Trigonometry
Trigonometry studies the relationships between angles and sides of triangles. Right-triangle definitions extend to all real angles via the unit circle, giving the six trigonometric functions, laws of sines and cosines, and the area formula.
\[\sin\theta=\frac{\text{opp}}{hyp},\quad\cos\theta=\frac{\text{adj}}{hyp},\quad\tan\theta=\frac{\text{opp}}{\text{adj}},\]\[\csc\theta=\frac{1}{\sin\theta},\quad\sec\theta=\frac{1}{\cos\theta},\quad\cot\theta=\frac{1}{\tan\theta}.\] Memory aid for signs by quadrant (ASTC): Q1 All positive; Q2 Sin positive; Q3 Tan positive; Q4 Cos positive.
\[180°=\pi\text{ rad}.\quad\text{Arc length: }s=r\theta\;(\theta\text{ in radians}).\quad\text{Sector area: }A=\tfrac{1}{2}r^2\theta.\]Degrees Radians \(\sin\) \(\cos\) \(\tan\) 0° 0 0 1 0 30° \(\pi/6\) \(\frac{1}{2}\) \(\frac{\sqrt3}{2}\) \(\frac{1}{\sqrt3}\) 45° \(\pi/4\) \(\frac{\sqrt2}{2}\) \(\frac{\sqrt2}{2}\) 1 60° \(\pi/3\) \(\frac{\sqrt3}{2}\) \(\frac{1}{2}\) \(\sqrt3\) 90° \(\pi/2\) 1 0 undef
\[\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}.\] Use when: AAS, ASA, or SSA (ambiguous case — may give 0, 1, or 2 triangles). SSA ambiguous case: given \(a, b, A\). If \(a < b\sin A\) → no triangle; if \(a=b\sin A\) → right triangle; if \(b\sin A < a < b\) → two triangles; if \(a\ge b\) → one triangle.
\[c^2=a^2+b^2-2ab\cos C;\qquad\cos C=\frac{a^2+b^2-c^2}{2ab}.\]\[\text{Area}=\tfrac{1}{2}ab\sin C.\quad\text{Heron's: }A=\sqrt{s(s-a)(s-b)(s-c)},\;s=\tfrac{a+b+c}{2}.\] Use Law of Cosines when: SAS (two sides + included angle) or SSS (three sides).
Problem. In \(\triangle ABC\), \(A=40°\), \(B=75°\), \(b=12\). Find \(a\) and the area.
\(C=180-40-75=65°.\quad\dfrac{a}{\sin40°}=\dfrac{12}{\sin75°}.\)
\(a=\dfrac{12\sin40°}{\sin75°}=\dfrac{12(0.6428)}{0.9659}\approx7.99.\)
Area \(=\dfrac{1}{2}ab\sin C=\dfrac{1}{2}(7.99)(12)\sin65°\approx43.5.\)
Problem. A triangle has sides \(a=7, b=9, c=5\). Find angle \(B\).
\(\cos B=\dfrac{a^2+c^2-b^2}{2ac}=\dfrac{49+25-81}{2(7)(5)}=\dfrac{-7}{70}=-0.1.\)
\(B=\arccos(-0.1)\approx95.7°.\)
Problem. Given \(a=6,\;b=8,\;A=30°\). Find all possible triangles.
\(b\sin A=8\sin30°=4.\) Since \(b\sin A=4 < a=6 < b=8\): two triangles possible.
\(\sin B=\dfrac{b\sin A}{a}=\dfrac{8(0.5)}{6}=\dfrac{2}{3}.\quad B_1=\arcsin(2/3)\approx41.8°\) or \(B_2=180°-41.8°=138.2°.\)
Triangle 1: \(B_1\approx41.8°,\;C_1\approx108.2°.\) Triangle 2: \(B_2\approx138.2°,\;C_2\approx11.8°.\) Both valid since all angles positive and sum to 180°.
