Electromagnetic waves are used to describe a wide variety of phenomena, including radio waves. These waves are intercepted by antennas and the information contained in the wave is processed, giving us means of communication.
The periodic behavior of waves is often described using trigonometric functions such as sine, cosine, and tangent. The derivatives of trigonometric functions are also required if we were to study the propagation of waves.
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 the sine function, the cosine function, and the tangent function are given as follows:
\[\dfrac{d}{dx}\sin(x)=\cos(x)\]
\[\dfrac{d}{dx}\cos(x)=-\sin(x)\]
\[\dfrac{d}{dx}\tan(x)=\sec^2(x)\]
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.
The functions are also known as sin, cos and tan. The article now will use this convention.
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:
The derivate of the sine function is a cosine function with the same sign.
The derivate of the cosine function is a sine function with a different sign.
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.
Consider the function \(f(x)=\sin(x^2)\). We will find its derivative using the derivative of the sine function, the Chain Rule, and the Power Rule.
Let \(u=x^2\) ![]()
and differentiate using the Chain Rule.
\[\dfrac{df}{dx}=\dfrac{d}{du}\sin(u)\dfrac{du}{dx}\]
Differentiate the sine function.
\[\dfrac{df}{dx}=\cos(u)\dfrac{du}{dx}\]
Find \(\dfrac{du}{dx}\) using the Power Rule.
\[\dfrac{du}{dx}=2x\]
Substitute back \(u=x^2\) and \(\dfrac{du}{dx}=2x\).
\[\dfrac{df}{dx}=\left( \cos(x^2) \right) \cdot 2x\]
Rearrange the equation.
\[\dfrac{df}{dx}=2x \cdot cos(x^")\]
We will now find the derivative of a function involving the cosine function.
Consider the function \(g(x)=\cos^4(x)\). We will find its derivative using the derivative of the cosine function, the Power Rule, and the Chain Rule. Do not forget that the derivative of the cosine function is the negative of the sine function!
Let \(u=\cos(x)\) and differentiate using the chain rule.
\[\dfrac{dg}{dx}=\dfrac{d}{du}u^4\dfrac{du}{dx}\]
Differentiate using the Power Rule.
\[\dfrac{dg}{dx}=4u^3 \dfrac{du}{dx}\]
Find \(\\dfrac{du}{dx}) by differentiating the cosine function.
\[\dfrac{du}{dx}=-\sin(x)\]
Substitute back \(u=\cos(x)\) and \(\dfrac{du}{dx}=-\sin(x)\).
\[\dfrac{dg}{dx}=(4\cos^3(x))(-\sin(x))\]
Rearrange.
\[\dfrac{dg}{dx}=-4\sin(x)\cos^3(x)\]
The derivate of a function involving the tangent function is straightforward. Let's take a look at one more example.
Consider the function \(r(x)=tan(x^2-1)\). We will find its derivative using the derivative of the tangent function, the Chain Rule, and the Power Rule.
Let \(u=x^2-1\) and differentiate using the chain rule.
\[\dfrac{dr}{dx}=\dfrac{d}{du}\tan(u)\dfrac{du}{dx}\]
Differentiate the tangent function.
\[\dfrac{dr}{dx}=\sec^2(u)\dfrac{du}{dx}\]
Find \(\dfrac{du}{dx}\) using the Power Rule.
\[\dfrac{du}{dx}=2x\]
Substitute back \(u=x^2-1\) and \(\dfrac{du}{dx}=2x\).
\[\dfrac{dr}{dx}=\left( \sec^2(x^2-1) \right) \cdot (2x)\]
Rearrange.
\[\dfrac{dr}{dx}=2x\sec^2(x^2-1)\]
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!
We will first prove the limit \(lim_{h \rightarrow 0} \dfrac{sin(h)}{h}\). Consider the unit circle and the triangles in the following diagram.
Fig. 1. Diagram showing different areas related to angle \(h\).
Let \(S_1\) be the area of the isosceles triangle \(OAC\), \(S_2\) the area of the circular sector \(OAC\), and \(S_3\) the area of the right triangle \(OAB\). The area of the triangles can be found by noting that their base is equal to \(1\), the height of the triangle \(OAC\) is equal to \(\sin(h)\) and the height of the triangle \(OAB\) is equal to \(\tan(h)\).
