Chapter 4

Exercise 7.1

Q1

(a)\[ 270^{\circ}\times\frac{\pi}{180}=\frac{3\pi}{2} \]

(b)\[ 210^{\circ}\times\frac{\pi}{180}=\frac{7\pi}{6} \]

(c)\[ 315^{\circ}\times\frac{\pi}{180}=\frac{7\pi}{4} \]

(d)\[ 750^{\circ}\times\frac{\pi}{180}=\frac{25\pi}{6} \]

Q2

(a)\[ \frac{\pi}{6} \div \frac{\pi}{180}=30^{\circ} \]

(b)\[ \frac{3\pi}{4} \div \frac{\pi}{180}=135^{\circ} \]

(c)\[ \frac{5\pi}{2} \div \frac{\pi}{180}=450^{\circ} \]

(d)\[ 7\pi \div \frac{\pi}{180}=1260^{\circ} \]

Exercise 7.2

Q1

(a)\[ sin{\frac{2\pi}{3}}=sin({\pi-\frac{\pi}{3}})\\ =sin{\frac{\pi}{3}}\\ =\frac{\sqrt{3}}{2}\] (b)\[ cos{\frac{2\pi}{3}}=cos({\pi-\frac{\pi}{3}})\\ =-cos{\frac{\pi}{3}}\\ =-\frac{1}{2}\] (c)\[ tan{\frac{2\pi}{3}}=tan({-\frac{\pi}{3}+\pi})\\ =tan({-\frac{\pi}{3}})\\ =-tan{\frac{\pi}{3}}\\ =-\sqrt{3}\] (d)\[ sin{\frac{5\pi}{4}}=sin({\pi+\frac{\pi}{4}})\\ =-sin{\frac{\pi}{4}}\\ =-\frac{1}{\sqrt{2}}\] (e)\[ cos{\frac{5\pi}{4}}=cos({\pi+\frac{\pi}{4}})\\ =-cos{\frac{\pi}{4}}\\ =-\frac{1}{\sqrt{2}}\] (f)\[ tan{\frac{5\pi}{4}}=tan({\pi+\frac{\pi}{4}})\\ =tan{\frac{\pi}{4}}\\ =1\]

Q2

(a)\[ \lim_{x\to 0}\frac{tan{x}}{x}=\lim_{x\to 0}\frac{sin{x}}{x \ cos{x}}\\ = \lim_{x\to 0}\frac{sin{x}}{x} \times \lim_{x\to 0}\frac{1}{cos{x}} \\ = 1\times \frac{1}{1}\\ = 1 \] (b)

method 1:

\[ \lim_{x\to 0}\frac{sin{2x}}{x}=2\lim_{x\to 0}\frac{sin{2x}}{2x}\\ =2\lim_{2x\to 0}\frac{sin{2x}}{2x}\\ =2\times 1\\=2 \]

method 2:

\[ sin{2x}=2sin{x}cos{x}\\ \lim_{x\to 0}\frac{sin{2x}}{x}=\lim_{x\to 0}\frac{2sin{x}cos{x}}{x}\\ =2\lim_{x\to 0}\frac{sin{x}}{x} \times \lim_{x\to 0}cos{x} \\ =2\times 1 \times 1\\ =2 \]

Exercise 7.3

Q1

(g)\[ \frac{dy}{dx}=\frac{d}{dx}(sin{x} \times cos{x})\\ =cos{x}\frac{d}{dx}sin{x}+sin{x} \frac{d}{dx}cos{x}\\ =cos{x} \times cos{x} + sin{x} \times (-sin{x})\\ =cos^2{x}-sin^2{x}\\ =cos{2x}\] (h)\[ \frac{dy}{dx}=\frac{d}{dx}(\frac{cos{x}}{x^3+1})\\ =\frac{(x^3+1)\frac{d}{dx}cos{x}-cos{x}\frac{d}{dx}(x^3+1)}{(x^3+1)^2} \\ =\frac{-(x^3+1)sin{x}-cos{x}\times 3x^2}{(x^3+1)^2} \\ =\frac{-(x^3+1)sin{x}-3x^2cos{x}}{(x^3+1)^2} \\\] (i)\[ \frac{dy}{dx}=\frac{d}{dx}(x+cos{x})^2\\ =2(x+cos{x}) \times (1-sin{x}) \] (j)\[ \frac{dy}{dx}=\frac{d}{dx}(sin{x}+cos{x})^2\\ =2(sin{x}+cos{x}) \times (cos{x}-sin{x})\\ =2(cos^2{x}-sin^2{x})\\ =2cos{2x} \]

Q2

\[ for\ n=2,y=sin^2{x}\\ \frac{dy}{dx}=\frac{d}{dx}sin^2{x}\\ =2sin{x}\times \frac{d}{dx}sin{x}\\ =2sin{x}\ cos{x}\\\]

\[for\ n=3,y=sin^3{x}\\ \frac{dy}{dx}=\frac{d}{dx}sin^3{x}\\ =3sin^2{x}\times \frac{d}{dx}sin{x}\\ =3sin^2{x}\ cos{x}\\\]

\[for\ n=4,y=sin^4{x}\\ \frac{dy}{dx}=\frac{d}{dx}sin^4{x}\\ =4sin^3{x}\times \frac{d}{dx}sin{x}\\ =4sin^3{x}\ cos{x}\\\]

