Week 7 - Derivatives of Trigonometric Functions

1. Trigonometry

  • Trigonometric Functions
      • sinθ=opphyp\sin{\theta} = \frac{opp}{hyp}, cscθ=hypopp\csc{\theta} = \frac{hyp}{opp}
      • cosθ=adjhyp\cos{\theta} = \frac{adj}{hyp}, secθ=hypadj\sec{\theta} = \frac{hyp}{adj}
      • tanθ=oppadj\tan{\theta} = \frac{opp}{adj}, cotθ=adjopp\cot{\theta} = \frac{adj}{opp}
      • sinθ=yr\sin{\theta} = \frac{y}{r}, cscθ=ry\csc{\theta} = \frac{r}{y}
      • cosθ=xr\cos{\theta} = \frac{x}{r}, secθ=rx\sec{\theta} = \frac{r}{x}
      • tanθ=yx\tan{\theta} = \frac{y}{x}, cotθ=xy\cot{\theta} = \frac{x}{y}

1.1. Angles

  • πrad=180\pi rad = 180^{\circ}
    • 1rad=(180π)57.31 rad = (\frac{180}{\pi})^{\circ} \approx 57.3^{\circ}
Degree 00^{\circ} 3030^{\circ} 4545^{\circ} 6060^{\circ} 9090^{\circ} 120120^{\circ} 135135^{\circ} 150150^{\circ} 180180^{\circ} 270270^{\circ} 360360^{\circ}
Radians 00 π6\frac{\pi}{6} π4\frac{\pi}{4} π3\frac{\pi}{3} π2\frac{\pi}{2} 2π3\frac{2\pi}{3} 3π4\frac{3\pi}{4} 5π6\frac{5\pi}{6} π\pi 3π2\frac{3\pi}{2} 2π2\pi
  • θ=ar\theta = \frac{a}{r}, a=rθa = r\theta

1.2. Some of Trigonometric Identities

  • sin2x+cos2x=1\sin^2{x} + \cos^2{x} = 1
  • sinx=sinx\sin{-x} = -\sin{x}
  • cosx=cosx\cos{-x} = \cos{x}
  • sinx+2π=sinx\sin{x+2\pi} = \sin{x}, cosx+2π=cosx\cos{x+2\pi} = \cos{x}

1.3. Graphs of the Trigonometric Functions

2. Differentiate Trig Functions

  • To prove: if f(x)=sinxf(x) = \sin{x}, then f(x)=cosxf'(x) = \cos{x}
    • f(x)=limh0sin(x+h)sin(x)h=limh0sinxcosh+cosxsinhsin(x)h=limh0[sinx(cosh1h)+cosx(sinhh)]=sinxlimh0cosh1h+cosxlimh0sinhh\begin{aligned} f'(x) &= \lim_{h \to 0}\frac{\sin(x+h) - \sin(x)}{h} \\ &= \lim_{h \to 0}\frac{\sin{x}\cos{h} + \cos{x}\sin{h} - \sin(x)}{h} \\ &= \lim_{h \to 0}[\sin{x}(\frac{\cos{h} - 1}{h}) + \cos{x}(\frac{\sin{h}}{h})] \\ &= \sin{x}\lim_{h \to 0}\frac{\cos{h} - 1}{h} + \cos{x}\lim_{h \to 0}\frac{\sin{h}}{h} \end{aligned}
    • Recall the Squeeze Theorem, we know limx0sin(x)x=1\displaystyle\lim_{x\to{0}}\frac{\sin(x)}{x}=1, then,
    • limθ0cosθ1θ=limθ0cosθ1θcosθ+1cosθ+1=limθ0cos2θ1θ(cosθ+1)=limθ0sin2θθ(cosθ+1)=limθ0sinθθsinθcosθ+1=limθ0sinθθlimθ0sinθcosθ+1=1(01+1)limθ0cosθ1θ=0\begin{aligned} \lim_{\theta \to 0}\frac{\cos{\theta} - 1}{\theta} &= \lim_{\theta \to 0}{\frac{\cos{\theta} - 1}{\theta} \cdot \frac{\cos{\theta} + 1}{\cos{\theta} + 1}} = \lim_{\theta \to 0}\frac{\cos^2{\theta} - 1}{\theta(\cos{\theta} + 1)} \\ &= \lim_{\theta \to 0}\frac{-\sin^2{\theta}}{\theta(\cos{\theta} + 1)} = -\lim_{\theta \to 0}{\frac{\sin{\theta}}{\theta} \cdot \frac{\sin{\theta}}{\cos{\theta}+1}} \\ &= -\lim_{\theta \to 0}\frac{\sin{\theta}}{\theta} \cdot \lim_{\theta \to 0}\frac{\sin{\theta}}{\cos{\theta}+1} \\ &= - 1 \cdot (\frac{0}{1 + 1}) \\ \lim_{\theta \to 0}\frac{\cos{\theta} - 1}{\theta} &= 0 \end{aligned}
    • So, f(x)=sinxlimh0cosh1h+cosxlimh0sinhh=(sinx)0+(cosx)1=cosx\begin{aligned} f'(x) &= \sin{x}\lim_{h \to 0}\frac{\cos{h} - 1}{h} + \cos{x}\lim_{h \to 0}\frac{\sin{h}}{h} \\ &= (\sin{x}) \cdot 0 + (\cos{x}) \cdot 1 \\ &= \cos{x} \end{aligned}

