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Tangent line to cos(x) at x = pi/3 via limit definition

Let f(x)=cos⁡(x)f(x) = \cos(x).

Use the limit definition of the derivative, f′(a)=lim⁡h→0f(a+h)−f(a)h,f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}, to compute f′(π/3)f'(\pi/3) for f(x)=cos⁡(x)f(x) = \cos(x). Then use this slope together with the point (π/3,f(π/3))(\pi/3, f(\pi/3)) to find the equation of the tangent line to f(x)=cos⁡(x)f(x) = \cos(x) at x=π/3x = \pi/3. What is the equation of this tangent line?

Hint

Write f′(π/3)=lim⁡h→0cos⁡(π/3+h)−cos⁡(π/3)hf'(\pi/3) = \lim_{h\to 0}\frac{\cos(\pi/3+h)-\cos(\pi/3)}{h}, expand cos⁡(π/3+h)\cos(\pi/3+h) using the angle addition formula cos⁡(A+h)=cos⁡Acos⁡h−sin⁡Asin⁡h\cos(A+h) = \cos A \cos h - \sin A \sin h, then split the limit using the known limits lim⁡h→0cos⁡h−1h=0\lim_{h\to0}\frac{\cos h - 1}{h}=0 and lim⁡h→0sin⁡hh=1\lim_{h\to0}\frac{\sin h}{h}=1.

Step-by-step solution

Expand cos⁡(π/3+h)=cos⁡(π/3)cos⁡h−sin⁡(π/3)sin⁡h\cos(\pi/3 + h) = \cos(\pi/3)\cos h - \sin(\pi/3)\sin h. Substituting into the difference quotient gives cos⁡(π/3)cos⁡h−sin⁡(π/3)sin⁡h−cos⁡(π/3)h=cos⁡(π/3)⋅cos⁡h−1h−sin⁡(π/3)⋅sin⁡hh.\frac{\cos(\pi/3)\cos h - \sin(\pi/3)\sin h - \cos(\pi/3)}{h} = \cos(\pi/3)\cdot\frac{\cos h - 1}{h} - \sin(\pi/3)\cdot\frac{\sin h}{h}.

Taking h→0h \to 0, the first term vanishes since lim⁡h→0cos⁡h−1h=0\lim_{h\to0}\frac{\cos h-1}{h}=0, and the second term gives −sin⁡(π/3)⋅1=−sin⁡(π/3)-\sin(\pi/3)\cdot 1 = -\sin(\pi/3). So f′(π/3)=−sin⁡(π/3)=−32f'(\pi/3) = -\sin(\pi/3) = -\frac{\sqrt{3}}{2}.

Since f(π/3)=cos⁡(π/3)=12f(\pi/3) = \cos(\pi/3) = \frac{1}{2}, the tangent line passes through (π/3,1/2)(\pi/3, 1/2) with slope −32-\frac{\sqrt{3}}{2}. Using point-slope form: y−12=−32(x−π3).y - \frac{1}{2} = -\frac{\sqrt{3}}{2}\left(x - \frac{\pi}{3}\right).

Sign of the tangent slope at pi over three

Hint
Step-by-step solution

The tangent slope found in part a was negative sine of pi over three, which equals negative square root of three over two. Since square root of three over two is a positive number, the negative sign makes the overall slope negative. This makes sense since cosine is decreasing near pi over three.

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