Problem. A sector has radius 10 cm and central angle 120°. Find the arc length and area.
\(\theta=120°=\dfrac{2\pi}{3}\) rad. Arc length: \(s=r\theta=10\cdot\dfrac{2\pi}{3}=\dfrac{20\pi}{3}\approx20.9\) cm.
Area: \(A=\dfrac{1}{2}r^2\theta=\dfrac{1}{2}(100)\dfrac{2\pi}{3}=\dfrac{100\pi}{3}\approx104.7\text{ cm}^2.\)
Unit Circle & Trigonometric Identities
The unit circle extends trigonometry to all real numbers. Trigonometric identities are equations true for all values of the variable — they allow simplification, equation solving, and calculus manipulations.
For angle \(\theta\) in standard position, terminal point on unit circle: \((\cos\theta,\sin\theta).\) Even/Odd: \(\cos(-\theta)=\cos\theta\) (even); \(\sin(-\theta)=-\sin\theta\) (odd); \(\tan(-\theta)=-\tan\theta\) (odd).
\[\sin^2\theta+\cos^2\theta=1,\quad1+\tan^2\theta=\sec^2\theta,\quad1+\cot^2\theta=\csc^2\theta.\]\[\tan\theta=\frac{\sin\theta}{\cos\theta},\quad\cot\theta=\frac{\cos\theta}{\sin\theta}.\]
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The point \(P=(\cos\theta,\sin\theta)\) always lies on \(x^2+y^2=1\) — that's exactly the Pythagorean identity \(\sin^2\theta+\cos^2\theta=1\) above.
\[\sin(A\pm B)=\sin A\cos B\pm\cos A\sin B,\]\[\cos(A\pm B)=\cos A\cos B\mp\sin A\sin B,\]\[\tan(A\pm B)=\frac{\tan A\pm\tan B}{1\mp\tan A\tan B}.\]
\[\sin2\theta=2\sin\theta\cos\theta,\]\[\cos2\theta=\cos^2\theta-\sin^2\theta=1-2\sin^2\theta=2\cos^2\theta-1,\]\[\tan2\theta=\frac{2\tan\theta}{1-\tan^2\theta}.\]\[\sin\frac{\theta}{2}=\pm\sqrt{\frac{1-\cos\theta}{2}},\quad\cos\frac{\theta}{2}=\pm\sqrt{\frac{1+\cos\theta}{2}},\quad\tan\frac{\theta}{2}=\frac{1-\cos\theta}{\sin\theta}.\]
\[\sin A\cos B=\tfrac{1}{2}[\sin(A+B)+\sin(A-B)],\]\[\cos A\cos B=\tfrac{1}{2}[\cos(A-B)+\cos(A+B)],\]\[\sin A+\sin B=2\sin\!\left(\frac{A+B}{2}\right)\cos\!\left(\frac{A-B}{2}\right).\]
Problem. Find the exact value of \(\cos(75°)\).
\(\cos75°=\cos(45°+30°)=\cos45°\cos30°-\sin45°\sin30°\)
\(=\dfrac{\sqrt2}{2}\cdot\dfrac{\sqrt3}{2}-\dfrac{\sqrt2}{2}\cdot\dfrac{1}{2}=\dfrac{\sqrt6-\sqrt2}{4}.\)
Problem. Prove: \(\dfrac{\sin2\theta}{1+\cos2\theta}=\tan\theta.\)
LHS: \(\dfrac{2\sin\theta\cos\theta}{1+(2\cos^2\theta-1)}=\dfrac{2\sin\theta\cos\theta}{2\cos^2\theta}=\dfrac{\sin\theta}{\cos\theta}=\tan\theta=\) RHS. \(\checkmark\)
Problem. Solve \(\cos2\theta+\cos\theta=0\) on \([0°,360°)\).
