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Tony's Intertwingled Memex • Reference Manual

Derivatives of Trigonometric Functions

Formulas & Definitions
Core Definition

The derivatives of the six fundamental trigonometric functions (with angles in radians) are:

>ddx[sin⁡x]=cos⁡xddx[cos⁡x]=−sin⁡x>ddx[tan⁡x]=sec⁡2xddx[cot⁡x]=−csc⁡2x>ddx[sec⁡x]=sec⁡xtan⁡xddx[csc⁡x]=−csc⁡xcot⁡x>\begin{aligned} > \frac{d}{dx}[\sin x] &= \cos x &\quad \frac{d}{dx}[\cos x] &= -\sin x \\ > \frac{d}{dx}[\tan x] &= \sec^2 x &\quad \frac{d}{dx}[\cot x] &= -\csc^2 x \\ > \frac{d}{dx}[\sec x] &= \sec x \tan x &\quad \frac{d}{dx}[\csc x] &= -\csc x \cot x > \end{aligned}>dxd​[sinx]>dxd​[tanx]>dxd​[secx]​=cosx=sec2x=secxtanx​dxd​[cosx]dxd​[cotx]dxd​[cscx]​=−sinx=−csc2x=−cscxcotx>​
Mnemonic Pattern: Every "co-" function (cos⁡,cot⁡,csc⁡\cos, \cot, \csccos,cot,csc) produces a negative derivative.

Concept Explanation

1. Proofs of Fundamental Trigonometric Derivatives

The derivatives of sin⁡x\sin xsinx and cos⁡x\cos xcosx are established from the limit definition using the addition angle formulas and special trigonometric limits lim⁡θ→0sin⁡θθ=1\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1limθ→0​θsinθ​=1 and lim⁡θ→0cos⁡θ−1θ=0\lim_{\theta \to 0} \frac{\cos \theta - 1}{\theta} = 0limθ→0​θcosθ−1​=0:

ddx[sin⁡x]=lim⁡h→0sin⁡(x+h)−sin⁡xh=lim⁡h→0[sin⁡x(cos⁡h−1h)+cos⁡x(sin⁡hh)]=cos⁡x\frac{d}{dx}[\sin x] = \lim_{h \to 0} \frac{\sin(x+h) - \sin x}{h} = \lim_{h \to 0} \left[ \sin x \left(\frac{\cos h - 1}{h}\right) + \cos x \left(\frac{\sin h}{h}\right) \right] = \cos xdxd​[sinx]=h→0lim​hsin(x+h)−sinx​=h→0lim​[sinx(hcosh−1​)+cosx(hsinh​)]=cosx

The remaining four trigonometric derivatives are derived directly via the Quotient Rule:

  • Derivative of tan⁡x=sin⁡xcos⁡x\tan x = \frac{\sin x}{\cos x}tanx=cosxsinx​:
    ddx[tan⁡x]=(cos⁡x)(cos⁡x)−(sin⁡x)(−sin⁡x)cos⁡2x=cos⁡2x+sin⁡2xcos⁡2x=1cos⁡2x=sec⁡2x\frac{d}{dx}[\tan x] = \frac{(\cos x)(\cos x) - (\sin x)(-\sin x)}{\cos^2 x} = \frac{\cos^2 x + \sin^2 x}{\cos^2 x} = \frac{1}{\cos^2 x} = \sec^2 xdxd​[tanx]=cos2x(cosx)(cosx)−(sinx)(−sinx)​=cos2xcos2x+sin2x​=cos2x1​=sec2x
  • Derivative of sec⁡x=1cos⁡x\sec x = \frac{1}{\cos x}secx=cosx1​:
    ddx[sec⁡x]=(0)(cos⁡x)−(1)(−sin⁡x)cos⁡2x=sin⁡xcos⁡2x=1cos⁡x⋅sin⁡xcos⁡x=sec⁡xtan⁡x\frac{d}{dx}[\sec x] = \frac{(0)(\cos x) - (1)(-\sin x)}{\cos^2 x} = \frac{\sin x}{\cos^2 x} = \frac{1}{\cos x} \cdot \frac{\sin x}{\cos x} = \sec x \tan xdxd​[secx]=cos2x(0)(cosx)−(1)(−sinx)​=cos2xsinx​=cosx1​⋅cosxsinx​=secxtanx
  • Derivative of cot⁡x=cos⁡xsin⁡x\cot x = \frac{\cos x}{\sin x}cotx=sinxcosx​:
    ddx[cot⁡x]=(−sin⁡x)(sin⁡x)−(cos⁡x)(cos⁡x)sin⁡2x=−(sin⁡2x+cos⁡2x)sin⁡2x=−1sin⁡2x=−csc⁡2x\frac{d}{dx}[\cot x] = \frac{(-\sin x)(\sin x) - (\cos x)(\cos x)}{\sin^2 x} = \frac{-(\sin^2 x + \cos^2 x)}{\sin^2 x} = -\frac{1}{\sin^2 x} = -\csc^2 xdxd​[cotx]=sin2x(−sinx)(sinx)−(cosx)(cosx)​=sin2x−(sin2x+cos2x)​=−sin2x1​=−csc2x
  • Derivative of csc⁡x=1sin⁡x\csc x = \frac{1}{\sin x}cscx=sinx1​:
    ddx[csc⁡x]=(0)(sin⁡x)−(1)(cos⁡x)sin⁡2x=−cos⁡xsin⁡2x=−csc⁡xcot⁡x\frac{d}{dx}[\csc x] = \frac{(0)(\sin x) - (1)(\cos x)}{\sin^2 x} = -\frac{\cos x}{\sin^2 x} = -\csc x \cot xdxd​[cscx]=sin2x(0)(sinx)−(1)(cosx)​=−sin2xcosx​=−cscxcotx

2. Periodicity of Higher-Order Derivatives of Sine and Cosine

Successive derivatives of sin⁡(x)\sin(x)sin(x) and cos⁡(x)\cos(x)cos(x) cycle in a closed loop of period 4:

Derivative Order nf(x)=sin⁡xg(x)=cos⁡xn=0 (Original)sin⁡xcos⁡xn=1 (1st)cos⁡x−sin⁡xn=2 (2nd)−sin⁡x−cos⁡xn=3 (3rd)−cos⁡xsin⁡xn=4 (4th)sin⁡xcos⁡x\begin{array}{c|c|c} \text{Derivative Order } n & f(x) = \sin x & g(x) = \cos x \\ \hline n = 0 \text{ (Original)} & \sin x & \cos x \\ n = 1 \text{ (1st)} & \cos x & -\sin x \\ n = 2 \text{ (2nd)} & -\sin x & -\cos x \\ n = 3 \text{ (3rd)} & -\cos x & \sin x \\ n = 4 \text{ (4th)} & \sin x & \cos x \\ \end{array}Derivative Order nn=0 (Original)n=1 (1st)n=2 (2nd)n=3 (3rd)n=4 (4th)​f(x)=sinxsinxcosx−sinx−cosxsinx​g(x)=cosxcosx−sinx−cosxsinxcosx​​

To evaluate any arbitrarily high nthn^{\text{th}}nth derivative, compute n mod 4n \bmod 4nmod4 (the remainder after dividing by 4):

f(n)(x)=f(n mod 4)(x)f^{(n)}(x) = f^{(n \bmod 4)}(x)f(n)(x)=f(nmod4)(x)


Engineering Applications

  • Mechanical & Civil Engineering (Harmonic Oscillation): Mass-spring-damper systems and structural vibration modes follow sinusoidal displacements x(t)=Asin⁡(ωt+ϕ)x(t) = A \sin(\omega t + \phi)x(t)=Asin(ωt+ϕ), where velocity is v(t)=Aωcos⁡(ωt+ϕ)v(t) = A \omega \cos(\omega t + \phi)v(t)=Aωcos(ωt+ϕ) and acceleration is a(t)=−Aω2sin⁡(ωt+ϕ)a(t) = -A \omega^2 \sin(\omega t + \phi)a(t)=−Aω2sin(ωt+ϕ).
  • Electrical & Computer Engineering (AC Signal Analysis): Alternating current voltage waveforms v(t)=Vpeakcos⁡(ωt)v(t) = V_{\text{peak}} \cos(\omega t)v(t)=Vpeak​cos(ωt) are differentiated to evaluate inductor voltages and reactive power dissipation.
  • Entertainment Engineering (Robotic Stage Motion): Automated stage lifts and rotary actuators utilize sinusoidal velocity profiles to achieve smooth starting and stopping without mechanical shock.