2.1. Derivatives of Trigonometric Functions

  • ddx(sinx)=cosx\frac{d}{dx}(\sin{x}) = \cos{x}, ddx(cscx)=cscxcotx\frac{d}{dx}(\csc{x}) = -\csc{x}\cot{x}
  • ddx(cosx)=sinx\frac{d}{dx}(\cos{x}) = -\sin{x}, ddx(secx)=secxtanx\frac{d}{dx}(\sec{x}) = \sec{x}\tan{x}
  • ddx(tanx)=sec2x\frac{d}{dx}(\tan{x}) = \sec^2{x}, ddx(cotx)=csc2x\frac{d}{dx}(\cot{x}) = -\csc^2{x}

3. Inverse Trigonometric Functions

  • y=sinxy = \sin{x}
  • y=sin1x=arcsinxsiny=x and π2xπ2y = \sin^{-1}{x} = \arcsin{x} \Leftrightarrow \sin{y} = x\ and\ -\frac{\pi}{2} \le x \le \frac{\pi}{2}
    • we can conclude this: arcsin(sinx)=x\arcsin{(\sin{x})} = x
  • Sample: Simplify the expression cos(tan1x)\cos{(\tan^{-1}x)}:
    • If y=tan1xy=\tan^{-1}x, then tany=x\tan{y}=x, we can use a diagram to express this:
    • Then cos(tan1x)=cosy=11+x2\cos{(\tan^{-1}x)} = \cos{y} = \frac{1}{\sqrt{1+x^2}}

3.1. The Derivatives of Inverse Trig Functions

  • ddxarcsinx=?\frac{d}{dx}\arcsin{x} = ?
    • Let's say f(x)=arcsinxf(x) = \arcsin{x}
      • We know f(sinx)=xf(\sin{x}) = x
      • So use the chain rules, derivative the both sides, we got: f(sinx)cosx=1f(sinx)=1cosxf(sinx)=11sin2xf(x)=11x2\begin{aligned} f'(\sin{x}) \cdot \cos{x} &= 1 \\ f'(\sin{x}) &= \frac{1}{\cos{x}} \\ f'(\sin{x}) &= \frac{1}{\sqrt{1-\sin^2{x}}} \\ f'(x) &= \frac{1}{\sqrt{1-x^2}} \\ \end{aligned}
  • ddxarccosx=?\frac{d}{dx}\arccos{x} = ?:
    • In a right triangle, the other two angles are αβ\alpha\, \beta, we know:
    • (sinα=y cosβ=x\sin{\alpha} = y\,\ \cos{\beta} = x) and the hypotenuse equals 1, then we got:
    • α+β=π2\alpha + \beta = \frac{\pi}{2}
    • arcsiny=α arccosy=β\arcsin{y} = \alpha\,\ \arccos{y} = \beta
    • arcsiny+arccosy=π2\arcsin{y} + \arccos{y} = \frac{\pi}{2}
    • ddyarcsiny+ddyarccosy=ddyπ2\frac{d}{dy}\arcsin{y} + \frac{d}{dy}\arccos{y} = \frac{d}{dy} \frac{\pi}{2}
    • ddyarcsiny+ddyarccosy=0\frac{d}{dy}\arcsin{y} + \frac{d}{dy}\arccos{y} = 0
    • ddyarccosy=11x2\frac{d}{dy}\arccos{y} = - \frac{1}{\sqrt{1-x^2}}
  • ddxarctanx=?\frac{d}{dx}\arctan{x} = ?
    • set f(x)=arctanxf(x) = \arctan{x},
      • then f(tanx)=xf(\tan{x}) = x
    • use the chain rules: f(tanx)sec2x=1f(tanx)=1sec2xf(tanx)=1tan2x+1f(x)=1x2+1\begin{aligned} f'(\tan{x}) \cdot \sec^2{x} &= 1 \\ f'(\tan{x}) &= \frac{1}{\sec^2{x}} \\ f'(\tan{x}) &= \frac{1}{\tan^2{x}+1} \\ f'(x) &= \frac{1}{x^2+1} \\ \end{aligned}