Replace \(\cos2\theta=2\cos^2\theta-1\): \(2\cos^2\theta-1+\cos\theta=0.\)
Factor: \((2\cos\theta-1)(\cos\theta+1)=0.\)
\(\cos\theta=\frac{1}{2}\Rightarrow\theta=60°,300°.\quad\cos\theta=-1\Rightarrow\theta=180°.\)
Problem. If \(\sin\theta=\frac{5}{13}\) and \(\theta\) is in Q2, find \(\sin2\theta\) and \(\cos\frac{\theta}{2}.\)
Q2: \(\cos\theta=-\frac{12}{13}.\quad\sin2\theta=2\sin\theta\cos\theta=2\!\left(\frac{5}{13}\right)\!\left(-\frac{12}{13}\right)=-\frac{120}{169}.\)
\(\cos\frac{\theta}{2}=+\sqrt{\frac{1+\cos\theta}{2}}=\sqrt{\frac{1-12/13}{2}}=\sqrt{\frac{1/13}{2}}=\sqrt{\frac{1}{26}}=\frac{1}{\sqrt{26}}.\) (Q2 \(\Rightarrow\theta/2\in\) Q1, so cosine positive.)
Conic Sections
Conic sections are the curves formed by intersecting a double cone with a plane. Each conic has a focus–directrix geometric definition and appears in physics (planetary orbits, optics, projectile paths).
Conic Standard equation Key features Circle \(x^2+y^2=r^2\) Centre \((0,0)\), radius \(r\) Parabola (opens up) \(x^2=4py\) Focus \((0,p)\), directrix \(y=-p\) Ellipse (\(a>b>0\)) \(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1\) Vertices \((\pm a,0)\), foci \((\pm c,0)\), \(c^2=a^2-b^2\) Hyperbola (horiz.) \(\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1\) Vertices \((\pm a,0)\), foci \((\pm c,0)\), \(c^2=a^2+b^2\), asymptotes \(y=\pm\frac{b}{a}x\)
Replace \(x\) with \((x-h)\) and \(y\) with \((y-k)\) in the standard form: To find centre: complete the square in both \(x\) and \(y\).
\[e=\frac{c}{a}.\] Circle: \(e=0.\) Ellipse: \(0
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Same \(a=3\) throughout. \(e=0\) is a circle, \(0
Problem. Write \(x^2+y^2-6x+8y+9=0\) in standard form. Find centre, radius.
Group: \((x^2-6x)+(y^2+8y)=-9.\) Complete the square:
\((x^2-6x+9)+(y^2+8y+16)=-9+9+16=16.\)
\((x-3)^2+(y+4)^2=16.\) Centre \((3,-4)\), radius 4.
Problem. Find the foci and eccentricity of \(\dfrac{(x-1)^2}{25}+\dfrac{(y+2)^2}{16}=1.\)
\(a^2=25,\;b^2=16,\;c^2=25-16=9,\;c=3.\) Centre \((1,-2).\)
Foci: \((1\pm3,\,-2)=(4,-2)\) and \((-2,-2).\) \(e=c/a=3/5.\)
Problem. Find the focus and directrix of \((y-3)^2=-12(x+1).\)
Form: \((y-k)^2=4p(x-h)\) with \(4p=-12\Rightarrow p=-3.\) Vertex \((-1,3).\)
Opens left (since \(p<0\)). Focus: \((-1+(-3),3)=(-4,3).\) Directrix: \(x=-1-(-3)=2.\)
Problem. Write \(4x^2-9y^2-8x+36y-68=0\) in standard form and identify foci.
\(4(x^2-2x)-9(y^2-4y)=68.\) Complete squares: \(4(x-1)^2-4-9(y-2)^2+36=68.\)
\(4(x-1)^2-9(y-2)^2=36.\) Divide by 36: \(\dfrac{(x-1)^2}{9}-\dfrac{(y-2)^2}{4}=1.\)
\(a^2=9,\;b^2=4,\;c^2=13,\;c=\sqrt{13}.\) Centre \((1,2).\) Foci: \((1\pm\sqrt{13},\,2).\)
Vectors & Parametric Equations
A vector has both magnitude and direction. Vectors model force, velocity, and displacement. Parametric equations describe motion and curves by expressing both coordinates as functions of a parameter \(t\).
