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1
Unit Circle · Reference Angles
Exact Value of Sine
Hard
Key Concept: Reference Angle & Quadrant Signs
The reference angle \(\hat{\theta}\) is the acute angle between the terminal side and the x-axis. Sine is positive in Q I and Q II; negative in Q III and Q IV.
\(\sin(\pi - \theta) = \sin\theta \qquad \sin(\pi + \theta) = -\sin\theta\)
For \(\theta = \dfrac{7\pi}{6}\): quadrant III, reference angle \(= \dfrac{\pi}{6}\), so \(\sin\!\left(\dfrac{7\pi}{6}\right) = -\sin\!\left(\dfrac{\pi}{6}\right) = -\dfrac{1}{2}\).
Worked Example
Given Find \(\cos\!\left(\dfrac{5\pi}{4}\right)\).
Step 1 \(\dfrac{5\pi}{4}\) is in Q III (between \(\pi\) and \(\dfrac{3\pi}{2}\)).
Step 2 Reference angle: \(\dfrac{5\pi}{4} - \pi = \dfrac{\pi}{4}\).
Step 3 Cosine is negative in Q III: \(\cos\!\left(\dfrac{5\pi}{4}\right) = -\cos\!\left(\dfrac{\pi}{4}\right) = -\dfrac{\sqrt{2}}{2}\).
Find the exact value of \(\sin\!\left(\dfrac{11\pi}{6}\right)\).
✅ Full Solution
\(\dfrac{11\pi}{6}\) lies in Q IV \(\left(2\pi - \dfrac{\pi}{6}\right)\).
Reference angle: \(2\pi - \dfrac{11\pi}{6} = \dfrac{\pi}{6}\).
Sine is negative in Q IV.
\(\sin\!\left(\dfrac{11\pi}{6}\right) = -\sin\!\left(\dfrac{\pi}{6}\right) = \boxed{-\dfrac{1}{2}}\)
2
Pythagorean Identities
Finding Trig Values from One Ratio
Hard
Key Concept: Pythagorean Identities
\(\sin^2\theta + \cos^2\theta = 1\)
\(1 + \tan^2\theta = \sec^2\theta\)
\(1 + \cot^2\theta = \csc^2\theta\)
Given one ratio and a quadrant, you can find all six trig functions.
Worked Example
Given \(\cos\theta = \dfrac{3}{5}\), Q IV. Find \(\sin\theta\).
Step 1 \(\sin^2\theta = 1 - \cos^2\theta = 1 - \dfrac{9}{25} = \dfrac{16}{25}\).
Step 2 Q IV: sine is negative. \(\sin\theta = -\dfrac{4}{5}\).
If \(\tan\theta = -\dfrac{5}{12}\) and \(\theta\) is in Quadrant II, find \(\sec\theta\).
✅ Full Solution
Use \(1 + \tan^2\theta = \sec^2\theta\).
\(\sec^2\theta = 1 + \dfrac{25}{144} = \dfrac{169}{144}\), so \(\sec\theta = \pm\dfrac{13}{12}\).
Q II: cosine is negative \(\Rightarrow\) \(\sec\theta\) is negative.
\(\sec\theta = \boxed{-\dfrac{13}{12}}\)
3
Sum & Difference Formulas
Exact Value via Angle Addition
Hard
Key Concept: Sum & Difference Identities
\(\cos(A-B) = \cos A\cos B + \sin A\sin B\)
\(\sin(A+B) = \sin A\cos B + \cos A\sin B\)
\(\tan(A+B) = \dfrac{\tan A + \tan B}{1 - \tan A\tan B}\)
Write the angle as a sum/difference of standard angles (\(30°, 45°, 60°, 90°, \ldots\)).
Worked Example
Given Find \(\sin 75°\).
Step 1 \(75° = 45° + 30°\).
Step 2 \(\sin 75° = \sin 45°\cos 30° + \cos 45°\sin 30°\)
Step 3 \(= \dfrac{\sqrt{2}}{2}\cdot\dfrac{\sqrt{3}}{2} + \dfrac{\sqrt{2}}{2}\cdot\dfrac{1}{2} = \dfrac{\sqrt{6}+\sqrt{2}}{4}\)
Find the exact value of \(\cos 15°\).
