## Answers to exercises

##### Exercise 1
$$\lim\limits_{x\to 0} \dfrac{\tan x}{x} = \lim\limits_{x\to 0}\Bigl(\dfrac{\sin x}{x}\times \dfrac{1}{\cos x}\Bigr) = 1\times 1 = 1$$.
##### 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
1. Multiplying top and bottom by $$1+\cos x$$ gives $\lim_{x\to 0} \dfrac{1-\cos x}{x^2} = \lim_{x\to 0} \dfrac{\sin^2x}{x^2(1+\cos x)} = \Bigl(\lim_{x\to 0}\dfrac{\sin x}{x}\Bigr)^2 \times \Bigl(\lim_{x\to 0} \dfrac{1}{1+\cos x}\Bigr) = \dfrac{1}{2}.$
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
1. 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 $\cos C - \cos D = -2\,\sin A\,\sin B = -2\,\sin\Bigl(\dfrac{C+D}{2}\Bigr)\,\sin\Bigl(\dfrac{C-D}{2}\Bigr).$
2. \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
1. $$\dfrac{d}{dx} (\operatorname{cosec} x) = \dfrac{d}{dx}(\sin x)^{-1} = -1(\sin x)^{-2} \times \cos x = -\dfrac{\cos x}{\sin^2 x} = -\operatorname{cosec} x\,\cot x$$.
2. $$\dfrac{d}{dx} (\sec x) = \dfrac{d}{dx}(\cos x)^{-1} = -1(\cos x)^{-2}\times -\sin x = \dfrac{\sin x}{\cos^2 x} = \tan x\,\sec x$$.
3. $$\dfrac{d}{dx} (\cot x) = \dfrac{d}{dx}\Bigl(\dfrac{\cos x}{\sin x}\Bigr) = \dfrac{-\sin^2x-\cos^2x}{\sin^2x}= -\operatorname{cosec}^2 x$$.
##### 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
1. $$\displaystyle \int_{\frac{\pi}{6}}^\frac{\pi}{3} (\sin 2x + \cos 3x)\, dx = \Bigl[-\dfrac{1}{2}\cos 2x + \dfrac{1}{3} \sin 3x \Bigr]_{\frac{\pi}{6}}^\frac{\pi}{3} = \dfrac{1}{6}$$.
2. We have $$\dfrac{d}{dx} (x\,\sin x) = \sin x + x\,\cos x$$. Hence $\int_0^{\frac{\pi}{2}} x\,\cos x\, dx = \Bigl[x\,\sin x\Bigr]_{0}^\frac{\pi}{2} - \int_0^{\frac{\pi}{2}} \sin x\, dx = \dfrac{\pi}{2}-1.$
3. $$\displaystyle \int \tan^2 x\, dx = \int \bigl(\sec^2 x - 1\bigr)\, dx = \tan x - x + C$$.
##### 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

1. This restricted function has domain $$[0,\pi]$$ and range $$[-1,1]$$. So its inverse has domain $$[-1,1]$$ and range $$[0,\pi]$$.
2. Graphs of inverse cosine of minus x and minus inverse cosine of x showing one is a translation of the other.

3. Detailed description
Since $$\cos \dfrac{\pi}{3} = \dfrac{1}{2}$$, we have $$\cos^{-1}(\dfrac{1}{2}) = \dfrac{\pi}{3}$$ and so $$\cos^{-1}(-\dfrac{1}{2}) = \pi - \cos^{-1}(\dfrac{1}{2}) = \pi - \dfrac{\pi}{3} = \dfrac{2\pi}{3}$$.
4. 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, $\dfrac{dy}{dx} = -\dfrac{1}{\sqrt{1 - x^2}}.$
5. 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 $$\sin^{-1} x + \cos^{-1} x = \dfrac{\pi}{2}$$.