Complete collection of examples for graphs, identities, and formulas
← Back to CourseProblem: Find amplitude, period, phase shift, and vertical shift.
Solution:
General form: y = A sin(B(x - h)) + k
Rewrite: y = 3sin(2(x - π/8)) + 1
Problem: A cosine function has max = 5, min = 1, period = 4π. Write the equation.
Solution:
Amplitude = (max - min)/2 = (5 - 1)/2 = 2
Vertical shift = (max + min)/2 = (5 + 1)/2 = 3
B = 2π/period = 2π/(4π) = 1/2
$$y = 2\cos\left(\frac{1}{2}x\right) + 3$$
Problem: Find the asymptotes of y = tan(3x).
Solution:
Tangent has asymptotes where cos = 0, i.e., at 3x = π/2 + nπ
$$x = \frac{\pi}{6} + \frac{n\pi}{3}, \quad n \in \mathbb{Z}$$
First few asymptotes: x = π/6, π/2, 5π/6, 7π/6, ...
Problem: A Ferris wheel has diameter 40m, center 25m high, and rotates every 2 minutes. Model height vs time.
Solution:
Amplitude = radius = 20m
Vertical shift = center height = 25m
Period = 2 min, so B = 2π/2 = π
$$h(t) = -20\cos(\pi t) + 25$$
(Negative cosine because rider starts at bottom)
Problem: Find the exact value of sin(75°).
Solution:
$$\sin 75° = \sin(45° + 30°) = \sin 45°\cos 30° + \cos 45°\sin 30°$$
$$= \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} + \frac{\sqrt{2}}{2} \cdot \frac{1}{2}$$
$$= \frac{\sqrt{6}}{4} + \frac{\sqrt{2}}{4} = \frac{\sqrt{6} + \sqrt{2}}{4}$$
Answer: (√6 + √2)/4 ≈ 0.9659
Problem: Find the exact value of cos(15°).
Solution:
$$\cos 15° = \cos(45° - 30°) = \cos 45°\cos 30° + \sin 45°\sin 30°$$
$$= \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} + \frac{\sqrt{2}}{2} \cdot \frac{1}{2}$$
$$= \frac{\sqrt{6} + \sqrt{2}}{4}$$
Answer: (√6 + √2)/4
Problem: If tan(A) = 2 and tan(B) = 3, find tan(A + B).
Solution:
$$\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} = \frac{2 + 3}{1 - 2 \times 3} = \frac{5}{1 - 6} = \frac{5}{-5} = -1$$
Answer: tan(A + B) = -1
Problem: Simplify: sin(x + π/6) + sin(x - π/6)
Solution:
Expand each term:
sin(x + π/6) = sin(x)cos(π/6) + cos(x)sin(π/6) = (√3/2)sin(x) + (1/2)cos(x)
sin(x - π/6) = sin(x)cos(π/6) - cos(x)sin(π/6) = (√3/2)sin(x) - (1/2)cos(x)
Adding: (√3/2)sin(x) + (1/2)cos(x) + (√3/2)sin(x) - (1/2)cos(x) = √3 sin(x)
Problem: If sin(θ) = 3/5 and θ is in Q1, find sin(2θ).
Solution:
First find cos(θ): cos²θ = 1 - sin²θ = 1 - 9/25 = 16/25, so cos(θ) = 4/5 (positive in Q1)
$$\sin(2\theta) = 2\sin\theta\cos\theta = 2 \times \frac{3}{5} \times \frac{4}{5} = \frac{24}{25}$$
Answer: sin(2θ) = 24/25
Problem: If cos(θ) = -1/3 and θ is in Q2, find cos(2θ).
Solution:
Using cos(2θ) = 2cos²θ - 1:
$$\cos(2\theta) = 2\left(-\frac{1}{3}\right)^2 - 1 = 2 \times \frac{1}{9} - 1 = \frac{2}{9} - 1 = -\frac{7}{9}$$
Answer: cos(2θ) = -7/9
Problem: Solve sin(2x) = cos(x) for 0 ≤ x < 2π.
Solution:
Replace sin(2x) = 2sin(x)cos(x):
2sin(x)cos(x) = cos(x)
cos(x)(2sin(x) - 1) = 0
Either cos(x) = 0 or sin(x) = 1/2
cos(x) = 0: x = π/2, 3π/2
sin(x) = 1/2: x = π/6, 5π/6
Answer: x = π/6, π/2, 5π/6, 3π/2
Problem: Find sin(150°).
Solution:
$$\sin 150° = \sin(180° - 30°) = \sin 30° = \frac{1}{2}$$
Answer: sin(150°) = 1/2
Problem: Find cos(210°).
Solution:
$$\cos 210° = \cos(180° + 30°) = -\cos 30° = -\frac{\sqrt{3}}{2}$$
Answer: cos(210°) = -√3/2
Problem: Find tan(315°).
Solution:
315° = 360° - 45°, so it's in Q4 where tan is negative.
$$\tan 315° = -\tan 45° = -1$$
Or: tan(315°) = tan(-45°) = -tan(45°) = -1
Answer: tan(315°) = -1
Problem: Simplify: sin(π/2 - x) + cos(π - x) + sin(3π/2 + x)
Solution:
sin(π/2 - x) = cos(x) (cofunction identity)
cos(π - x) = -cos(x)
sin(3π/2 + x) = -cos(x)
Sum = cos(x) + (-cos(x)) + (-cos(x)) = -cos(x)