Circular Functions

# Circular Functions ## Objectifs et prérequis This chapter aims to develop a deep understanding of circular functions (trigonometric functions) as they relate to the unit circle and real-world applications. Students will learn to convert between degrees and radians, evaluate the six trigonometric ratios for special and general angles, apply trigonometric laws to solve triangles, compute areas and lengths related to circles, analyze trigonometric graphs, solve trigonometric equations, and use identities to simplify expressions and find extreme values. Mastery of algebraic manipulation, basic geometry (properties of triangles and circles), and familiarity with the Cartesian plane are essential prerequisites. ## Introduction Circular functions, also known as trigonometric functions, are fundamental in mathematics and its applications in physics, engineering, and navigation. They arise naturally from the study of circles and right triangles, and their periodic nature makes them indispensable for modeling oscillatory phenomena. In this chapter, we extend the trigonometric ratios beyond acute angles to any angle using the unit circle, enabling the analysis of general angles and their properties. We explore the relationships between angles and arc lengths, derive and apply the sine and cosine rules for solving triangles, investigate the graphs of sine, cosine, and tangent functions, and develop techniques for solving trigonometric equations and simplifying expressions using identities. The chapter culminates in applications such as finding maximum and minimum values, working with small angles, and understanding inverse trigonometric functions. ## I. Converting angles in degrees to radians and vice versa ### Definition An angle measured in radians is defined as the ratio of the arc length subtended by the angle to the radius of the circle. One radian is the angle subtended at the center of a circle by an arc equal in length to the radius. > To convert degrees to radians: multiply by $\frac{\pi}{180}$. > To convert radians to degrees: multiply by $\frac{180}{\pi}$. ### Examples **Example 1:** Convert $150^\circ$ to radians. $$ 150^\circ \times \frac{\pi}{180} = \frac{150\pi}{180} = \frac{5\pi}{6} \text{ radians} $$ **Example 2:** Convert $\frac{3\pi}{4}$ radians to degrees. $$ \frac{3\pi}{4} \times \frac{180}{\pi} = \frac{3 \times 180}{4} = 135^\circ $$ ### Exercises **Exercise 1:** Convert $225^\circ$ to radians. *Solution:* $225 \times \frac{\pi}{180} = \frac{5\pi}{4}$ radians. **Exercise 2:** Convert $\frac{7\pi}{6}$ radians to degrees. *Solution:* $\frac{7\pi}{6} \times \frac{180}{\pi} = 210^\circ$. ## II. The six trigonometric ratios, ratios of special angles 1 ### Definition For an angle $\theta$ in standard position, with point $(x, y)$ on the terminal