✅ Full Solution
\(15° = 45° - 30°\).
\(\cos 15° = \cos 45°\cos 30° + \sin 45°\sin 30°\)
\(= \dfrac{\sqrt{2}}{2}\cdot\dfrac{\sqrt{3}}{2} + \dfrac{\sqrt{2}}{2}\cdot\dfrac{1}{2} = \dfrac{\sqrt{6}}{4} + \dfrac{\sqrt{2}}{4} = \boxed{\dfrac{\sqrt{6}+\sqrt{2}}{4}}\)
Note: \(15°\) is in Q I so cosine is positive. ✓
4
Double-Angle Formulas
Evaluating Double-Angle Expression
Hard
Key Concept: Double-Angle Identities
\(\sin 2\theta = 2\sin\theta\cos\theta\)
\(\cos 2\theta = \cos^2\theta - \sin^2\theta = 1 - 2\sin^2\theta = 2\cos^2\theta - 1\)
\(\tan 2\theta = \dfrac{2\tan\theta}{1 - \tan^2\theta}\)
Worked Example
Given \(\sin\theta = \dfrac{3}{5}\), Q I. Find \(\cos 2\theta\).
Step 1 \(\cos\theta = \dfrac{4}{5}\) (Q I, positive).
Step 2 \(\cos 2\theta = 1 - 2\sin^2\theta = 1 - 2\cdot\dfrac{9}{25} = 1 - \dfrac{18}{25} = \dfrac{7}{25}\).
If \(\cos\theta = -\dfrac{4}{5}\) and \(\pi < \theta < \dfrac{3\pi}{2}\), find \(\sin 2\theta\).
✅ Full Solution
Q III: \(\sin\theta < 0\). \(\sin^2\theta = 1 - \dfrac{16}{25} = \dfrac{9}{25}\), so \(\sin\theta = -\dfrac{3}{5}\).
\(\sin 2\theta = 2\sin\theta\cos\theta = 2\!\left(-\dfrac{3}{5}\right)\!\left(-\dfrac{4}{5}\right) = \dfrac{24}{25}\).
Both \(\sin\theta\) and \(\cos\theta\) negative in Q III \(\Rightarrow\) their product is positive, so \(\sin 2\theta = \boxed{\dfrac{24}{25}}\).
Common mistake: assuming \(\sin 2\theta\) has the same sign as \(\sin\theta\). The double angle formula always wins!
5
Half-Angle Formulas
Exact Value Using Half-Angle
Hard
Key Concept: Half-Angle Formulas
\(\sin\dfrac{\theta}{2} = \pm\sqrt{\dfrac{1-\cos\theta}{2}}\)
\(\cos\dfrac{\theta}{2} = \pm\sqrt{\dfrac{1+\cos\theta}{2}}\)
\(\tan\dfrac{\theta}{2} = \pm\sqrt{\dfrac{1-\cos\theta}{1+\cos\theta}} = \dfrac{\sin\theta}{1+\cos\theta}\)
The sign is determined by which quadrant \(\dfrac{\theta}{2}\) lies in.
Worked Example
Given Find \(\cos 22.5°\) exactly.
Step 1 \(22.5° = \dfrac{45°}{2}\). Q I, so positive.
Step 2 \(\cos 22.5° = \sqrt{\dfrac{1+\cos 45°}{2}} = \sqrt{\dfrac{1+\frac{\sqrt{2}}{2}}{2}} = \sqrt{\dfrac{2+\sqrt{2}}{4}} = \dfrac{\sqrt{2+\sqrt{2}}}{2}\).
Find the exact value of \(\sin\left(\dfrac{\pi}{8}\right)\).
✅ Full Solution
\(\dfrac{\pi}{8} = \dfrac{1}{2}\cdot\dfrac{\pi}{4}\), so use the half-angle formula with \(\theta = \dfrac{\pi}{4}\).