For \(\mathbf{u}=\langle u_1,u_2\rangle\) and \(\mathbf{v}=\langle v_1,v_2\rangle\), scalar \(k\): Component form from magnitude/angle: \(\mathbf{v}=\langle|\mathbf{v}|\cos\theta,\;|\mathbf{v}|\sin\theta\rangle.\)
\[\mathbf{u}\cdot\mathbf{v}=u_1v_1+u_2v_2=|\mathbf{u}||\mathbf{v}|\cos\theta.\] Perpendicular: \(\mathbf{u}\cdot\mathbf{v}=0.\quad\) Parallel: \(\mathbf{v}=k\mathbf{u}.\) Vector projection of \(\mathbf{u}\) onto \(\mathbf{v}\):
Curve: \(x=f(t),\;y=g(t).\) Method 1: solve for \(t\) from one equation, substitute into the other. Method 2 (trig): if \(x=a\cos t\) and \(y=b\sin t\), use \(\cos^2t+\sin^2t=1\Rightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1.\) Projectile motion: \(x=v_0\cos\alpha\cdot t,\quad y=v_0\sin\alpha\cdot t-\tfrac{1}{2}g t^2.\)
Problem. Find the angle between \(\mathbf{u}=\langle4,-3\rangle\) and \(\mathbf{v}=\langle2,5\rangle\).
\(\mathbf{u}\cdot\mathbf{v}=4(2)+(-3)(5)=8-15=-7.\)
\(|\mathbf{u}|=\sqrt{16+9}=5,\quad|\mathbf{v}|=\sqrt{4+25}=\sqrt{29}.\)
\(\cos\theta=\dfrac{-7}{5\sqrt{29}}\approx\dfrac{-7}{26.93}\approx-0.260.\quad\theta\approx105.1°.\)
Problem. Project \(\mathbf{u}=\langle6,2\rangle\) onto \(\mathbf{v}=\langle3,4\rangle\).
\(\mathbf{u}\cdot\mathbf{v}=18+8=26.\quad|\mathbf{v}|^2=9+16=25.\)
\(\text{proj}_{\mathbf{v}}\mathbf{u}=\dfrac{26}{25}\langle3,4\rangle=\left\langle\dfrac{78}{25},\dfrac{104}{25}\right\rangle.\)
Problem. Convert \(x=3\cos t,\;y=5\sin t,\;t\in[0,2\pi]\) to Cartesian form.
\(\dfrac{x}{3}=\cos t,\quad\dfrac{y}{5}=\sin t.\quad\cos^2t+\sin^2t=1\Rightarrow\dfrac{x^2}{9}+\dfrac{y^2}{25}=1.\)
An ellipse with semi-axes \(a=5\) (vertical), \(b=3\) (horizontal).
Problem. A ball is kicked at 20 m/s at 60° to the horizontal. Find: (a) max height, (b) range, (c) time of flight.
\(v_x=20\cos60°=10\text{ m/s},\quad v_y=20\sin60°=10\sqrt3\text{ m/s}.\)
(a) At max height \(v_y=0\): \(t_{\text{top}}=10\sqrt3/9.8\approx1.77\text{ s}.\quad H=10\sqrt3(1.77)-\frac12(9.8)(1.77)^2\approx15.3\text{ m}.\)
(b) Time of flight: \(T=2t_{\text{top}}\approx3.53\text{ s}.\quad\text{Range}=10(3.53)\approx35.3\text{ m}.\)
Polar Coordinates
Polar coordinates \((r,\theta)\) locate a point by distance from the origin and angle from the positive \(x\)-axis. Many curves that are complicated in Cartesian form have elegant polar equations.
