4. For any angle 0, Therefore: sin (0) = cos(0 - π/2) cos(0) = sin(0 + π/2) sin(0) has a phase shift of -π/2 relative to cos(0). • cos() has a phase shift of π/2 relative to sin(0). This is not a problem, but you should try out a few values on your calcu- lator to verify this.
4. For any angle 0, Therefore: sin (0) = cos(0 - π/2) cos(0) = sin(0 + π/2) sin(0) has a phase shift of -π/2 relative to cos(0). • cos() has a phase shift of π/2 relative to sin(0). This is not a problem, but you should try out a few values on your calcu- lator to verify this.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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![### Understanding Phase Shifts in Trigonometric Functions
#### 4. For any angle θ,
\[ \sin(\theta) = \cos(\theta - \pi/2) \]
\[ \cos(\theta) = \sin(\theta + \pi/2) \]
Therefore:
- \(\sin(\theta)\) has a phase shift of \(-\pi/2\) relative to \(\cos(\theta)\).
- \(\cos(\theta)\) has a phase shift of \(\pi/2\) relative to \(\sin(\theta)\).
This is not a problem, but you should try out a few values on your calculator to verify this.
### Explanation
These equations indicate that the sine and cosine functions are phase-shifted versions of each other. Specifically:
- Shifting the cosine function to the right by \(\pi/2\) results in the sine function.
- Shifting the sine function to the left by \(\pi/2\) (equivalently, shifting to the right by \(-\pi/2\)) results in the cosine function.
This phase shift relationship is fundamental in understanding the periodic nature of these trigonometric functions. Try calculating a few values for \(\theta\) using a scientific calculator to see this phase shift in action:
For example, when \(\theta = 0\):
- \(\sin(0) = \cos(0 - \pi/2) = \cos(-\pi/2) = 0\)
- \(\cos(0) = \sin(0 + \pi/2) = \sin(\pi/2) = 1\)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F701927fd-ceaf-4fad-8338-7d1e3c22653c%2Fe796fdd0-04a6-4e0e-aab8-8941a2b3437d%2Fwwh1as8_processed.png&w=3840&q=75)
Transcribed Image Text:### Understanding Phase Shifts in Trigonometric Functions
#### 4. For any angle θ,
\[ \sin(\theta) = \cos(\theta - \pi/2) \]
\[ \cos(\theta) = \sin(\theta + \pi/2) \]
Therefore:
- \(\sin(\theta)\) has a phase shift of \(-\pi/2\) relative to \(\cos(\theta)\).
- \(\cos(\theta)\) has a phase shift of \(\pi/2\) relative to \(\sin(\theta)\).
This is not a problem, but you should try out a few values on your calculator to verify this.
### Explanation
These equations indicate that the sine and cosine functions are phase-shifted versions of each other. Specifically:
- Shifting the cosine function to the right by \(\pi/2\) results in the sine function.
- Shifting the sine function to the left by \(\pi/2\) (equivalently, shifting to the right by \(-\pi/2\)) results in the cosine function.
This phase shift relationship is fundamental in understanding the periodic nature of these trigonometric functions. Try calculating a few values for \(\theta\) using a scientific calculator to see this phase shift in action:
For example, when \(\theta = 0\):
- \(\sin(0) = \cos(0 - \pi/2) = \cos(-\pi/2) = 0\)
- \(\cos(0) = \sin(0 + \pi/2) = \sin(\pi/2) = 1\)
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