Transformations of Trigonometric Functions (B) Worksheet | Algebra ... - Free Printable

Looking at the worksheet, I can see that questions 1-3 already have graphs drawn and we need to find their periods. Questions 4-9 are blank and need both graphing and period calculation.

Since the user wants me to solve the problem accurately and the options for explaining or teaching are false, I'll focus on providing correct answers for all 9 problems.

Let me work through each one:

Problem 1: y = sin 2x
- Standard sine function has period 360°
- When there's a coefficient inside (like 2x), period = 360° / |coefficient|
- Period = 360° / 2 = 180°
- The graph shows one complete cycle from 0° to 180°, confirming this

Problem 2: y = tan 2x
- Standard tangent function has period 180°
- Period = 180° / |coefficient| = 180° / 2 = 90°
- The graph shows asymptotes every 90°, confirming this

Problem 3: y = cos 2x
- Standard cosine function has period 360°
- Period = 360° / |coefficient| = 360° / 2 = 180°
- The graph shows one complete cycle from 0° to 180°, confirming this

Problem 4: y = 2 sin 2
- This appears to be a typo - it should likely be y = 2 sin 2x
- Assuming y = 2 sin 2x: amplitude is 2, period = 360°/2 = 180°
- If literally y = 2 sin 2, this would be a constant value (2 × sin(2 radians) ≈ 2 × 0.909 = 1.818)
- Given context, I'll assume it's y = 2 sin 2x with period 180°

Problem 5: y = 2 cos x
- Amplitude is 2 (vertical stretch)
- No coefficient inside, so period remains 360°
- Graph should show cosine wave with peaks at ±2 instead of ±1

Problem 6: y = ½ sin x
- Amplitude is ½ (vertical compression)
- No coefficient inside, so period remains 360°
- Graph should show sine wave with peaks at ±0.5 instead of ±1

Problem 7: y = tan(x - 30°)
- Phase shift of 30° to the right
- No coefficient inside, so period remains 180°
- Asymptotes will occur where x - 30° = 90° + 180°n, so x = 120° + 180°n

Problem 8: y = cos(x + 60°)
- Phase shift of 60° to the left
- No coefficient inside, so period remains 360°
- Maximum occurs when x + 60° = 0°, so x = -60° (or 300°)

Problem 9: y = 2 sin ½x
- Amplitude is 2
- Coefficient inside is ½, so period = 360° / ½ = 720°
- The graph shows x-axis from -180° to 0°, which makes sense since the full period is 720°

Final Answer:
1) 180°
2) 90°
3) 180°
4) 180° (assuming y = 2 sin 2x)
5) 360°
6) 360°
7) 180°
8) 360°
9) 720°