Derivatives of Sin, Cos and Tan

Derivation of the trigonometric functions Sine, Cosine, and Tangent

The derivatives of the sine function, the cosine function, and the tangent function all involve more trigonometric functions.

The derivatives of these trigonometric functions, along with basic differentiation rules, can be used to find the derivatives of the other trigonometric functions: secant, cosecant, and cotangent. Let's first take a look at some examples involving the trigonometric functions sine, cosine, and tangent.

Derivatives of sin, cos and tan rules and tricks

If you see the derivates and their formulas it is easy to see a pattern. The derivates of the first two functions \(sin\) and \(cos\) are the opposite of the original functions.

In this case, there is a simple rule or trick you can memorise:

1. The derivate of the sine function is a cosine function with the same sign.

2. The derivate of the cosine function is a sine function with a different sign.

3. The derivate of a tangent function is the \(\dfrac{1}{\cos^2(x)}\)

Derivatives of Sin, Cos, and Tan: Examples

Let's start with the derivative of a function involving the sine function.

We will now find the derivative of a function involving the cosine function.

The derivate of a function involving the tangent function is straightforward. Let's take a look at one more example.

We have been using the differentiation rules for these trigonometric functions without proving them. Let's now take a look at how to find the derivative of each function.

Proof of the derivates of Sin, Cos and Tan functions

You can prove the derivates of each trigonometric function using the definition of a limit. With this, you will have a better understanding of the derivates of each one.

Differentiating the Sine Function

The derivative of the sine function can be found by using the definition of the derivative of a function.

\[\dfrac{d}{dx}\sin(x) = lim_{h \rightarrow 0 \dfrac{\sin(x+h)-\sin(x)}{h}}\]

We can now use the identity for the sine of the sum of two angles to rewrite the above expression.

\[\dfrac{d}{dx}\sin(x)=lim_{h \rightarrow 0} \dfrac{\sin(x)\cosh(h)+\sinh(h)\cos(x)-\sin(x)}{h}\]

This can be rewritten using algebra and the properties of limits

\[\dfrac{d}{dx}\sin(x)=\sin(x)lim_{h \rightarrow 0} \dfrac{\cos(h)-1}{h}+\cos(x)lim_{h \rightarrow 0} \dfrac{\sin(h)}{h}\]

The value of the involved limits can be found by using The Squeeze Theorem.

If we use the limits \(\lim_{h \rightarrow 0} \dfrac{sin(h)}{h}=1\) and then \(\lim_{h \rightarrow 0} \dfrac{cos(h)-1}{h}=1\)

We find the derivative of the sine function by substituting the above expressions.

\[\dfrac{d}{dx}\sin(x)=\cos(x)\]

For this derivation, we used the values of two limits without proving them. For the sake of completeness, let's dive into their proof!

Differentiating the Cosine Function

The derivative of the cosine function can be found in a similar way.

\[\dfrac{d}{dx}\cos(x)= lim_{h \rightarrow 0} \dfrac{\cos(x+h)-\cos(x)}{h}\]

We can now use the identity for the cosine of the sum of two angles to rewrite the above expression.

\[\dfrac{d}{dx}\cos(x)=lim_{h \rightarrow 0} \dfrac{\cos(x)scos(h)-\sin(x)\sin(h)-\cos(x)}{h}\]

Once again, we rewrite this with the help of some algebra and the properties of limits.

\[\dfrac{d}{dx}\cos(x)=cos(x)lim_{h \rightarrow 0} \dfrac{\cos(h)-1}{h}-\sin(x)lim_{h \rightarrow 0} \dfrac{\sin(h)}{h}\]

Next, we substitute the values of the above limits and find the derivative of the cosine function.

\[\dfrac{d}{dx}\cos(x)=-\sin(x)\]

Using the definition of derivative is not the only way to prove the derivative of the cosine function. We can use the derivative of the sine function along with trigonometric identities in our favor!

Differentiating the Tangent Function

We can also use the definition of a derivative to find the derivative of the tangent function. However, since we already know the derivatives of the sine and the cosine functions we can try using the quotient rule instead. We begin by writing the tangent function as the quotient of the sine function and the cosine function.

\[\dfrac{d}{dx} \tan(x)=\dfrac{d}{dx} \left( \dfrac{\sin(x)}{\cos(x)} \right)\]

Next, we use the quotient rule.

\[\dfrac{d}{dx}\tan(x)=\dfrac{ \left( \dfrac{d}{dx}\sin(x)\right)\cdot \cos(x)- \left( \dfrac{d}{dx} \cos(x) \right) \cdot \sin(x) }{\cos^2(x)}\]

Let's now substitute the derivatives of the sine and the cosine functions.

\[\dfrac{d}{dx}\tan(x) = \dfrac{\cos^2(x)+\sin^2(x)}{cos^2(x)}\]

The numerator can be simplified with the Pythagorean trigonometric identity.

\[\dfrac{d}{dx} \tan(x)=\dfrac{1}{\cos^2(x)}\]

This can be further simplified if we recall that the secant function is the reciprocal of the cosine function.

\[\dfrac{d}{dx}\tan(x)=\sec^2(x)\]

In this case, using the quotient rule is faster and easier than using the definition of a derivative!

Derivative of the inverse functions of Sin, Cos, and Tan

The functions sin, cos and tan also have an inverse\(f^1\). These functions are the next ones.

• \(arcsine\) or \(\sin^{-1}\).

• \(arccosine\) or \(\cos^{-1}\).

• \(arctangent\) or \(\tan^{-1}\).

Their derivates are found in the formulas below.

\[\dfrac{d}{dx} sin^{-1}(x)=\dfrac{1}{\sqrt{1-x^2}}\]

\[\dfrac{d}{dx} cos^{-1}(x)=\dfrac{-1}{\sqrt{1-x^2}}\]

\[\dfrac{d}{dx} tan^{-1}(x)=\dfrac{1}{1-x^2}\]