\[x=r\cos\theta,\quad y=r\sin\theta,\quad r^2=x^2+y^2,\quad\theta=\arctan\!\left(\frac{y}{x}\right)\text{ (adjust for quadrant)}.\] Note: A point has infinitely many polar representations: \((r,\theta)\equiv(r,\theta+2\pi k)\equiv(-r,\theta+\pi).\)
Name Equation Description Circle at origin \(r=a\) Radius \(|a|\) Circle through origin \(r=2a\cos\theta\) Centre \((a,0)\), radius \(|a|\) Cardioid \(r=a(1+\cos\theta)\) Heart-shaped; passes through origin Limaçon \(r=a+b\cos\theta\) Inner loop if \(a Rose (\(n\) odd) \(r=a\cos(n\theta)\) \(n\) petals Rose (\(n\) even) \(r=a\cos(n\theta)\) \(2n\) petals Lemniscate \(r^2=a^2\cos2\theta\) Figure-eight shape Spiral \(r=a\theta\) Archimedean spiral
About the polar axis (x-axis): replace \(\theta\) with \(-\theta\) — equation unchanged. About \(\theta=\pi/2\) (y-axis): replace \(\theta\) with \(\pi-\theta\) — equation unchanged. About the pole (origin): replace \(r\) with \(-r\) or \(\theta\) with \(\theta+\pi\) — equation unchanged.
Problem. Convert \(r=4\cos\theta\) to Cartesian and identify the curve.
Multiply both sides by \(r\): \(r^2=4r\cos\theta\Rightarrow x^2+y^2=4x.\)
Complete the square: \((x-2)^2+y^2=4.\) Circle with centre \((2,0)\), radius 2.
Problem. Convert \(r=\dfrac{3}{2-\cos\theta}\) and identify the conic.
\(r(2-\cos\theta)=3\Rightarrow2r-r\cos\theta=3\Rightarrow2\sqrt{x^2+y^2}-x=3.\)
\(2\sqrt{x^2+y^2}=x+3\Rightarrow4(x^2+y^2)=(x+3)^2=x^2+6x+9.\)
\(3x^2-6x+4y^2=9.\) Divide: \(\dfrac{(x-1)^2}{4}+\dfrac{y^2}{3}=1.\) Ellipse.
Problem. Sketch \(r=2+2\cos\theta\). Find max \(r\), min \(r\), and symmetry.
Max: \(\theta=0\Rightarrow r=4.\) Min: \(\theta=\pi\Rightarrow r=0.\) Midline: \(\theta=\pi/2\) or \(3\pi/2\Rightarrow r=2.\)
Symmetric about the polar axis (replacing \(\theta\) with \(-\theta\) gives same equation). This is a cardioid pointing right.
Exponential & Logarithmic Functions
Exponential functions model multiplicative change — growth and decay. Logarithms are their inverses and appear naturally in measurement scales (Richter, decibel, pH) and in solving exponential equations.
Domain: \(\mathbb{R}\). Range: \((0,\infty)\) when \(a>0.\) Horizontal asymptote: \(y=0.\) Natural exponential: \(f(x)=e^x,\;e\approx2.71828.\)
\[\log_b(MN)=\log_b M+\log_b N,\quad\log_b\!\left(\frac{M}{N}\right)=\log_b M-\log_b N,\]\[\log_b(M^p)=p\log_b M,\quad\log_b x=\frac{\log x}{\log b}=\frac{\ln x}{\ln b}.\] \(\log_b 1=0,\;\log_b b=1,\;b^{\log_b x}=x,\;\log_b(b^x)=x.\)
\[P(t)=P_0\,b^t,\quad P(t)=P_0\,e^{rt}\quad(r>0:\text{growth};\;r<0:\text{decay}).\]\[\text{Half-life: }A(t)=A_0\!\left(\tfrac{1}{2}\right)^{t/h}.\quad\text{Compound: }A=P\!\left(1+\frac{r}{n}\right)^{nt}.\] Logistic growth: \(P(t)=\dfrac{K}{1+Ae^{-bt}}\) where \(K\) is the carrying capacity.
