Revision:Differentiation of Trigonometry Functions

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Revision:Differentiation of Trigonometry Functions

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TSR Wiki > Study Help > Subjects and Revision > Revision Notes > Mathematics > Differentiation of trig functions

1 What you need to know
2 Proof
3 Applying the Chain Rule: An example
4 Comments

What you need to know

It is possible to find the derivative of trigonometric functions. Here is a list of the standard forms that you need to know:

$\displaystyle \frac{d (\sin x)}{dx} = \cos x$

$\displaystyle \frac{d (\cos x)}{dx} = -\sin x$

$\displaystyle \frac{d (\sec x)}{dx} = \sec x \tan x$

$\displaystyle \frac{d (\csc x)}{dx} = -\csc x \cot x$

$\displaystyle \frac{d (\tan x)}{dx} = \sec ^2 x$

$\displaystyle \frac{d (\cot x)}{dx} = -\csc ^2 x$

One condition upon these results is that x must be measured in radians.

Proof

To differentiate sin x from first principles, let us consider the gradient m of the line between the points (x, sin x) and (x+o, sin(x+o)), where o is small.

$\displaystyle m = \frac{\sin (x+o) - \sin x}{x+o - x} = \frac{\sin (x+o) - \sin x}{o}$

$\displaystyle = \frac{2\cos \tfrac{2x+o}{2} \sin \tfrac{o}{2}}{o} = \frac{\cos \tfrac{2x+o}{2} \sin \tfrac{o}{2}}{\tfrac{o}{2}}$

$\displaystyle =\cos (x + \tfrac{o}{2} ) \frac{\sin (o/2)}{o/2}$

As o tends to 0, [sin(o/2)]/(o/2) tends to 1, and cos(x + o/2) tends to cos x, and so m tends to cos x; that is,

$\dfrac{\text{d}}{\text{d} x} \sin x = \cos x.$

Remember that cos x = sin(π/2 - x). Replacing x with (π/2 - x), and using the chain rule, in the above:

$\dfrac{\text{d}}{\text{d} x} \sin ({\pi\over 2} - x) = -\cos ({\pi\over 2} - x).$

That is,

$\dfrac{\text{d}}{\text{d} x} \cos x = -\sin x.$

Using the quotient rule:

$\displaystyle \frac{\text{d}}{\text{d} x} \tan x = \frac{\text{d}}{\text{d} x} \frac{\sin x}{\cos x} = \frac{\cos^2 x + \sin^2 x}{\cos^2 x} = \sec^2 x.$