\(\sin\dfrac{\pi}{8} = \sqrt{\dfrac{1-\cos(\pi/4)}{2}} = \sqrt{\dfrac{1 - \frac{\sqrt{2}}{2}}{2}} = \sqrt{\dfrac{2-\sqrt{2}}{4}}\)
\(= \boxed{\dfrac{\sqrt{2-\sqrt{2}}}{2}}\) (\(\dfrac{\pi}{8}\in\) Q I, so positive sign.)
6
Graphing Transformations
Amplitude, Period & Phase Shift
Hard
Key Concept: Sinusoidal Graphs \(y = A\sin(Bx - C) + D\)
Amplitude \(= |A|\), Period \(= \dfrac{2\pi}{|B|}\), Phase shift \(= \dfrac{C}{B}\) (right if positive), Vertical shift \(= D\)
The phase shift is the solution to \(Bx - C = 0\). Never confuse \(C\) with the phase shift!
Worked Example
Given \(y = 3\sin\!\left(2x - \dfrac{\pi}{3}\right)\)
A Amplitude \(= 3\).
Period \(= \dfrac{2\pi}{2} = \pi\).
Phase shift \(= \dfrac{\pi/3}{2} = \dfrac{\pi}{6}\) (right).
For \(y = -4\cos\!\left(3x + \dfrac{\pi}{2}\right) + 1\), what is the phase shift?
✅ Full Solution
Rewrite: \(3x + \dfrac{\pi}{2} = 3\!\left(x + \dfrac{\pi}{6}\right)\), so \(B=3, C=-\dfrac{\pi}{2}\) (when written as \(Bx-C\): \(3x-(-\tfrac{\pi}{2})\)).
Phase shift \(= \dfrac{C}{B} = \dfrac{-\pi/2}{3} = -\dfrac{\pi}{6}\). Negative \(\Rightarrow\) left.
Phase shift: \(\boxed{\dfrac{\pi}{6} \text{ to the left}}\)
Trap: many students read the "\(+\dfrac{\pi}{2}\)" and say "right"; always factor out \(B\) first!
7
Inverse Trigonometric Functions
Composition of Inverse Trig
Hard
Key Concept: Inverse Trig Domains & Ranges
\(\arcsin\): domain \([-1,1]\), range \(\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]\)
\(\arccos\): domain \([-1,1]\), range \([0,\pi]\)
\(\arctan\): domain \(\mathbb{R}\), range \(\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)\)
For \(\cos\!\left(\arcsin x\right)\): let \(\theta = \arcsin x\), draw a right triangle, find the adjacent side.
Worked Example
Given Evaluate \(\sin\!\left(\arccos\dfrac{1}{3}\right)\).
Step 1 Let \(\theta = \arccos\dfrac{1}{3}\), so \(\cos\theta = \dfrac{1}{3}\) and \(\theta\in[0,\pi]\).
Step 2 \(\sin\theta = \sqrt{1 - \tfrac{1}{9}} = \dfrac{\sqrt{8}}{3} = \dfrac{2\sqrt{2}}{3}\). (Positive in \([0,\pi]\).)
Find the exact value of \(\tan\!\left(\arcsin\dfrac{5}{13}\right)\).
✅ Full Solution
Let \(\theta = \arcsin\dfrac{5}{13}\): opposite \(= 5\), hypotenuse \(= 13\).
Adjacent \(= \sqrt{169 - 25} = \sqrt{144} = 12\).
\(\tan\theta = \dfrac{5}{12}\). Since \(\arcsin\) has range \(\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]\) and \(\dfrac{5}{13}>0\), \(\theta\in\) Q I, so tangent is positive.
\(\tan\!\left(\arcsin\dfrac{5}{13}\right) = \boxed{\dfrac{5}{12}}\)
8
Solving Trig Equations
General Solution on \([0, 2\pi)\)
Hard
Key Concept: Solving \(2\sin^2\theta - \sin\theta - 1 = 0\)
Treat \(\sin\theta = u\): factor the quadratic in \(u\), solve for \(u\), then find all \(\theta\) in \([0,2\pi)\).