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\(P(t)=P_0e^{rt}\). Same shape as Worked Example 6.3 (half-life) when \(r<0\), and 6.4 (compound growth) when \(r>0\).
Exponential: (1) isolate base power; (2) apply \(\ln\) both sides; (3) solve. Logarithmic: (1) combine using log properties; (2) exponentiate; (3) check domain. Warning: always verify solutions — \(\log\) of a negative number is undefined.
Problem. Solve \(5^{x+1}=3^{2x-1}\). Give exact and decimal answers.
Take \(\ln\): \((x+1)\ln5=(2x-1)\ln3.\)
\(x\ln5+\ln5=2x\ln3-\ln3.\quad x(\ln5-2\ln3)=-\ln3-\ln5.\)
\(x=\dfrac{-\ln3-\ln5}{\ln5-2\ln3}=\dfrac{-\ln15}{\ln5-\ln9}=\dfrac{\ln15}{\ln9-\ln5}=\dfrac{\ln15}{\ln(9/5)}\approx\dfrac{2.708}{0.588}\approx4.61.\)
Problem. Solve \(\log(x+4)-\log(x-1)=1.\)
\(\log\!\left(\dfrac{x+4}{x-1}\right)=1\Rightarrow\dfrac{x+4}{x-1}=10\Rightarrow x+4=10x-10\Rightarrow9x=14\Rightarrow x=\dfrac{14}{9}.\)
Check: \(x=14/9>1\) ✓, and \(x+4>0\) ✓.
Problem. Strontium-90 has a half-life of 28.8 years. How long until only 15% of an initial sample remains?
\(A(t)=A_0(0.5)^{t/28.8}.\) Set \(A=0.15\,A_0:\quad0.5^{t/28.8}=0.15.\)
\(\dfrac{t}{28.8}\ln0.5=\ln0.15\Rightarrow t=28.8\cdot\dfrac{\ln0.15}{\ln0.5}=28.8\cdot\dfrac{-1.897}{-0.693}\approx78.8\text{ years}.\)
Problem. \$5000 is invested at 6% p.a. compounded monthly. Find the amount after 10 years, and compare with continuous compounding.
Monthly: \(A=5000\!\left(1+\dfrac{0.06}{12}\right)^{120}=5000(1.005)^{120}\approx5000(1.8194)\approx\$9097.\)
Continuous: \(A=5000\,e^{0.06\times10}=5000\,e^{0.6}\approx5000(1.8221)\approx\$9110.\) Difference: \$13 over 10 years.
Rational Functions & Introduction to Limits
Rational functions are quotients of polynomials. Their graphs exhibit asymptotic behaviour — the conceptual bridge to AP Calculus. Limits describe what a function approaches, even where it is not defined.
For \(f(x)=\dfrac{p(x)}{q(x)}\) (fully reduced), \(n=\deg p,\;d=\deg q\): Vertical asymptote: \(x=a\) where \(q(a)=0\) after cancelling common factors. Hole (removable discontinuity): at \(x=a\) where a factor \((x-a)\) cancels from both \(p\) and \(q.\) Horizontal asymptote:
\(\displaystyle\lim_{x\to a}f(x)=L\) means \(f(x)\) can be made arbitrarily close to \(L\) by taking \(x\) close (not equal) to \(a.\) One-sided limits: \(\displaystyle\lim_{x\to a^-}f(x)\) (from left), \(\displaystyle\lim_{x\to a^+}f(x)\) (from right). The two-sided limit exists iff both one-sided limits are equal. Limits at infinity: \(\displaystyle\lim_{x\to\pm\infty}f(x)=\) horizontal asymptote (if finite).
