Answers to exercises
Exercise 1
lim.Exercise 2
\lim\limits_{x\to 0} \dfrac{\sin 3x+\sin 7x}{5x} = \lim\limits_{x\to 0} \Bigl(\dfrac{3}{5}\times\dfrac{\sin 3x}{3x}\Bigr) + \lim\limits_{x\to 0} \Bigl(\dfrac{7}{5}\times\dfrac{\sin 7x}{7x}\Bigr) = \dfrac{3}{5}+\dfrac{7}{5} = 2.Exercise 3
- Multiplying top and bottom by 1+\cos x gives \lim_{x\to 0} \dfrac{1-\cos x}{x^2} = \lim_{x\to 0} \dfrac{\sin^2x}{x^2(1+\cos x)} = \Bigl(\lim_{x\to 0}\dfrac{\sin x}{x}\Bigr)^2 \times \Bigl(\lim_{x\to 0} \dfrac{1}{1+\cos x}\Bigr) = \dfrac{1}{2}.
- Hence, for small x, we have \dfrac{1-\cos x}{x^2} \approx \dfrac{1}{2}, and so \cos x \approx 1-\dfrac{1}{2}x^2.
Exercise 4
- We use two compound-angle formulas: \begin{align*} \cos(A-B) &= \cos A\,\cos B + \sin A\,\sin B \\ \cos(A+B) &= \cos A\,\cos B - \sin A\,\sin B. \end{align*} Let C=A+B and D=A-B. Then A=\dfrac{1}{2}(C+D) and B=\dfrac{1}{2}(C-D). So it follows that \cos C - \cos D = -2\,\sin A\,\sin B = -2\,\sin\Bigl(\dfrac{C+D}{2}\Bigr)\,\sin\Bigl(\dfrac{C-D}{2}\Bigr).
- \begin{align*} \dfrac{d}{dx} (\cos x) &= \lim_{h\to 0} \dfrac{\cos(x+h)-\cos x}{h} \\ &= \lim_{h\to 0} \dfrac{-2\,\sin(x+\dfrac{h}{2})\,\sin\dfrac{h}{2}}{h} \\ &= -\Bigl(\lim_{h\to 0} \,\sin\bigl(x+\dfrac{h}{2}\bigr)\Bigr) \times \Bigl(\lim_{h\to 0} \dfrac{\sin\dfrac{h}{2}}{\dfrac{h}{2}}\Bigr) = -\sin x. \end{align*}
Exercise 5
\dfrac{d}{dx}(\tan x) = \dfrac{d}{dx}\Bigl( \dfrac{\sin x}{\cos x} \Bigr) = \dfrac{\cos^2x +\sin^2x}{\cos^2x} = \dfrac{1}{\cos^2 x} = \sec^2 x.Exercise 6
- \dfrac{d}{dx} (\operatorname{cosec} x) = \dfrac{d}{dx}(\sin x)^{-1} = -1(\sin x)^{-2} \times \cos x = -\dfrac{\cos x}{\sin^2 x} = -\operatorname{cosec} x\,\cot x.
- \dfrac{d}{dx} (\sec x) = \dfrac{d}{dx}(\cos x)^{-1} = -1(\cos x)^{-2}\times -\sin x = \dfrac{\sin x}{\cos^2 x} = \tan x\,\sec x.
- \dfrac{d}{dx} (\cot x) = \dfrac{d}{dx}\Bigl(\dfrac{\cos x}{\sin x}\Bigr) = \dfrac{-\sin^2x-\cos^2x}{\sin^2x}= -\operatorname{cosec}^2 x.
Exercise 7
\begin{align*} \dfrac{d}{dx} \log_e\Bigl(\dfrac{1+\sin x}{\cos x}\Bigr) &= \dfrac{d}{dx}\bigl(\log_e(1+\sin x) - \log_e(\cos x)\bigr) \\ &= \dfrac{\cos x}{1+\sin x} + \dfrac{\sin x}{\cos x} = \dfrac{\cos^2x + \sin^2x + \sin x}{(1+\sin x)\,\cos x} \\ &= \dfrac{1+\sin x}{(1+\sin x)\,\cos x} = \dfrac{1}{\cos x} = \sec x. \end{align*}Exercise 8
The derivative of the function is
\dfrac{dy}{dx} = \dfrac{d}{dx} \Bigl(\dfrac{\sin x}{3+4\cos x}\Bigr) = \dfrac{3\cos x + 4}{(3+4\cos x)^2}.Since \cos x \geq -1, it follows that \dfrac{dy}{dx} > 0 whenever 3 + 4\cos x \neq 0. Hence, y is an increasing function wherever it is defined.
