above steps, we just follow the same method to convert “cos to sin” for making an angel in terms of “x”.

As cos and sin are complement to each other so,

$$ \implies \frac{d}{dx}cos x = sin(x) \times (-1) $$ $$ \implies \frac{d}{dx}cos x = -sin(x) $$

Surely, it's the simplest way to prove it.

Derivative of tan x:

For proving the derivative of tan x, we will use the trigonometric identity

Since,

$$ tan x = \frac{sinx}{cosx} $$

if we take the derivative on the both sides of it

$$ \frac{d}{dx}tan x = \frac{d}{dx} \left( \frac{sinx}{cosx} \right) $$

Here, We will use the Quotient Rule of Differentiation which is stated as:

$$ \frac{d}{dx} (\frac{f}{g} ) = \frac{gf'-fg'}{g^2}$$

In our case

$$ \frac{d}{dx}tan x = \frac{cosx \times cosx + sinx \times sinx}{cos^2 x} $$ $$ \frac{d}{dx}tan x = \frac{cos^2x + sin^2x}{cos^2 x} $$

Now, we use again that identity

cos²(x) + sin²(x) = 1

This will implies as:

$$ \implies \frac{d}{dx}tan x = \frac{1}{cos^2 x} $$

Now, it's all done but its look more appropriate if we use

sec(x) = 1/cos(x)

$$ \implies \frac{d}{dx}tan x = sec^2x $$

So, these are some general trigonometric functions(sin cos tan) which we proved in terms of definition of derivative and also in terms of trigonometric identity.