1. Find domain (exclude zeros of \(q\)). 2. Factorise; cancel; identify holes. 3. Find vertical asymptotes. 4. Find horizontal/oblique asymptote. 5. Find intercepts. 6. Analyse sign; sketch.
Problem. Analyse \(f(x)=\dfrac{2x^2-8}{x^2-x-6}.\)
Factor: \(\dfrac{2(x-2)(x+2)}{(x-3)(x+2)}.\) Cancel \((x+2)\): hole at \(x=-2,\;y=\frac{2(-2-2)}{-2-3}=\frac{-8}{-5}=\frac{8}{5}.\)
Simplified: \(\dfrac{2(x-2)}{x-3},\;x\neq-2.\) VA: \(x=3.\) HA: \(n=d\Rightarrow y=2.\) \(x\)-intercept: \(x=2.\) \(y\)-intercept: \(f(0)=\frac{-8}{-6}=\frac{4}{3}.\)
Problem. Find all asymptotes of \(f(x)=\dfrac{x^2+2x+3}{x-1}.\)
VA: \(x=1.\) \(n=2>d=1\): oblique asymptote — divide:
\(x^2+2x+3=(x-1)(x+3)+6.\) So oblique asymptote: \(y=x+3.\)
Problem. Find \(\displaystyle\lim_{x\to2}\frac{x^2-4}{x-2}\) and \(\displaystyle\lim_{x\to\infty}\frac{3x^2-1}{2x^2+5}.\)
(a) Factor: \(\dfrac{(x-2)(x+2)}{x-2}=x+2\to4\) as \(x\to2.\) (Removable discontinuity.)
(b) Divide numerator and denominator by \(x^2\): \(\dfrac{3-1/x^2}{2+5/x^2}\to\dfrac{3}{2}\) as \(x\to\infty.\)
Problem. Solve \(\dfrac{x+1}{x-2}>0.\)
Critical values: \(x=-1\) (numerator zero) and \(x=2\) (denominator zero). Test sign in each interval:
\((-\infty,-1)\): both negative → positive ✓. \((-1,2)\): \(+/-\) → negative ✗. \((2,\infty)\): both positive ✓.
Solution: \(x\in(-\infty,-1)\cup(2,\infty).\)
Sequences & Series
A sequence is an ordered list of numbers; a series is the sum of a sequence's terms. Arithmetic and geometric types appear in financial models, while infinite series connect to calculus through limits.
Common difference \(d\); \(n\)-th term and sum:
Common ratio \(r\); \(n\)-th term, finite sum, and infinite sum:
\[\sum_{k=1}^{n}k=\frac{n(n+1)}{2},\quad\sum_{k=1}^{n}k^2=\frac{n(n+1)(2n+1)}{6},\quad\sum_{k=1}^{n}k^3=\left[\frac{n(n+1)}{2}\right]^2.\]\[\sum_{k=0}^{n}ar^k=\frac{a(1-r^{n+1})}{1-r}\quad(\text{geometric partial sum}).\]
\[(a+b)^n=\sum_{k=0}^{n}\binom{n}{k}a^{n-k}b^k,\quad\binom{n}{k}=\frac{n!}{k!(n-k)!}.\] Pascal's triangle rows give binomial coefficients for small \(n.\) Finding a specific term: the \((r+1)\)-th term of \((a+b)^n\) is \(\dbinom{n}{r}a^{n-r}b^r.\)
To prove a statement \(P(n)\) for all \(n\ge1\): (1) Base case: verify \(P(1)\). (2) Inductive step: assume \(P(k)\) true, prove \(P(k+1)\) follows.