If \(\sin\theta = k\), solutions: \(\theta = \arcsin k\) and \(\theta = \pi - \arcsin k\) (plus \(+2\pi n\)).
Never forget both reference-angle solutions!
Worked Example
Given Solve \(2\cos^2\theta - \cos\theta = 0\) on \([0,2\pi)\).
Factor \(\cos\theta(2\cos\theta - 1)=0\).
Case 1 \(\cos\theta = 0 \Rightarrow \theta = \dfrac{\pi}{2}, \dfrac{3\pi}{2}\).
Case 2 \(\cos\theta = \dfrac{1}{2} \Rightarrow \theta = \dfrac{\pi}{3}, \dfrac{5\pi}{3}\).
Solve \(2\sin^2\theta - \sin\theta - 1 = 0\) on \([0, 2\pi)\). How many solutions are there?
✅ Full Solution
Let \(u = \sin\theta\): \(2u^2 - u - 1 = (2u+1)(u-1)=0\).
Case 1: \(\sin\theta = 1 \Rightarrow \theta = \dfrac{\pi}{2}\). (1 solution)
Case 2: \(\sin\theta = -\dfrac{1}{2} \Rightarrow \theta = \dfrac{7\pi}{6}, \dfrac{11\pi}{6}\). (2 solutions in Q III, IV)
Total: \(\boxed{3}\) solutions: \(\dfrac{\pi}{2},\ \dfrac{7\pi}{6},\ \dfrac{11\pi}{6}\).
9
Law of Sines
Ambiguous Case (SSA)
Hard
Key Concept: Law of Sines & Ambiguous Case
\(\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C}\)
For SSA with acute \(A\): let \(h = b\sin A\).
If \(a < h\): no triangle. If \(a = h\): one right triangle. If \(h < a < b\): two triangles. If \(a \geq b\): one triangle.
Worked Example
Given \(A=30°, a=4, b=6\). How many triangles exist?
h \(h = b\sin A = 6\sin 30° = 3\).
Check \(h=3 < a=4 < b=6\): two triangles.
In triangle \(ABC\), \(A = 35°\), \(a = 9\), \(b = 12\). How many distinct triangles are possible?
✅ Full Solution
\(h = b\sin A = 12\sin 35° \approx 12(0.5736) \approx 6.88\).
Since \(h \approx 6.88 < a = 9 < b = 12\), we are in the ambiguous case.
Condition \(h < a < b\) \(\Rightarrow\) \(\boxed{\text{two triangles}}\).
Both \(B_1 \approx 49.9°\) and \(B_2 = 180° - 49.9° = 130.1°\) give valid triangles (each has \(A + B < 180°\)).
10
Law of Cosines
Finding a Missing Side
Medium-Hard
Key Concept: Law of Cosines
\(c^2 = a^2 + b^2 - 2ab\cos C\)
Use when you have SAS or SSS. To find an angle: \(\cos C = \dfrac{a^2+b^2-c^2}{2ab}\).
Worked Example
Given \(a=5, b=7, C=60°\). Find \(c\).
Apply \(c^2 = 25 + 49 - 2(5)(7)\cos 60° = 74 - 70\cdot\tfrac{1}{2} = 74-35 = 39\).
Answer \(c = \sqrt{39}\).
In triangle \(ABC\), \(a = 8\), \(b = 11\), and \(C = 120°\). Find side \(c\).
✅ Full Solution
\(c^2 = 8^2 + 11^2 - 2(8)(11)\cos 120°\)
\(= 64 + 121 - 176\!\left(-\dfrac{1}{2}\right) = 185 + 88 = 273\).
\(c = \boxed{\sqrt{273}}\approx 16.52\).
Key: \(\cos 120° = -\frac{1}{2}\), so the \(-2ab\cos C\) term becomes positive — making \(c\) larger, as expected for an obtuse angle.
11
Verifying Trig Identities
Simplification Strategy
Hard
Key Concept: Identity Simplification Tips
Strategy: convert everything to \(\sin\) and \(\cos\), combine fractions, factor, and use Pythagorean identities.
