# Derivatives of Trigonometric functions.

As you know,

The functions SINE x(sin x) , CO-SECANT x(cos x) , TANGENT x(tan x), CO-SECANT x(csc x), SECANT x(sec x) and COTANGENT x(cot x)  are called trigonometrical functions.

We are going to learn and prove ,what are the Derivatives of these Trigonometrical Functions , here.

a> Derivative of sin x:

The derivative of sin x is: $\dfrac{d}{dx}\sin x = \cos x$

Proof:

Ley y=SIN x and let this be equation (i)

and let $\Delta x$ be a small increment in x and $\Delta y$ be the corresponding small increment in y,

Then we can write: $y + \Delta y = \sin (x+ \Delta x)$ and let it be equation (ii).

Now if we subtract equation (i) from equation (ii), we can get: $\Delta y = \sin (x+ \Delta x) - \sin x$

Now using the trigonometrical formula ( sin c – sin d = 2.sin((c-d)/2).cos((c+d)/2)

We get: $\Delta y = 2 \times \sin \frac{\Delta x}{2} \times \cos \frac{2x+\Delta x}{2}$

Now dividing both side of above equation by $\Delta x$ we get: $\dfrac{\Delta y}{\Delta x} = \dfrac{\sin\frac{\Delta x}{2}}{\frac{\Delta x}{2}}\times\cos\frac{2x+\Delta x}{2}$

Thus: $\dfrac{d}{dx}y = \dfrac{d}{dx}\sin x = \displaystyle\lim_{{\Delta x}\to 0}\left(\dfrac{\sin \frac{\Delta x}{2}}{\frac{\Delta x}{2}}\times\cos\frac{2x+\Delta x}{2}\right)$

or, $\dfrac{d}{dx}\sin x = 1 \times \cos \frac{2x}{2} = \cos x$

b> Derivative of cos x.

The derivative of cos x is: $\frac{d}{d x}\cos x = -\sin x$

Proof:

Let y=cos x

and from trigonometric relations we also know: $\cos x = \sqrt{1-\sin ^{2}x}$

So, $\frac{d }{d x} y = \frac{d }{d x}\left(\sqrt{1-\sin ^{2}x}\right)$

Now using chain rule : $\frac{d }{d x} y = \frac{d }{d \left(1-\sin ^{2}x\right)} \left(1-\sin ^{2}x\right)^{\frac{1}{2}} \times \frac{d }{d x}\left(1-\sin ^{2}x\right)$

or, $\frac{d }{d x} y = \frac{0-2\times\sin x\times\cos x}{2\times \sqrt{1-\sin ^{2}x}} = -\sin x$

Thus , $\frac{d }{d x} \cos x = -\sin x$

c> Derivative of  tan x.

The derivative of  tan x is: $\dfrac{d}{dx}\tan x = \sec ^{2}x$

Proof: $\dfrac{d}{dx}\tan x = \dfrac{\sin x}{\cos x}$

Now using the quotient rule : or, $\dfrac{d }{dx}\tan x =\dfrac{\cos ^2 x + \sin ^2 x}{\cos ^{2}x}$

Thus, $\dfrac{d}{dx}\tan x = \sec ^{2}x$

d> Derivative of csc x , secx and cot x.

Using the relation:

csc x= 1/sin x

sec x= 1/cos x

and

cot x = 1/tan x

and then using the quotient rule, we can find the:

Derivative of csc x: $\frac{d }{d x}\csc x = - \csc x \times \cot x$

Derivative of sec x: $\frac{d }{d x}\sec x = \sec x \times \tan x$

Derivative of cot x: $\frac{d }{d x}\cot x = -\csc ^{2} x$

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