Problem. Find the sum of the arithmetic series \(5+9+13+\cdots+101.\)
\(d=4,\;a_1=5,\;a_n=101.\quad n=\dfrac{101-5}{4}+1=25.\)
\(S_{25}=\dfrac{25}{2}(5+101)=\dfrac{25}{2}(106)=1325.\)
Problem. Find \(\sum_{n=1}^{\infty}3\!\left(\dfrac{2}{5}\right)^n.\)
\(a_1=3(2/5)=6/5,\;r=2/5,\;|r|<1.\quad S_\infty=\dfrac{6/5}{1-2/5}=\dfrac{6/5}{3/5}=\dfrac{6}{3}=2.\)
Problem. Find the term containing \(x^3\) in the expansion of \(\left(2x-\dfrac{1}{x}\right)^7.\)
General term: \(\binom{7}{k}(2x)^{7-k}\!\left(-\dfrac{1}{x}\right)^k=\binom{7}{k}2^{7-k}(-1)^k\,x^{7-k-k}=\binom{7}{k}2^{7-k}(-1)^k\,x^{7-2k}.\)
Set \(7-2k=3\Rightarrow k=2.\) Term: \(\binom{7}{2}2^5(-1)^2x^3=21\cdot32\cdot x^3=672\,x^3.\)
Problem. Prove \(1+2+3+\cdots+n=\dfrac{n(n+1)}{2}\) for all \(n\ge1.\)
Base case \(n=1\): LHS\(=1\), RHS\(=\dfrac{1\cdot2}{2}=1.\;\checkmark\)
Inductive step: Assume true for \(n=k\): \(1+\cdots+k=\dfrac{k(k+1)}{2}.\)
For \(n=k+1\): \(1+\cdots+k+(k+1)=\dfrac{k(k+1)}{2}+(k+1)=(k+1)\!\left(\dfrac{k}{2}+1\right)=\dfrac{(k+1)(k+2)}{2}.\;\checkmark\)
Practice Set
If \(\sin\theta=\dfrac{3}{5}\) and \(\theta\) is in Quadrant I, find \(\cos\theta\), \(\tan\theta\), and \(\sin 2\theta\).
\(\cos\theta=\sqrt{1-9/25}=4/5.\quad\tan\theta=(3/5)/(4/5)=3/4.\quad\sin2\theta=2(3/5)(4/5)=24/25.\)
Write the equation of the ellipse with foci at \((\pm 3,0)\) and vertices at \((\pm 5,0)\).
\(c=3,\;a=5,\;b^2=a^2-c^2=16.\quad\dfrac{x^2}{25}+\dfrac{y^2}{16}=1.\)
Convert \(r=4\cos\theta\) to Cartesian form and identify the curve.
\(r^2=4r\cos\theta\Rightarrow x^2+y^2=4x\Rightarrow(x-2)^2+y^2=4.\) Circle, centre \((2,0)\), radius 2.
Find the sum of the geometric series \(12+8+\tfrac{16}{3}+\cdots\)
\(r=2/3,\;|r|<1.\quad S_\infty=12/(1-2/3)=36.\)
Solve \(\log_2(x+3)+\log_2(x-3)=4\).
\(\log_2[(x+3)(x-3)]=4\Rightarrow x^2-9=16\Rightarrow x=5.\) (Reject \(-5\): \(\log\) undefined.)
Find the vertical and horizontal asymptotes of \(f(x)=\dfrac{3x^2+1}{x^2-4}\) and describe end behaviour.
Vertical: \(x=\pm2.\) Horizontal: degrees equal, \(y=3/1=3.\) As \(x\to\pm\infty\), \(f(x)\to 3\) from above (since numerator grows slightly faster for large \(|x|\)).
Find the angle between \(\mathbf{u}=\langle 1,\sqrt{3}\rangle\) and \(\mathbf{v}=\langle 2,0\rangle\).
\(\mathbf{u}\cdot\mathbf{v}=2.\;|\mathbf{u}|=2,\;|\mathbf{v}|=2.\;\cos\theta=2/4=1/2.\;\theta=60°.\)
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