Exercise 9
The perimeter of the triangle is P = 2a + 2a\cos \theta = 2a(1+\cos\theta), so
a = \dfrac{P}{2(1+\cos\theta)}.The area of the triangle is
A = \dfrac{1}{2}\,\bigl(2a\cos \theta\bigr)\,\bigl(a\,\sin \theta\bigr) = a^2\cos\theta\,\sin\theta = \dfrac{1}{2}a^2\sin 2\theta.Substituting for a gives
A = \dfrac{P^2\sin2\theta}{8(1+\cos\theta)^2}.We want to find \theta in the range 0 to \dfrac{\pi}{2} such that \dfrac{dA}{d\theta} = 0. By the quotient rule, if \dfrac{dA}{d\theta} = 0, then
16P^2\cos2\theta\,(1+\cos\theta)^2 + 16P^2\sin2\theta\,(1 + \cos\theta)\,\sin\theta = 0.It follows after some calculation that (1 + \cos\theta)^2(2\cos\theta - 1) = 0. So \cos\theta = -1 or \cos\theta = \dfrac{1}{2}. Hence, for 0 < \theta < \dfrac{\pi}{2}, the only solution is \theta = \dfrac{\pi}{3}. The triangle is equilateral.
Exercise 10
- \displaystyle \int_{\frac{\pi}{6}}^\frac{\pi}{3} (\sin 2x + \cos 3x)\, dx = \Bigl[-\dfrac{1}{2}\cos 2x + \dfrac{1}{3} \sin 3x \Bigr]_{\frac{\pi}{6}}^\frac{\pi}{3} = \dfrac{1}{6}.
- We have \dfrac{d}{dx} (x\,\sin x) = \sin x + x\,\cos x. Hence \int_0^{\frac{\pi}{2}} x\,\cos x\, dx = \Bigl[x\,\sin x\Bigr]_{0}^\frac{\pi}{2} - \int_0^{\frac{\pi}{2}} \sin x\, dx = \dfrac{\pi}{2}-1.
- \displaystyle \int \tan^2 x\, dx = \int \bigl(\sec^2 x - 1\bigr)\, dx = \tan x - x + C.
Exercise 11
We have \sin^2\theta = \dfrac{1}{2}(1-\cos 2\theta), and so
\int \sin^2\theta\, d\theta = \dfrac{1}{2}\Bigl(\theta - \dfrac{1}{2}\sin 2\theta\Bigr) + C.Exercise 12
Since the period is \pi \sqrt{2}, we have \dfrac{2\pi}{n} = \pi\sqrt{2}, so n=\sqrt{2}. The amplitude is C=2. Hence, from v^2=n^2(C^2-x^2), when x=0, v=\pm 2\sqrt{2}. Thus the speed is 2\sqrt{2} m/s.
Exercise 13
This restricted function has domain [0,\pi] and range [-1,1]. So its inverse has domain [-1,1] and range [0,\pi].
Graphs of inverse cosine of minus x and minus inverse cosine of x showing one is a translation of the other.
Detailed description
Since \cos \dfrac{\pi}{3} = \dfrac{1}{2}, we have \cos^{-1}(\dfrac{1}{2}) = \dfrac{\pi}{3} and so \cos^{-1}(-\dfrac{1}{2}) = \pi - \cos^{-1}(\dfrac{1}{2}) = \pi - \dfrac{\pi}{3} = \dfrac{2\pi}{3}.- Let y = \cos^{-1}x. Then x = \cos y, and so \dfrac{dx}{dy} = -\sin y = -\sqrt{1 - \cos^2 y} = -\sqrt{1 - x^2}. (Note that 0 \leq y \leq \pi and so 0 \leq \sin y \leq 1.) Hence, \dfrac{dy}{dx} = -\dfrac{1}{\sqrt{1 - x^2}}.
- Let f(x) = \sin^{-1}x + \cos^{-1}x, for x \in [-1, 1]. Then f'(x) = \dfrac{1}{\sqrt{1-x^2}}- \dfrac{1}{\sqrt{1-x^2}} = 0. So f(x) = C, for some constant C. Now f(0)=\dfrac{\pi}{2} and so \sin^{-1} x + \cos^{-1} x = \dfrac{\pi}{2}.