(b)\[ \frac{dy}{dx}=\frac{d}{dx}sin{x}\\ =nsin^{n-1}{x}\times \frac{d}{dx}sin{x}\\ =nsin^{n-1}{x}\ cos{x}\\ \]

Q3

method 1:Derivation of compound function

\[ \frac{dy}{dx}=\frac{d}{dx}sin{2x}\\ =cos{2x}\times \frac{d}{dx}2x\\ =2cos{2x}\\ \] \[ \frac{dz}{dx}=\frac{d}{dx}cos{2x}\\ =-sin{2x}\times \frac{d}{dx}2x\\ =-2sin{2x}\\ \]

method 2:use compound angle formulas(just give an example)

\[ sin{2x}=2sin{x}cos{x}\\ \frac{dy}{dx}=\frac{d}{dx}sin{2x}\\ =\frac{d}{dx}(2sin{x} \ cos{x})\\ =2(cos{x}\frac{d}{dx}sin{x}+sin{x} \frac{d}{dx}cos{x})\\ =2cos{x}cos{x}-2sin{x}sin{x}\\ =2(cos^2{x}-sin^2{x})\\ =2cos{2x}\]

method 1:Derivation of compound function

\[ \frac{dy}{dx}=\frac{d}{dx}sin{3x}\\ =cos{3x}\times \frac{d}{dx}3x\\ =3cos{3x}\\ \] \[ \frac{dz}{dx}=\frac{d}{dx}cos{3x}\\ =-sin{3x}\times \frac{d}{dx}3x\\ =-3sin{3x}\\ \]

method 2:use compound angle formulas(just give an example)

\[sin{3x}=sin{2x+x}=sin{2x}\ cos{x}+cos{2x}sin{x}\\ \frac{dy}{dx}=\frac{d}{dx}sin{3x}\\ \frac{d}{dx}(sin{2x}\ cos{x}+cos{2x}sin{x})\\ =(cos{x}\frac{d}{dx}sin{2x}+sin{2x} \frac{d}{dx}cos{x})+(sin{x}\frac{d}{dx}cos{2x}+cos{2x} \frac{d}{dx}sin{x}) \\ =cos{x}\times 2cos{2x}-sin{2x}\times sin{x}-2sin{x}\times sin{2x}+cos{2x} \times cos{x} \\ =3cos{2x}\times cos{x}-3sin{2x} \times sin{x} \\ =3cos({2x+x})\\ =3cos{3x} \]

(c)\[ \frac{dy}{dx}=\frac{d}{dx}sin{nx}\\ =cos{nx}\times \frac{d}{dx}nx\\ =ncos{nx}\\ \] \[ \frac{dz}{dx}=\frac{d}{dx}cos{nx}\\ =-sin{nx}\times \frac{d}{dx}nx\\ =-nsin{nx}\\ \]

Q4

(a)\[ f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h} \\ =\lim_{h\to 0}\frac{sin[{a(x+h)+b}]-sin({ax+b})}{h} \\ =\lim_{h\to 0}\frac{sin({ax+b+ah})-sin({ax+b})}{h} \\ =\lim_{h\to 0}\frac{sin({ax+b})cos{ah}+cos({ax+b})sin{ah}-sin({ax+b})}{h} \\ =\lim_{h\to 0}\frac{sin({ax+b})(cos{ah}-1)+cos({ax+b})sin{ah}}{h} \\ =a\lim_{ah\to 0}\frac{sin({ax+b})(cos{ah}-1)+cos({ax+b})sin{ah}}{ah} \\ =a[sin({ax+b})\times 0+cos({ax+b})\times 1] \\ =acos({ax+b})\]

\[ g'(x)=\lim_{h\to 0}\frac{g(x+h)-g(x)}{h} \\ =\lim_{h\to 0}\frac{cos[{a(x+h)+b}]-cos({ax+b})}{h} \\ =\lim_{h\to 0}\frac{cos({ax+b+ah})-cos({ax+b})}{h} \\ =\lim_{h\to 0}\frac{cos({ax+b})cos{ah}-sin({ax+b})sin{ah}-cos({ax+b})}{h} \\ =\lim_{h\to 0}\frac{cos({ax+b})(cos{ah}-1)-sin({ax+b})sin{ah}}{h} \\ =a\lim_{ah\to 0}\frac{cos({ax+b})(cos{ah}-1)-sin({ax+b})sin{ah}}{ah} \\ =a[cos({ax+b})\times 0-sin({ax+b})\times 1] \\ =-asin({ax+b})\]

(b)\[ f''(x)= [acos({ax+b})]'\\ = a[g(x)]'\\ = a \times (-asin({ax+b})) \\ = -a^2 sin({ax+b})\] \[ g''(x)= [-asin({ax+b})]'\\ = -a[f(x)]'\\ = -a \times (acos({ax+b})) \\ = -a^2 cos({ax+b})\]

3. \[ f^{(n)}(x)=\left\{ \begin{aligned} (-1)^{\frac{n-1}{2}}a^ncos({ax+b}), n~is~odd \\ (-1)^{\frac{n}{2}}a^nsin({ax+b}), n~is~even \end{aligned} \right. \]

\[ g^{(n)}(x)=\left\{ \begin{aligned} (-1)^{\frac{n+1}{2}}a^nsin({ax+b}), n~ is~ odd \\ (-1)^{\frac{n}{2}}a^ncos({ax+b}), n~ is~ even \end{aligned} \right. \]