\(\dfrac{1-\cos^2\theta}{\sin\theta} = \dfrac{\sin^2\theta}{\sin\theta} = \sin\theta\)
Common trick: multiply numerator and denominator by conjugate \((1\pm\cos\theta)\).
Worked Example
Simplify \(\dfrac{\tan\theta}{\sec\theta}\).
Convert \(= \dfrac{\sin\theta/\cos\theta}{1/\cos\theta} = \dfrac{\sin\theta}{\cos\theta}\cdot\cos\theta = \sin\theta\).
Simplify: \(\dfrac{\sin\theta}{1-\cos\theta} - \dfrac{\sin\theta}{1+\cos\theta}\).
✅ Full Solution
Common denominator: \((1-\cos\theta)(1+\cos\theta) = 1 - \cos^2\theta = \sin^2\theta\).
Numerator: \(\sin\theta(1+\cos\theta) - \sin\theta(1-\cos\theta) = \sin\theta[(1+\cos\theta)-(1-\cos\theta)] = 2\sin\theta\cos\theta\).
Result: \(\dfrac{2\sin\theta\cos\theta}{\sin^2\theta} = \dfrac{2\cos\theta}{\sin\theta} = 2\cot\theta\)... let me recheck: \(\dfrac{2\cos\theta}{\sin\theta} = 2\cdot\dfrac{1}{\tan\theta}\). Actually result is \(\dfrac{2\cos\theta}{\sin\theta}\).
Wait — re-examine: numerator \(= \sin\theta(1+\cos\theta) - \sin\theta(1-\cos\theta) = 2\sin\theta\cos\theta\) is wrong. Correct numerator: \(\sin\theta(1+\cos\theta) - \sin\theta(1-\cos\theta) = \sin\theta + \sin\theta\cos\theta - \sin\theta + \sin\theta\cos\theta = 2\sin\theta\cos\theta\)... Denominator \(= \sin^2\theta\). So \(\dfrac{2\sin\theta\cos\theta}{\sin^2\theta} = \dfrac{2\cos\theta}{\sin\theta}\). But answer B says \(2\csc\theta = \dfrac{2}{\sin\theta}\). Let's recheck numerator: \(\sin\theta(1+\cos\theta) - \sin\theta(1-\cos\theta)\). That equals \(\sin\theta + \sin\theta\cos\theta - \sin\theta + \sin\theta\cos\theta = 2\sin\theta\cos\theta\). Hmm. Actually hold on: \(-(sin\theta)(1-\cos\theta) = -\sin\theta + \sin\theta\cos\theta\). So total \(= \sin\theta + \sin\theta\cos\theta - \sin\theta + \sin\theta\cos\theta = 2\sin\theta\cos\theta\). That gives \(2\cot\theta\), not \(2\csc\theta\). The correct answer to this problem is actually \(2\cot\theta\). The answer listed as B should read \(2\cot\theta\). \(\boxed{2\cot\theta}\)
12
Product-to-Sum Formulas
Converting Products to Sums
Hard
Key Concept: Product-to-Sum Identities
\(\sin A\cos B = \tfrac{1}{2}[\sin(A+B)+\sin(A-B)]\)
\(\cos A\cos B = \tfrac{1}{2}[\cos(A-B)+\cos(A+B)]\)
\(\sin A\sin B = \tfrac{1}{2}[\cos(A-B)-\cos(A+B)]\)
Worked Example
Express \(2\sin 3\theta\cos\theta\) as a sum.
Formula \(2\sin A\cos B = \sin(A+B)+\sin(A-B)\).
Result \(= \sin 4\theta + \sin 2\theta\).
Express \(\cos 5\theta \cos 3\theta\) as a sum or difference of trigonometric functions.
✅ Full Solution
Use \(\cos A\cos B = \dfrac{1}{2}[\cos(A-B)+\cos(A+B)]\).
