In Exercises 17–30, determine the amplitude, period, and phase shift of each function. Then graph one period of the function. y = −2 sin(2x + π/2)
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- 0. Review of College Algebra4h 45m
- 1. Measuring Angles40m
- 2. Trigonometric Functions on Right Triangles2h 5m
- 3. Unit Circle1h 19m
- 4. Graphing Trigonometric Functions1h 19m
- 5. Inverse Trigonometric Functions and Basic Trigonometric Equations1h 41m
- 6. Trigonometric Identities and More Equations2h 34m
- 7. Non-Right Triangles1h 38m
- 8. Vectors2h 25m
- 9. Polar Equations2h 5m
- 10. Parametric Equations1h 6m
- 11. Graphing Complex Numbers1h 7m
4. Graphing Trigonometric Functions
Graphs of the Sine and Cosine Functions
Multiple Choice
Determine the value of y=sin(−2π)+50 without using a calculator or the unit circle.

A
y=50
B
y=51
C
y=49
D
y=50+23
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Verified step by step guidance1
Understand the problem: We need to find the value of y = \(\sin\[\left\)(-\(\frac{\pi}{2}\]\right\)) + 50 without using a calculator or the unit circle.
Recall the property of the sine function: \(\sin\)(-x) = -\(\sin\)(x). This means \(\sin\[\left\)(-\(\frac{\pi}{2}\]\right\)) = -\(\sin\[\left\)(\(\frac{\pi}{2}\]\right\)).
Know the value of \(\sin\[\left\)(\(\frac{\pi}{2}\]\right\)): The sine of \(\frac{\pi}{2}\) is 1, so \(\sin\[\left\)(-\(\frac{\pi}{2}\]\right\)) = -1.
Substitute the value back into the equation: y = -1 + 50.
Simplify the expression: y = 49.
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