side and $r = \sqrt{x^2 + y^2} \neq 0$, the six trigonometric ratios are: - $\sin \theta = \frac{y}{r}$ - $\cos \theta = \frac{x}{r}$ - $\tan \theta = \frac{y}{x}$ ($x \neq 0$) - $\csc \theta = \frac{r}{y}$ ($y \neq 0$) - $\sec \theta = \frac{r}{x}$ ($x \neq 0$) - $\cot \theta = \frac{x}{y}$ ($y \neq 0$) > The six trigonometric ratios are defined using coordinates on the terminal side of an angle in standard position. ### Special Angles (First Quadrant) We consider angles $0^\circ$, $30^\circ$, $45^\circ$, $60^\circ$, $90^\circ$ and their radian equivalents. | Angle (°) | Angle (rad) | $\sin \theta$ | $\cos \theta$ | $\tan \theta$ | $\csc \theta$ | $\sec \theta$ | $\cot \theta$ | |-----------|-------------|----------------|----------------|----------------|----------------|----------------|----------------| | $0$ | $0$ | $0$ | $1$ | $0$ | undefined | $1$ | undefined | | $30$ | $\frac{\pi}{6}$ | $\frac{1}{2}$ | $\frac{\sqrt{3}}{2}$ | $\frac{1}{\sqrt{3}}$ | $2$ | $\frac{2}{\sqrt{3}}$ | $\sqrt{3}$ | | $45$ | $\frac{\pi}{4}$ | $\frac{\sqrt{2}}{2}$ | $\frac{\sqrt{2}}{2}$ | $1$ | $\sqrt{2}$ | $\sqrt{2}$ | $1$ | | $60$ | $\frac{\pi}{3}$ | $\frac{\sqrt{3}}{2}$ | $\frac{1}{2}$ | $\sqrt{3}$ | $\frac{2}{\sqrt{3}}$ | $2$ | $\frac{1}{\sqrt{3}}$ | | $90$ | $\frac{\pi}{2}$ | $1$ | $0$ | undefined | $1$ | undefined | $0$ | > The values of trigonometric ratios for $30^\circ$, $45^\circ$, and $60^\circ$ can be derived from special right triangles. ### Examples **Example 1:** Find $\sin 120^\circ$. $120^\circ$ is in QII, reference angle $60^\circ$. $\sin$ is positive in QII. $\sin 120^\circ = \sin 60^\circ = \frac{\sqrt{3}}{2}$ **Example 2:** Find $\sec 225^\circ$. $225^\circ$ is in QIII, reference angle $45^\circ$. $\sec = \frac{1}{\cos}$, $\cos$ negative in QIII. $\cos 225^\circ = -\cos 45^\circ = -\frac{\sqrt{2}}{2}$, so $\sec 225^\circ = -\frac{2}{\sqrt{2}} = -\sqrt{2}$ ### Exercises **Exercise 1:** Find $\cot 300^\circ$. *Solution:* $300^\circ$ in QIV, reference $60^\circ$, $\cot$ negative in QIV. $\cot 60^\circ = \frac{1}{\sqrt{3}}$, so $\cot 300^\circ = -\frac{1}{\sqrt{3}}$ **Exercise 2:** Evaluate $\csc \left(-\frac{\pi}{3}\right)$. *Solution:* $-\frac{\pi}{3}$ is coterminal with $\frac{5\pi}{3}$ (QIV), reference $\frac{\pi}{3}$. $\csc$ negative in QIV. $\csc \frac{\pi}{3} = \frac{2}{\sqrt{3}}$, so result is $-\frac{2}{\sqrt{3}}$ ## III. The six trigonometric ratios, ratios of special angles 2 ### Extension to All Quadrants The signs of trigonometric ratios depend on the quadrant: - QI: all positive - QII: $\sin$, $\csc$ positive - QIII: $\tan$, $\cot$ positive - QIV: $\cos$, $\sec$ positive > The mnemonic "ALL STUDENTS TAKE CALCULUS" helps recall the signs: All (QI), Sine (QII), Tangent (QIII), Cosine (QIV). ### Examples **Example 1:** Determine the sign of $\tan 210^\circ$. $210^\circ$ is in QIII → $\tan$ positive. **Example 2:** If $\cos \theta = -\frac{3}{5}$ and $\theta$ is in QII, find $\sin \theta$. Use $\sin^2 \theta + \cos^2 \theta = 1$: $\sin^2 \theta = 1 - \left(-\frac{3}{5}\right)^2 = 1 - \frac{9}{25} = \frac{16}{25}$ $\sin \theta = \pm \frac{4}{5}$ → positive in QII → $\sin \theta = \frac{4}{5}$ ### Exercises **Exercise 1:** Given $\sin \theta = \frac{5}{13}$ and $\theta$ in QII, find $\tan \theta$. *Solution:* $\cos^2 \theta = 1 - \left(\frac{5}{13}\right)^2 = \frac{144}{169}$ → $\cos \theta = -\frac{12}{13}$ (QII) → $\tan \theta = \frac{5/13}{-12/13} = -\frac{5}{12}$ **Exercise 2:** Find $\sec 330^\circ$. *Solution:* $330^\circ$ in QIV, reference $30^\circ$, $\sec$ positive in QIV. $\sec 30^\circ = \frac{2}{\sqrt{3}}$ → $\sec 330^\circ = \frac{2}{\sqrt{3}}$ ## IV. Area and Perimeter of a circle, arc length ### Definitions - Circumference: $C = 2\pi r$ or $C = \pi d$ - Area: $A = \pi r^2$ - Arc length: $s = r\theta$, where $\theta$ is in radians > The arc length subtended by an angle $\theta$ (in radians) in a circle of radius $r$ is $s = r\theta$. ### Examples **Example 1:** Find the arc length of a circle with radius $5$ cm subtended by an angle of $\frac{\pi}{3}$ radians. $s = 5 \times \frac{\pi}{3} = \frac{5\pi}{3}$ cm **Example 2:** A circular garden has a diameter of $14$ m. Find its area. Radius $r = 7$ m → $A = \pi (7)^2 = 49\pi$ m² ### Exercises **Exercise 1:** Find the radius of a circle if an arc of length $8\pi$ cm subtends an angle of $\frac{2\pi}{3}$ radians. *Solution:* $s = r\theta$ → $8\pi = r \times \frac{2\pi}{3}$ → $r = 8\pi \times \frac{3}{2\pi} = 12$ cm **Exercise 2:** Calculate the area of a circle with circumference $20\pi$ m. *Solution:* $C = 2\pi r = 20\pi$ → $r = 10$ m → $A = \pi (10)^2 = 100\pi$ m² ## V. Area of sector and Area of segment ### Definitions - Area of sector: $A = \frac{1}{2} r^2 \theta$ ($\theta$ in radians) or $A = \frac{\theta}{360} \pi r^2$ ($\theta$ in degrees) - Area of segment: $A_{\text{segment}} = A_{\text{sector}} - A_{\text{triangle}}$ > The area of a sector is proportional to the angle it subtends at the center. ### Examples **Example 1:** Find the area of a sector with radius $6$ cm and central angle $45^\circ$. $A = \frac{45}{360} \pi (6)^2 = \frac{1}{8} \pi \times 36 = \frac{36\pi}{8} = \frac{9\pi}{2}$ cm² **Example 2:** Find the area of a segment formed by a chord subtending $60^\circ$ in a circle of radius $10$ cm. Sector area: $\frac{60}{360} \pi (10)^2 = \frac{1}{6} \times 100\pi = \frac{50\pi}{3}$ cm² Triangle area: $\frac{1}{2} r^2 \sin \theta = \frac{1}{2} \times 100 \times \sin 60^\circ = 50 \times \frac{\sqrt{3}}{2} = 25\sqrt{3}$ cm² Segment area: $\frac{50\pi}{3} - 25\sqrt{3}$ cm² ### Exercises **Exercise 1:** Find the area of a sector with radius $8$ cm and angle $\frac{2\pi}{3}$ radians. *Solution:* $A = \frac{1}{2} r^2 \theta = \frac{1}{2} \times 64 \times \frac{2\pi}{3} = \frac{64\pi}{3}$ cm² **Exercise 2:** A segment is formed by a chord of length $12$ cm in a circle of radius $10$ cm. Find the area of the segment. *Solution:* First, find central angle $\theta$ using chord length: $c = 2r \sin(\theta/2)$ → $12 = 20 \sin(\theta/2)$ → $\sin(\theta/2) = 0.6$ → $\theta/2 \approx 36.87^\circ$ → $\theta \approx 73.74^\circ$ Sector area: $\frac{73.74}{360} \pi (10)^2 \approx 64.25$ cm² Triangle area: $\frac{1}{2} \times 10^2 \times \sin 73.74^\circ \approx 50 \times 0.96 = 48$ cm² Segment area ≈ $64.25 - 48 = 16.25$ cm² ## VI. Sine Rule ### Statement In any triangle $ABC$, with sides $a$, $b$, $c$ opposite angles $A$, $B$, $C$ respectively: $$ \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R $$ where $R$ is the radius of the circumcircle. > The Sine Rule relates the sides of a triangle to the sines of their opposite angles. ### Applications Used when: - Two angles and one side are known (AAS or ASA) - Two sides and a non-included angle are known (SSA – ambiguous case) ### Examples **Example 1:** In $\triangle ABC$, $A = 45^\circ$, $B = 60^\circ$, $a = 10$ cm. Find $b$. First, $C = 180^\circ - 45^\circ - 60^\circ = 75^\circ$ $\frac{a}{\sin A} = \frac{b}{\sin B}$ → $\frac{10}{\sin 45^\circ} = \frac{b}{\sin 60^\circ}$ $\frac{10}{\sqrt{2}/2} = \frac{b}{\sqrt{3}/2}$ → $10 \cdot \frac{2}{\sqrt{2}} = b \cdot \frac{2}{\sqrt{3}}$ → $\frac{20}{\sqrt{2}} = \frac{2b}{\sqrt{3}}$ → $b = \frac{20}{\sqrt{2}} \cdot \frac{\sqrt{3}}{2} = 10 \cdot \frac{\sqrt{3}}{\sqrt{2}} = 5\sqrt{6}$ cm **Example 2:** In $\triangle ABC$, $a = 8$ cm, $b = 10$ cm, $A = 30^\circ$. Find possible values of $B$. $\frac{a}{\sin A} = \frac{b}{\sin B}$ → $\frac{8}{\sin 30^\circ} = \frac{10}{\sin B}$ → $\frac{8}{0.5} = 16 = \frac{10}{\sin B}$ → $\sin B = \frac{10}{16} = \frac{5}{8} = 0.625$ $B = \sin^{-1}(0.625) \approx 38.68^\circ$ or $180^\circ - 38.68^\circ = 141.32^\circ$ Check: if $B = 141.32^\circ$, then $A + B = 171.32^\circ < 180^\circ$ → valid. Two possible triangles. ### Exercises **Exercise 1:** In $\triangle PQR$, $p = 7$ cm, $q = 9$ cm, $P = 50^\circ$. Find $Q$. *Solution:* $\frac{7}{\sin 50^\circ} = \frac{9}{\sin Q}$ → $\sin Q = \frac{9 \sin 50^\circ}{7} \approx \frac{9 \times 0.766}{7} \approx 0.985$ → $Q \approx 80.0^\circ$ or $99.9^\circ$ (both valid since $50 + 80 < 180$, $50 + 99.9 < 180$) **Exercise 2:** In $\triangle XYZ$, $X = 40^\circ$, $Y = 70^\circ$, $z = 12$ cm. Find $x$. *Solution:* $Z = 180 - 40 - 70 = 70^\circ$ → $\frac{x}{\sin 40^\circ} = \frac{12}{\sin 70^\circ}$ → $x = \frac{12 \sin 40^\circ}{\sin 70^\circ} \approx \frac{12 \times 0.6428}{0.9397} \approx 8.20$ cm ## VII. Cosine Rule ### Statement In any triangle $ABC$: $$ a^2 = b^2 + c^2 - 2bc \cos A $$ $$ b^2 = a^2 + c^2 - 2ac \cos B $$ $$ c^2 = a^2 + b^2 - 2ab \cos C $$ > The Cosine Rule relates the lengths of the sides of a triangle to the cosine of one of its angles. ### Applications Used when: - Three sides are known (SSS) to find an angle - Two sides and the included angle are known (SAS) to find the third side ### Examples **Example 1:** In $\triangle ABC$, $b = 6$ cm, $c = 8$ cm, $A = 60^\circ$. Find $a$. $a^2 = 6^2 + 8^2 - 2 \cdot 6 \cdot 8 \cdot \cos 60^\circ = 36 + 64 - 96 \cdot 0.5 = 100 - 48 = 52$ $a = \sqrt{52} = 2\sqrt{13}$ cm **Example 2:** In $\triangle ABC$, $a = 9$ cm, $b = 10$ cm, $c = 11$ cm. Find angle $C$. $c^2 = a^2 + b^2 - 2ab \cos C$ $121 = 81 + 100 - 2 \cdot 9 \cdot 10 \cdot \cos C = 181 - 180 \cos C$ $180 \cos C = 181 - 121 = 60$ → $\cos C = \frac{60}{180} = \frac{1}{3}$ $C = \cos^{-1}\left(\frac{1}{3}\right) \approx 70.53^\circ$ ### Exercises **Exercise 1:** In $\triangle DEF$, $d = 5$ cm, $e = 7$ cm, $f = 8$ cm. Find angle $E$. *Solution:* $e^2 = d^2 + f^2 - 2df \cos E$ $49 = 25 + 64 - 2 \cdot 5 \cdot 8 \cdot \cos E = 89 - 80 \cos E$ $80 \cos E = 89 - 49 = 40$ → $\cos E = \frac{40}{80} = 0.5$ → $E = 60^\circ$ **Exercise 2:** In $\triangle ABC$, $AB = 12$ cm, $AC = 15$ cm, $\angle A = 120^\circ$. Find $BC$. *Solution:* $BC^2 = 12^2 + 15^2 - 2 \cdot 12 \cdot 15 \cdot \cos 120^\circ$ $\cos 120^\circ = -\frac{1}{2}$ → $-2 \cdot 12 \cdot 15 \cdot (-\frac{1}{2}) = +180$ $BC^2 = 144 + 225 + 180 = 549$ → $BC = \sqrt{549} = 3\sqrt{61}$ cm ## VIII. The general angle ### Definition A general angle is any angle, positive or negative, measured in standard position (vertex at origin, initial side on positive x-axis). It can be coterminal with an angle between $0^\circ$ and $360^\circ$ (or $0$ and $2\pi$ radians) by adding or subtracting multiples of $360^\circ$ ($2\pi$ radians). > Two angles are coterminal if they differ by an integer multiple of $360^\circ$ (or $2\pi$ radians). ### Reference Angle The reference angle $\theta'$ is the acute angle formed by the terminal side and the x-axis: - QI: $\theta' = \theta$ - QII: $\theta' = 180^\circ - \theta$ - QIII: $\theta' = \theta - 180^\circ$ - QIV: $\theta' = 360^\circ - \theta$ > The trigonometric ratio of any angle equals $\pm$ the trigonometric ratio of its reference angle, with sign determined by the quadrant. ### Examples **Example 1:** Find a coterminal angle for $-450^\circ$. $-450^\circ + 360^\circ = -90^\circ$; $-90^\circ + 360^\circ = 270^\circ$ → $270^\circ$ is coterminal. **Example 2:** Evaluate $\sin 480^\circ$. $480^\circ - 360^\circ = 120^\circ$ (QII) → reference angle $60^\circ$ → $\sin$ positive in QII → $\sin 480^\circ = \sin 60^\circ = \frac{\sqrt{3}}{2}$ ### Exercises **Exercise 1:** Find the reference angle for $-\frac{5\pi}{4}$ radians. *Solution:* $-\frac{5\pi}{4} + 2\pi = -\frac{5\pi}{4} + \frac{8\pi}{4} = \frac{3\pi}{4}$ (QII) → reference angle = $\pi - \frac{3\pi}{4} = \frac{\pi}{4}$ **Exercise 2:** Evaluate $\cos \left(-\frac{7\pi}{6}\right)$. *Solution:* $-\frac{7\pi}{6} + 2\pi = -\frac{7\pi}{6} + \frac{12\pi}{6} = \frac{5\pi}{6}$ (QII) → reference angle $\frac{\pi}{6}$ → $\cos$ negative in QII → $\cos \frac{5\pi}{6} = -\cos \frac{\pi}{6} = -\frac{\sqrt{3}}{2}$ ## IX. Pythagorean Identities and proofs using the Pythagorean identities ### Fundamental Identities From the unit circle ($x^2 + y^2 = 1$) and definitions $\cos \theta = x$, $\sin \theta = y$: $$ \sin^2 \theta + \cos^2 \theta = 1 $$ Dividing by $\cos^2 \theta$: $$ \tan^2 \theta + 1 = \sec^2 \theta $$ Dividing by $\sin^2 \theta$: $$ 1 + \cot^2 \theta = \csc^2 \theta $$ > The Pythagorean identities are derived from the Pythagorean theorem applied to the unit circle. ### Proofs **Proof of $\sin^2 \theta + \cos^2 \theta = 1$:** On the unit circle, $x = \cos \theta$, $y = \sin \theta$, and $x^2 + y^2 = 1$ by definition. Substituting gives the identity. **Proof of $1 + \tan^2 \theta = \sec^2 \theta$:** Start with $\sin^2 \theta + \cos^2 \theta = 1$. Divide both sides by $\cos^2 \theta$: $\frac{\sin^2 \theta}{\cos^2 \theta} + \frac{\cos^2 \theta}{\cos^2 \theta} = \frac{1}{\cos^2 \theta}$ → $\tan^2 \theta + 1 = \sec^2 \theta$ ### Examples **Example 1:** If $\sin \theta = \frac{3}{5}$ and $\theta$ in QII, find $\cos \theta$. $\sin^2 \theta + \cos^2 \theta = 1$ → $\left(\frac{3}{5}\right)^2 + \cos^2 \theta = 1$ → $\frac{9}{25} + \cos^2 \theta = 1$ → $\cos^2 \theta = \frac{16}{25}$ → $\cos \theta = -\frac{4}{5}$ (QII) **Example 2:** Simplify $\sec^2 \theta - \tan^2 \theta$. From identity: $\sec^2 \theta - \tan^2 \theta = 1$ ### Exercises **Exercise 1:** Given $\cot \theta = -\frac{4}{3}$ and $\theta$ in QIV, find $\csc \theta$. *Solution:* $1 + \cot^2 \theta = \csc^2 \theta$ → $1 + \left(-\frac{4}{3}\right)^2 = 1 + \frac{16}{9} = \frac{25}{9}$ → $\csc^2 \theta = \frac{25}{9}$ → $\csc \theta = -\frac{5}{3}$ (QIV, $\csc$ negative) **Exercise 2:** Prove that $(\sin \theta + \cos \theta)^2 = 1 + 2\sin \theta \cos \theta$. *Solution:* Expand left: $\sin^2 \theta + 2\sin \theta \cos \theta + \cos^2 \theta = (\sin^2 \theta + \cos^2 \theta) + 2\sin \theta \cos \theta = 1 + 2\sin \theta \cos \theta$ ## X. Graphs of Sine, Cosine and Tangent and their properties 1 ### Basic Graphs - $y = \sin x$: period $2\pi$, amplitude $1$, range $[-1, 1]$, odd function, passes through origin - $y = \cos x$: period $2\pi$, amplitude $1$, range $[-1, 1]$, even function, $y$-intercept at $1$ - $y = \tan x$: period $\pi$, range $(-\infty, \infty)$, odd function, vertical asymptotes at $x = \frac{\pi}{2} + k\pi$ > The sine and cosine functions are periodic with period $2\pi$; tangent has period $\pi$. ### Key Points for $y = \sin x$ over $[0, 2\pi]$: - $(0, 0)$, $\left(\frac{\pi}{2}, 1\right)$, $(\pi, 0)$, $\left(\frac{3\pi}{2}, -1\right)$, $(2\pi, 0)$ ### Key Points for $y = \cos x$ over $[0, 2\pi]$: - $(0, 1)$, $\left(\frac{\pi}{2}, 0\right)$, $(\pi, -1)$, $\left(\frac{3\pi}{2}, 0\right)$, $(2\pi, 1)$ ### Key Points for $y = \tan x$ over $(-\frac{\pi}{2}, \frac{\pi}{2})$: - $(0, 0)$, approaches $\infty$ as $x \to \frac{\pi}{2}^-$, approaches $-\infty$ as $x \to -\frac{\pi}{2}^+$ ### Examples **Example 1:** Sketch $y = 2\sin x$ over $[0, 2\pi]$. Amplitude $= 2$, period $= 2\pi$. Points: $(0,0)$, $\left(\frac{\pi}{2}, 2\right)$, $(\pi,0)$, $\left(\frac{3\pi}{2}, -2\right)$, $(2\pi,0)$ **Example 2:** Determine the period of $y = \cos(3x)$. Period of $\cos(bx)$ is $\frac{2\pi}{|b|}$ → $\frac{2\pi}{3}$ ### Exercises **Exercise 1:** State the amplitude and period of $y = -3\cos(2x)$. *Solution:* Amplitude $= |-3| = 3$, period $= \frac{2\pi}{2} = \pi$ **Exercise 2:** Sketch one cycle of $y = \tan\left(\frac{x}{2}\right)$. *Solution:* Period $= \frac{\pi}{|1/2|} = 2\pi$. Asymptotes at $\frac{x}{2} = \frac{\pi}{2} + k\pi$ → $x = \pi + 2k\pi$. So asymptotes at $x = \pi, 3\pi, ...$ and $x = -\pi, -3\pi, ...$. Zero at $x=0$. Shape: increasing from $-\infty$ to $\infty$ between asymptotes. ## XI. Graphs of Sine, Cosine and Tangent and their properties 2 ### Transformations For $y = A \sin(Bx - C) + D$ or $y = A \cos(Bx - C) + D$: - Amplitude: $|A|$ - Period: $\frac{2\pi}{|B|}$ - Phase shift: $\frac{C}{B}$ (right if $C>0$) - Vertical shift: $D$ > The constants $A$, $B$, $C$, $D$ in $y = A \sin(Bx - C) + D$ control amplitude, period, phase shift, and vertical shift. ### Examples **Example 1:** Analyze $y = 4 \sin\left(2x - \frac{\pi}{2}\right) + 1$. - Amplitude: $4$ - Period: $\frac{2\pi}{2} = \pi$ - Phase shift: $\frac{\pi/2}{2} = \frac{\pi}{4}$ to the right - Vertical shift: $1$ up **Example 2:** Write the equation of a sine wave with amplitude $3$, period $\pi$, phase shift $\frac{\pi}{6}$ left, vertical shift $-2$. Period $\pi = \frac{2\pi}{|B|}$ → $|B| = 2$ → take $B=2$ Phase shift $= -\frac{C}{B} = -\frac{\pi}{6}$ → $-\frac{C}{2} = -\frac{\pi}{6}$ → $C = \frac{\pi}{3}$ Equation: $y = 3 \sin\left(2x + \frac{\pi}{3}\right) - 2$ ### Exercises **Exercise 1:** For $y = -2 \cos\left(3x + \pi\right) - 1$, find amplitude, period, phase shift, vertical shift. *Solution:* Amplitude $= 2$, period $= \frac{2\pi}{3}$, phase shift $= -\frac{\pi}{3}$ (left), vertical shift $= -1$ **Exercise 2:** Determine the equation of a cosine function with maximum $5$, minimum $-3$, period $4\pi$, starting at maximum at $x=0$. *Solution:* Amplitude $= \frac{5 - (-3)}{2} = 4$, vertical shift $= \frac{5 + (-3)}{2} = 1$ Period $= 4\pi = \frac{2\pi}{|B|}$ → $|B| = \frac{1}{2}$ → $B = \frac{1}{2}$ Starts at max at $x=0$ → no phase shift → $y = 4 \cos\left(\frac{1}{2}x\right) + 1$ ## XII. General solution to trigonometric equations 1 ### Principle If $\sin \theta = \sin \alpha$, then $\theta = n\pi + (-1)^n \alpha$, $n \in \mathbb{Z}$ If $\cos \theta = \cos \alpha$, then $\theta = 2n\pi \pm \alpha$, $n \in \mathbb{Z}$ If $\tan \theta = \tan \alpha$, then $\theta = n\pi + \alpha$, $n \in \mathbb{Z}$ > The general solution accounts for all angles coterminal with the basic solutions due to periodicity. ### Examples **Example 1:** Solve $\sin \theta = \frac{1}{2}$. $\sin \alpha = \frac{1}{2}$ → $\alpha = \frac{\pi}{6}$ (or $30^\circ$) General solution: $\theta = n\pi + (-1)^n \frac{\pi}{6}$, $n \in \mathbb{Z}$ For $n=0$: $\theta = \frac{\pi}{6}$ $n=1$: $\theta = \pi - \frac{\pi}{6} = \frac{5\pi}{6}$ $n=2$: $\theta = 2\pi + \frac{\pi}{6} = \frac{13\pi}{6}$, etc. **Example 2:** Solve $\cos \theta = -\frac{\sqrt{3}}{2}$. $\cos \alpha = \frac{\sqrt{3}}{2}$ → $\alpha = \frac{\pi}{6}$, but cosine negative → angles in QII and QIII Basic solutions: $\theta = \pi - \frac{\pi}{6} = \frac{5\pi}{6}$, $\theta = \pi + \frac{\pi}{6} = \frac{7\pi}{6}$ Using $\cos \theta = \cos \alpha$ → $\theta = 2n\pi \pm \alpha$, but here $\alpha$ is the acute angle whose cosine is $\frac{\sqrt{3}}{2}$, so we use the form for negative: actually, better to say: if $\cos \theta = \cos \alpha$, then $\theta = 2n\pi \pm \alpha$. Here $\cos \theta = -\frac{\sqrt{3}}{2} = \cos \left(\pi - \frac{\pi}{6}\right) = \cos \frac{5\pi}{6}$, so $\alpha = \frac{5\pi}{6}$ Thus $\theta = 2n\pi \pm \frac{5\pi}{6}$ Check: $n=0$: $\frac{5\pi}{6}$, $-\frac{5\pi}{6}$ (equiv. to $\frac{7\pi}{6}$ mod $2\pi$) $n=1$: $2\pi + \frac{5\pi}{6} = \frac{17\pi}{6}$, $2\pi - \frac{5\pi}{6} = \frac{7\pi}{6}$ → correct. ### Exercises **Exercise 1:** Find the general solution for $\tan \theta = \sqrt{3}$. *Solution:* $\tan \alpha = \sqrt{3}$ → $\alpha = \frac{\pi}{3}$ → $\theta = n\pi + \frac{\pi}{3}$, $n \in \mathbb{Z}$ **Exercise 2:** Solve $\sin \theta = -\frac{1}{2}$ and give the general solution. *Solution:* $\sin \alpha = \frac{1}{2}$ → $\alpha = \frac{\pi}{6}$, but sine negative → $\theta = n\pi + (-1)^n \left(-\frac{\pi}{6}\right)$ or better: $\sin \theta = \sin \left(-\frac{\pi}{6}\right)$ → $\theta = n\pi + (-1)^n \left(-\frac{\pi}{6}\right)$ Alternatively: $\theta = n\pi - (-1)^n \frac{\pi}{6}$ → standard form: $\theta = n\pi + (-1)^{n+1} \frac{\pi}{6}$ Easier: solutions in QIII and QIV: $\theta = \pi + \frac{\pi}{6} = \frac{7\pi}{6}$, $\theta = 2\pi - \frac{\pi}{6} = \frac{11\pi}{6}$ → general: $\theta = 2n\pi + \frac{7\pi}{6}$ or $\theta = 2n\pi + \frac{11\pi}{6}$ But using formula: since $\sin \theta = \sin \alpha$ implies $\theta = n\pi + (-1)^n \alpha$, set $\alpha = -\frac{\pi}{6}$ → $\theta = n\pi + (-1)^n \left(-\frac{\pi}{6}\right) = n\pi - (-1)^n \frac{\pi}{6}$