\(A = 5\theta,\; B=3\theta\):
\(\cos 5\theta\cos 3\theta = \dfrac{1}{2}[\cos(5\theta-3\theta)+\cos(5\theta+3\theta)] = \boxed{\dfrac{1}{2}[\cos 2\theta + \cos 8\theta]}\)
13
Radian Measure & Arc Length
Sector Area & Arc
Medium-Hard
Key Concept: Arc Length & Sector Area
Arc length: \(s = r\theta\) (θ in radians)
Sector area: \(A = \dfrac{1}{2}r^2\theta\)
Always convert degrees to radians first!
Worked Example
Given \(r = 6\), central angle \(= 150°\). Find arc length.
Convert \(150° = \dfrac{5\pi}{6}\) rad.
Arc \(s = 6\cdot\dfrac{5\pi}{6} = 5\pi\).
A circle has radius \(9\) cm. A central angle of \(240°\) subtends a sector. What is the area of the sector?
✅ Full Solution
Convert: \(240° = 240\cdot\dfrac{\pi}{180} = \dfrac{4\pi}{3}\) rad.
\(A = \dfrac{1}{2}r^2\theta = \dfrac{1}{2}(9)^2\cdot\dfrac{4\pi}{3} = \dfrac{1}{2}\cdot 81\cdot\dfrac{4\pi}{3} = \dfrac{324\pi}{6} = \boxed{54\pi \text{ cm}^2}\).
14
Harmonic Form
Writing \(a\sin\theta + b\cos\theta\) in \(R\sin(\theta+\phi)\)
Hard
Key Concept: Harmonic (Combined) Form
\(a\sin\theta + b\cos\theta = R\sin(\theta + \phi)\)
where \(R = \sqrt{a^2+b^2}\) and \(\tan\phi = \dfrac{b}{a}\) (with \(a>0\) giving \(\phi\in Q I\)).
Maximum value is \(R\), minimum is \(-R\).
Worked Example
Write \(3\sin\theta + 4\cos\theta\) in harmonic form.
R \(R = \sqrt{9+16} = 5\).
φ \(\tan\phi = \dfrac{4}{3}\), so \(\phi = \arctan\dfrac{4}{3} \approx 53.1°\).
Result \(5\sin(\theta + 53.1°)\). Maximum value: \(5\).
Find the maximum value of \(f(\theta) = 5\sin\theta - 12\cos\theta\).
✅ Full Solution
\(R = \sqrt{5^2 + (-12)^2} = \sqrt{25+144} = \sqrt{169} = 13\).
The maximum value of \(R\sin(\theta+\phi)\) is \(R\).
Maximum value of \(f\) \(= \boxed{13}\).
Common error: adding \(5 + 12 = 17\) instead of using \(\sqrt{a^2+b^2}\).
15
Trig in Right Triangles
Elevation & Depression Applications
Medium-Hard
Key Concept: Angles of Elevation and Depression
Angle of elevation: measured upward from horizontal. \(\tan\alpha = \dfrac{\text{height}}{\text{horizontal distance}}\)
Two observers create two triangles — use both equations simultaneously to eliminate the unknown height or distance.
Worked Example
Given A tower casts a 40 m shadow when the sun's elevation is \(30°\). Height?
Solve \(\tan 30° = \dfrac{h}{40}\), so \(h = 40\tan 30° = \dfrac{40\sqrt{3}}{3} \approx 23.1\) m.
From point \(A\), the angle of elevation of a building's top is \(60°\). From point \(B\), 50 m farther from the base on the same line, the angle of elevation is \(45°\). Find the height of the building.
✅ Full Solution
Let \(h =\) height, \(d =\) horizontal distance from \(A\) to base.
From \(A\): \(\tan 60° = \dfrac{h}{d}\Rightarrow h = d\sqrt{3}\).
From \(B\): \(\tan 45° = \dfrac{h}{d+50}\Rightarrow h = d+50\).
Set equal: \(d\sqrt{3} = d+50 \Rightarrow d(\sqrt{3}-1) = 50 \Rightarrow d = \dfrac{50}{\sqrt{3}-1}\).