\(S_1=\dfrac{1}{2} \sin(h)\) and \(S_2=\dfrac{1}{2} \tan(h)\)
We can find the area \(S_2\)with the formula for the area of a circular sector.
\[S_2=\dfrac{1}{2}h\]
Note that \(S_3\) contains \(S_2\), which in turn contains \(S_1\). This means that we can set the following inequality:
\[S_3>S_2>S_1\]
By substituting the expressions for each area in the above inequality we can write the following:
\[\dfrac{1}{2}\tan(h)>\dfrac{1}{2}h>\dfrac{1}{2}\sin(h)\]
Next, we divide the whole inequality by \(\dfrac{1}{2}\sin(h)\):
\[\dfrac{1}{\cos(h)}>\dfrac{h}{\sin(h)}>1\]
We can take the reciprocal of each term of the inequality, reversing the inequality signs.
\[\cos(h)< \dfrac{\sin(h)}{h}<1\]
By The Squeeze Theorem, the values of \(\dfrac{\sin(h)}{h}\) are being squeezed between \(\cos(h)\) and 1\(1\) as \(h \rightarrow 0\).
Since \(\cos(0)=1\) we can conclude that \(lim_{h \rightarrow 0} \dfrac{\sin(h)}{h}=1\).
Let's now work on the second limit with some algebra.
\[lim_{h \rightarrow 0} \dfrac{\cos(h)-1}{h} = lim_{h \rightarrow 0} \dfrac{\cos(h)-1}{h}\left( \dfrac{cos(h)+1}{cos(h)+1} \right) = lim_{h \rightarrow 0} \dfrac{\cos^2(h)-1}{h(\cos(h)+1)}\]
We can now use the Pythagorean identity and the product of limits property.
\[lim_{h \rightarrow 0} \dfrac{\cos(h)-1}{h}=lim_{h \rightarrow 0} \dfrac{-\sin^2(h)}{h(\cos(h)+1)}= - \left( lim_{h \rightarrow 0} \dfrac{\sin(h)}{h} \right) \left( lim_{h \rightarrow 0} \dfrac{\sin(h)}{\cos(h)+1} \right)\]
The first limit is equal to 1 as we found previously. The second limit can be evaluated to find that it is equal to \(0\).
\[lim_{h \rightarrow 0} \dfrac{\cos(h)-1}{h}= 1\cdot(0)=0\]
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!
If we already know the derivative of the sine function we can use the Pythagorean trigonometric identity to find the derivative of the cosine function. Consider the following Pythagorean trigonometric identity:
\[\sin^2(x)+cos^2(x)=1\]
We can differentiate with respect to \(x\) both sides of the equation. Since the right-hand side of the equation is equal to a constant, its derivative is equal to \(0\).
\[\dfrac{d}{dx}\left( \sin^2(x)+cos^2(x)\right)=0\]
The chain rule can be used on the left-hand side of the equation.
\[2\sin(x)\dfrac{d}{dx}\sin(x)+2\cos(x)\dfrac{d}{dx}\cos(x)=0\]
We found previously that the derivative of the sine function is the cosine function, so we will substitute that result in the above equation.
\[2\sin(x)\cos(x)+2\cos(x)\dfrac{d}{dx}\cos(x)=0\]
Finally, we divide the equation by \(2\cos(x)\) and isolate the derivative of \(\cos(x)\).
\[\dfrac{d}{dx}\cos(x)=-\sin(x)\]
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.
\[\dfrac{d}{dx} sin^{-1}(x)=\dfrac{1}{\sqrt{1-x^2}}\]
\[\dfrac{d}{dx} cos^{-1}(x)=\dfrac{-1}{\sqrt{1-x^2}}\]
\[\dfrac{d}{dx} tan^{-1}(x)=\dfrac{1}{1-x^2}\]
Derivatives of Sin, Cos and Tan - Key takeaways
- The derivative of the sine function is the cosine function. That is, \(\dfrac{d}{dx}\sin(x)=\cos(x)\).
- The derivative of the cosine function is the negative of the sine function. That is,\(\dfrac{d}{dx}\cos(x)=-\sin(x)\).
- The derivative of the tangent function is the secant function squared. That is: \(\dfrac{d}{dx}\tan(x)=\sec^2(x)\).
- Two important limits are used for proving the derivatives of the sine function and cosine function. These are: \(lim_{h \rightarrow 0} \dfrac{\sin(h)}{h}=1\) and \(lim_{h \rightarrow 0} \dfrac{\cos(h)-1}{h}=0\).
- The derivative of the tangent function can be found using either the quotient rule or the definition of a derivative.
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