3.2. Why do Sine and Cosine Oscillate?

  • If f(x)=f(x)f''(x) = - f(x)
  • For example:
    • A running car, it's acceleration equals minor it's position, and the center point is 0.
      • Which means:
        • the velocity will be smaller and smaller when it's acceleration less than 0,
        • but when the position is less than 0 the acceleration will be positive, and the car will move backwards sooner.
      • So the car will be back and forward, just like the sine wave.
  • To Sine and Cosine, we know:
    • f(t)=sint, f(t)=sintf(t) = \sin{t},\ f''(t) = - \sin{t}
    • f(t)=cost, f(t)=costf(t) = \cos{t},\ f''(t) = - \cos{t}

3.3. Angle sum and difference identities

  • sin(x+y)=sinxcosy+cosxsiny\sin{(x+y)} = \sin{x}\cos{y} + \cos{x}\sin{y}
  • cos(x+y)=cosxcosysinxsiny\cos{(x+y)} = \cos{x}\cos{y} - \sin{x}\sin{y}
  • sin(xy)=sinxcosycosxsiny\sin{(x-y)} = \sin{x}\cos{y} - \cos{x}\sin{y}
  • cos(xy)=cosxcosy+sinxsiny\cos{(x-y)} = \cos{x}\cos{y} + \sin{x}\sin{y}
  • sin2x=2sinxcosx\sin{2x} = 2\sin{x}\cos{x}
  • cos2x=cos2xsin2x\cos{2x} = \cos^2{x} - \sin^2{x}
  • tan(x+y)=tan(x)+tany1tan(x)tan(y)\tan{(x+y)} = \frac{\tan(x) + \tan{y}}{1 - \tan(x)\tan(y)}
  • tan(xy)=tan(x)tany1+tan(x)tan(y)\tan{(x-y)} = \frac{\tan(x) - \tan{y}}{1 + \tan(x)\tan(y)}

3.4. Approximate sin 1

  • f(x)=sinx, f(0)=0f(x) = \sin{x},\ f(0) = 0
  • f(x)=cosx, f(0)=1f'(x) = \cos{x},\ f'(0) = 1
  • f(0+h)f(0)+hf(0)f(0+h) \approx f(0) + h \cdot f'(0), (when h approach to 0)
  • f(0+h)0+h1f(0+h) \approx 0 + h \cdot 1
  • f(h)hf(h) \approx h (when h approach to 0)
  • pick an x = 1/32, then sin132132\sin{\frac{1}{32} \approx \frac{1}{32}}
    • then use formula sin2x=2sinxcosx\sin{2x} = 2\sin{x}\cos{x} , we got sin2x=2sinx1sin2x\sin{2x} = 2\sin{x} \cdot \sqrt{1-\sin^2{x}}
    • So, sin(2132)=2sin(1/16)1sin2(1/32)\sin{(2 \cdot \frac{1}{32})} = 2\sin{(1/16)} \cdot \sqrt{1-\sin^2{(1/32)}}
    • We can keep multiple 2 to x until x = 1, then we got (sin1\sin{1})

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