\(h = d\sqrt{3} = \dfrac{50\sqrt{3}}{\sqrt{3}-1} \approx \dfrac{86.6}{0.732} \approx 118.3\) m. \(\boxed{\dfrac{50\sqrt{3}}{\sqrt{3}-1}}\)
16
Period & Frequency
Tangent Graph Asymptotes
Hard
Key Concept: Tangent Graph \(y = \tan(Bx+C)\)
Period of \(\tan\): \(\dfrac{\pi}{|B|}\)
Vertical asymptotes where \(\cos(Bx+C)=0\), i.e., \(Bx + C = \dfrac{\pi}{2} + n\pi\).
Unlike sine/cosine, the tangent period is \(\pi\) (not \(2\pi\)).
Worked Example
Find vertical asymptotes of \(y = \tan(2x)\).
Set \(2x = \dfrac{\pi}{2} + n\pi \Rightarrow x = \dfrac{\pi}{4} + \dfrac{n\pi}{2}\).
Period \(= \dfrac{\pi}{2}\).
What are the equations of the vertical asymptotes of \(y = \tan\!\left(2x - \dfrac{\pi}{3}\right)\)?
✅ Full Solution
Set \(2x - \dfrac{\pi}{3} = \dfrac{\pi}{2} + n\pi\).
\(2x = \dfrac{\pi}{2} + \dfrac{\pi}{3} + n\pi = \dfrac{3\pi+2\pi}{6} + n\pi = \dfrac{5\pi}{6} + n\pi\).
\(x = \dfrac{5\pi}{12} + \dfrac{n\pi}{2}\). \(\boxed{x = \dfrac{5\pi}{12} + \dfrac{n\pi}{2}}\)
17
Power-Reduction Formulas
Reducing Powers of Trig Functions
Hard
Key Concept: Power-Reduction Identities
\(\sin^2\theta = \dfrac{1-\cos 2\theta}{2}\)
\(\cos^2\theta = \dfrac{1+\cos 2\theta}{2}\)
\(\sin^2\theta\cos^2\theta = \dfrac{1-\cos 4\theta}{8}\)
Worked Example
Reduce \(\cos^4\theta\).
Step 1 \(\cos^4\theta = \left(\cos^2\theta\right)^2 = \left(\dfrac{1+\cos 2\theta}{2}\right)^2 = \dfrac{1+2\cos 2\theta+\cos^2 2\theta}{4}\).
Step 2 \(\cos^2 2\theta = \dfrac{1+\cos 4\theta}{2}\), so \(\cos^4\theta = \dfrac{3+4\cos 2\theta+\cos 4\theta}{8}\).
Rewrite \(\sin^4\theta\) in terms of cosines of multiple angles (no powers).
✅ Full Solution
\(\sin^4\theta = \left(\sin^2\theta\right)^2 = \left(\dfrac{1-\cos 2\theta}{2}\right)^2 = \dfrac{1-2\cos 2\theta+\cos^2 2\theta}{4}\).
Replace \(\cos^2 2\theta = \dfrac{1+\cos 4\theta}{2}\):
\(\sin^4\theta = \dfrac{1-2\cos 2\theta + \frac{1+\cos 4\theta}{2}}{4} = \dfrac{\frac{2-4\cos 2\theta+1+\cos 4\theta}{2}}{4} = \boxed{\dfrac{3-4\cos 2\theta+\cos 4\theta}{8}}\)
18
Cofunction Identities
Complementary Angle Relations
Medium-Hard
Key Concept: Cofunction & Even/Odd Identities
\(\sin\!\left(\dfrac{\pi}{2}-\theta\right)=\cos\theta,\quad \cos\!\left(\dfrac{\pi}{2}-\theta\right)=\sin\theta\)
\(\sin(-\theta)=-\sin\theta \text{ (odd)},\quad \cos(-\theta)=\cos\theta \text{ (even)}\)
\(\tan(-\theta) = -\tan\theta \text{ (odd)}\)
Worked Example
Simplify \(\sin\!\left(\dfrac{\pi}{2}+\theta\right)\).
Use sum formula: \(\sin\dfrac{\pi}{2}\cos\theta + \cos\dfrac{\pi}{2}\sin\theta = \cos\theta + 0 = \cos\theta\).
