Derivatives of Trigonometric Functions: Learn It 1
Calculate the derivatives of sine and cosine functions, including second derivatives and beyond
Determine the derivatives for basic trig functions like tangent, cotangent, secant, and cosecant
Derivatives of the Sine and Cosine Functions
Simple harmonic motion, a type of periodic motion where the restoring force is directly proportional to the displacement, is best described using trigonometric functions like sine and cosine. The behavior of these functions, particularly how they change over time, is crucial in understanding motion dynamics. The derivatives of sine and cosine functions help us compute velocity and acceleration at any point in the motion, linking theoretical physics closely with calculus.
We begin our exploration of the derivative for the sine function by using the limit definition to estimate its derivative.
For a function f(x), the derivative f′(x) is defined as:
f′(x)=limh→0f(x+h)−f(x)h
This allows us to approximate f′(x) for small values of h as:
f′(x)≈f(x+h)−f(x)h.
Using h=0.01, we estimate the derivative of the sine function as follows:
ddx(sinx)≈sin(x+0.01)−sinx0.01
By defining D(x)=sin(x+0.01)−sinx0.01 and plotting this using a graphing tool, we observe an approximation to the derivative of sinx.
Figure 1. The resulting graph of D(x) closely resembles the cosine curve, which supports the derivative relationship.
Upon examination, D(x) appears to be a close match to the graph of the cosine function. This graphical analysis provides a practical demonstration of the derivative, confirming that the derivative of sinx is indeed cosx.
If we were to follow the same steps to approximate the derivative of the cosine function, we would find that
ddx(cosx)=−sinx
derivatives of sinx and cosx
The derivative of the sine function sinx is the cosine function cosx.
ddx(sinx)=cosx
The derivative of the cosine function cosx is the negative sine function −sinx.
ddx(cosx)=−sinx
Proof
Because the proofs for ddx(sinx)=cosx and ddx(cosx)=−sinx use similar techniques, we provide only the proof for ddx(sinx)=cosx.
Before beginning, it is important to recall two important trigonometric limits:
limh→0sinhh=1 and limh→0cosh−1h=0
The graphs of y=(sinh)h and y=(cosh−1)h are shown in Figure 2.
Figure 2. These graphs show two important limits needed to establish the derivative formulas for the sine and cosine functions.
We also recall the following trigonometric identity for the sine of the sum of two angles:
sin(x+h)=sinxcosh+cosxsinh
Now that we have gathered all the necessary equations and identities, we proceed with the proof.
ddxsinx=limh→0sin(x+h)−sinxhApply the definition of the derivative.=limh→0sinxcosh+cosxsinh−sinxhUse trig identity for the sine of the sum of two angles.=limh→0(sinxcosh−sinxh+cosxsinhh)Regroup.=limh→0(sinx(cosh−1h)+cosx(sinhh))Factor outsinxandcosx.=sinx⋅0+cosx⋅1Apply trig limit formulas.=cosxSimplify.
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The figure below shows the relationship between the graph of f(x)=sinx and its derivative f′(x)=cosx. Notice that at the points where f(x)=sinx has a horizontal tangent, its derivative f′(x)=cosx takes on the value zero. We also see that where f(x)=sinx is increasing, f′(x)=cosx>0 and where f(x)=sinx is decreasing, f′(x)=cosx<0.
Figure 3. Where f(x) has a maximum or a minimum, f′(x)=0. That is, f′(x)=0 where f(x) has a horizontal tangent. These points are noted with dots on the graphs.
A particle moves along a coordinate axis in such a way that its position at time t is given by s(t)=2sint−t for 0≤t≤2π. At what times is the particle at rest?
To determine when the particle is at rest, set s′(t)=v(t)=0. Begin by finding s′(t). We obtain
s′(t)=2cost−1,
so we must solve
2cost−1=0 for 0≤t≤2π.
The solutions to this equation are t=π3 and t=5π3.
Thus the particle is at rest at times t=π3 and t=5π3.