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Chapter 7: Problem 18

Show from first principles that the function cos is differentiable at \(\pi / 2\) and its derivative at \(\pi / 2\) is \(-1\).

Short Answer

Expert verified
The function cos is differentiable at \(\pi/2\) because the limit of the difference quotient \(\frac{\cos(\frac{\pi}{2}+h)}{h}\) exists as \(h\) approaches \(0\). By using the cosine addition formula, we simplified the expression to \(\frac{-\sin(h)}{h}\). Evaluating this limit as \(h\) approaches \(0\) gives us \(-1\), which is the derivative of \(\cos\) at \(\pi/2\).

Step by step solution

01

Define the difference quotient

In order to show that the function is differentiable at \(\pi/2\), we need to evaluate the limit of the difference quotient at this point. The difference quotient is defined as: \[\frac{\cos(x+h) - \cos(x)}{h}\] where \(h\) is a small value that approaches \(0\).
02

Evaluate the difference quotient at \(\pi/2\)

Now, we need to evaluate the difference quotient at \(x = \pi/2\). This gives us: \[\frac{\cos(\dfrac{\pi}{2}+h) - \cos(\dfrac{\pi}{2})}{h}\] Notice that \(\cos(\pi/2) = 0\). So, the expression simplifies to: \[\frac{\cos(\dfrac{\pi}{2}+h)}{h}\]
03

Use trigonometric identity to simplify the expression

We can simplify this expression by using the cosine addition formula, which is: \[\cos(\alpha + \beta) = \cos(\alpha)\cos(\beta) - \sin(\alpha)\sin(\beta)\] In our case, we have: \[\cos(\dfrac{\pi}{2}+h) = \cos(\dfrac{\pi}{2})\cos(h) - \sin(\dfrac{\pi}{2})\sin(h)\] Since \(\cos(\pi/2) = 0\) and \(\sin(\pi/2) = 1\), the expression becomes: \[\cos(\dfrac{\pi}{2}+h) = -\sin(h)\] Now, our difference quotient is: \[\frac{-\sin(h)}{h}\]
04

Evaluate the limit as \(h\) approaches 0

To find the derivative, we need to evaluate the limit as \(h\) approaches 0: \[\lim_{h \to 0}\dfrac{-\sin(h)}{h}\] This is a well-known limit in calculus, and its value is \(-1\).
05

Conclusion

We have shown that the limit of the difference quotient exists at \(x = \pi/2\), which means that the function \(\cos\) is differentiable at this point. Furthermore, we found that the derivative of \(\cos\) at \(\pi/2\) is equal to \(-1\).

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Most popular questions from this chapter

Prove that \(|\sin x-\sin y| \leq|x-y|\) and \(|\cos x-\cos y| \leq|x-y|\) for all \(x, y \in \mathbb{R}\).

Find the polar coordinates of the points in \(\mathbb{R}^{2}\) whose Cartesian coordinates are as follows: (i) \((1,1)\), (ii) \((0,3)\), (iii) \((2,2 \sqrt{3})\), (iv) \((2 \sqrt{3}, 2)\).

Prove the following: (i) \(\left(\cot ^{-1}\right)^{\prime} y=-\frac{1}{1+y^{2}}\) for all \(y \in \mathbb{R}\), (ii) \(\left(\mathrm{csc}^{-1}\right)^{\prime} y=-\frac{1}{|y| \sqrt{y^{2}-1}}\) for all \(y \in \mathbb{R}\) with \(|y|>1\), (iii) \(\left(\mathrm{sec}^{-1}\right)^{\prime} y=\frac{1}{|y| \sqrt{y^{2}-1}}\) for all \(y \in \mathbb{R}\) with \(|y|>1\).

If \(y \in(1, \infty)\), then show that $$ \sec ^{-1} y=\lim _{a \rightarrow 1^{+}} \int_{a}^{y} \frac{1}{t \sqrt{t^{2}-1}} d t \quad \text { and } \quad \csc ^{-1} y=\frac{\pi}{2}-\lim _{a \rightarrow 1^{+}} \int_{a}^{y} \frac{1}{t \sqrt{t^{2}-1}} d t . $$

Prove that $$ \lim _{x \rightarrow \infty} \int_{1}^{x} \frac{1}{1+t^{2}} d t=\int_{0}^{1} \frac{1}{1+t^{2}} d t=\frac{\pi}{4} $$ that is, \(\lim _{x \rightarrow \infty} \arctan x=\arctan 1=\pi / 4\). Deduce that \(2.88<\pi<3.39\). (Hint: Substitute \(t=1 / s\) and use Proposition 6.20. Divide \([0,1]\) into subintervals of length \(\frac{1}{4}\).)
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