Simplify: \(\dfrac{\cos(-\theta)\cdot\sin\!\left(\dfrac{\pi}{2}-\theta\right)}{\tan(-\theta)}\).
✅ Full Solution
Apply identities: \(\cos(-\theta)=\cos\theta\), \(\sin(\pi/2-\theta)=\cos\theta\), \(\tan(-\theta)=-\tan\theta\).
Expression \(= \dfrac{\cos\theta\cdot\cos\theta}{-\tan\theta} = \dfrac{\cos^2\theta}{-\sin\theta/\cos\theta} = \dfrac{\cos^2\theta\cdot\cos\theta}{-\sin\theta} = -\dfrac{\cos^3\theta}{\sin\theta}\).
This simplifies to \(-\cos^2\theta\cdot\cot\theta\). \(\boxed{-\cos^2\theta\cot\theta}\)
19
Modeling with Sinusoids
Writing Equation from Graph Data
Hard
Key Concept: Finding Equation from Max/Min
\(A = \dfrac{\text{max}-\text{min}}{2},\quad D = \dfrac{\text{max}+\text{min}}{2}\)
\(\text{Period} = 2(\text{time from min to max})\), then \(B = \dfrac{2\pi}{\text{period}}\)
Phase shift: time of first maximum gives the horizontal shift.
Worked Example
Given Max \(= 10\) at \(t=1\), min \(= 2\). Period \(= 8\).
A \(= (10-2)/2 = 4\), \(D = 6\), \(B = 2\pi/8 = \pi/4\).
Equation \(y = 4\cos\!\left(\dfrac{\pi}{4}(t-1)\right)+6\).
A sinusoidal function has a maximum value of \(7\) at \(x = 1\), a minimum value of \(-3\), and a period of \(12\). Write its equation in the form \(y = A\cos\!\left(B(x-h)\right)+D\).
✅ Full Solution
\(A = \dfrac{7-(-3)}{2} = 5\), \(D = \dfrac{7+(-3)}{2} = 2\).
\(B = \dfrac{2\pi}{12} = \dfrac{\pi}{6}\).
Max at \(x=1\): phase shift \(h = 1\).
\(y = \boxed{5\cos\!\left(\dfrac{\pi}{6}(x-1)\right)+2}\)
20
Advanced Identity Proof
Complex Simplification
Hard
Key Concept: Sum-to-Product Formulas
\(\sin A + \sin B = 2\sin\!\left(\dfrac{A+B}{2}\right)\cos\!\left(\dfrac{A-B}{2}\right)\)
\(\cos A - \cos B = -2\sin\!\left(\dfrac{A+B}{2}\right)\sin\!\left(\dfrac{A-B}{2}\right)\)
Useful for solving equations and verifying identities with sums of trig functions.
Worked Example
Simplify \(\dfrac{\sin 5\theta + \sin 3\theta}{\cos 5\theta + \cos 3\theta}\).
Num \(= 2\sin 4\theta\cos\theta\).
Den \(\cos A + \cos B = 2\cos\!\left(\frac{A+B}{2}\right)\cos\!\left(\frac{A-B}{2}\right) = 2\cos 4\theta\cos\theta\).
Result \(= \tan 4\theta\).
Simplify: \(\dfrac{\sin 9\theta - \sin 5\theta}{\cos 9\theta - \cos 5\theta}\).
✅ Full Solution
Numerator: \(\sin 9\theta - \sin 5\theta = 2\cos\!\left(\dfrac{9\theta+5\theta}{2}\right)\sin\!\left(\dfrac{9\theta-5\theta}{2}\right) = 2\cos 7\theta\sin 2\theta\).
Denominator: \(\cos 9\theta - \cos 5\theta = -2\sin\!\left(\dfrac{9\theta+5\theta}{2}\right)\sin\!\left(\dfrac{9\theta-5\theta}{2}\right) = -2\sin 7\theta\sin 2\theta\).
Ratio: \(\dfrac{2\cos 7\theta\sin 2\theta}{-2\sin 7\theta\sin 2\theta} = \dfrac{\cos 7\theta}{-\sin 7\theta} = \boxed{-\cot 7\theta}\).
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