Example Problems

Example 1: Product with Trigonometric Terms

Find the equation of the line tangent to y=3xtan⁡(x)y = 3x \tan(x)y=3xtan(x) at x=πx = \pix=π.

Solution
  1. Find point of tangency (x1,y1)(x_1, y_1)(x1​,y1​):
    y1=y(π)=3(π)tan⁡(π)=3π(0)=0  ⟹  (π,0)y_1 = y(\pi) = 3(\pi)\tan(\pi) = 3\pi(0) = 0 \implies (\pi, 0)y1​=y(π)=3(π)tan(π)=3π(0)=0⟹(π,0)
  2. Find slope m=y′(π)m = y'(\pi)m=y′(π) using Product Rule:
    y′=ddx[3x]tan⁡x+3xddx[tan⁡x]=3tan⁡x+3xsec⁡2xy' = \frac{d}{dx}[3x]\tan x + 3x \frac{d}{dx}[\tan x] = 3\tan x + 3x \sec^2 xy′=dxd​[3x]tanx+3xdxd​[tanx]=3tanx+3xsec2x
    Evaluate at x=πx = \pix=π:
    m=y′(π)=3tan⁡(π)+3πsec⁡2(π)=3(0)+3π(−1)2=3πm = y'(\pi) = 3\tan(\pi) + 3\pi \sec^2(\pi) = 3(0) + 3\pi(-1)^2 = 3\pim=y′(π)=3tan(π)+3πsec2(π)=3(0)+3π(−1)2=3π
  3. Assemble tangent line equation:
    y−0=3π(x−π)  ⟹  y=3πx−3π2y - 0 = 3\pi(x - \pi) \implies y = 3\pi x - 3\pi^2y−0=3π(x−π)⟹y=3πx−3π2

Example 2: Arbitrary Higher-Order Derivative

Find the 97th97^{\text{th}}97th derivative of f(x)=sin⁡(x)f(x) = \sin(x)f(x)=sin(x) and the 54th54^{\text{th}}54th derivative of g(x)=cos⁡(x)g(x) = \cos(x)g(x)=cos(x).

Solution
  1. Evaluate 97 mod 497 \bmod 497mod4 for sin⁡(x)\sin(x)sin(x):
    97÷4=24 with remainder 1  ⟹  97≡1(mod4)97 \div 4 = 24 \text{ with remainder } 1 \implies 97 \equiv 1 \pmod 497÷4=24 with remainder 1⟹97≡1(mod4)
    Thus, f(97)(x)=f(1)(x)=cos⁡(x)f^{(97)}(x) = f^{(1)}(x) = \cos(x)f(97)(x)=f(1)(x)=cos(x).
  2. Evaluate 54 mod 454 \bmod 454mod4 for cos⁡(x)\cos(x)cos(x):
    54÷4=13 with remainder 2  ⟹  54≡2(mod4)54 \div 4 = 13 \text{ with remainder } 2 \implies 54 \equiv 2 \pmod 454÷4=13 with remainder 2⟹54≡2(mod4)
    Thus, g(54)(x)=g(2)(x)=−cos⁡(x)g^{(54)}(x) = g^{(2)}(x) = -\cos(x)g(54)(x)=g(2)(x)=−cos(x).

Connections & References

  • Parent Concepts: Derivatives, Trigonometric Identities
  • Prerequisites: Special Trigonometric Limits, Solving Right Triangles
  • Related Rules: Product Rule, Quotient Rule, Higher-Order Derivatives
  • Course Sessions: Calc-1 Session 07