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To compare \( \cos(A - B) \), \( \cos A - \cos B \), and \( \cos A \cdot \cos B + \sin A \cdot \sin B \):
1. **Cosine Subtraction Formula**:
\[
\cos(A - B) = \cos A \cdot \cos B + \sin A \cdot \sin B
\]
So, \( \cos(A - B) \) is equal to \( \cos A \cdot \cos B + \sin A \cdot \sin B \).
2. **Comparison with \( \cos A - \cos B \)**:
- \( \cos A - \cos B \) does not equal \( \cos(A - B) \) in general.
- It can be negative or positive depending on \( A \) and \( B \).
**General Conclusion**:
- \( \cos(A - B) \) is always equal to \( \cos A \cdot \cos B + \sin A \cdot \sin B \).
- \( \cos A - \cos B \) does not have a consistent relationship with \( \cos(A - B) \) or \( \cos A \cdot \cos B + \sin A \cdot \sin B \).
**Investigating Cosine Double Angle**:
- The double angle formula for cosine is:
\[
\cos(2A) = \cos^2 A - \sin^2 A = 2\cos^2 A - 1 = 1 - 2\sin^2 A
\]
- Use a calculator in TABLE mode to generate and plot values of \( \cos(2A) \) for various \( A \) to visualize its behavior.
Simplify answer:
\( \cos(A - B) \) is always equal to \( \cos A \cdot \cos B + \sin A \cdot \sin B \). The expression \( \cos A - \cos B \) does not consistently relate to \( \cos(A - B) \) or \( \cos A \cdot \cos B + \sin A \cdot \sin B \). For the cosine double angle, \( \cos(2A) = \cos^2 A - \sin^2 A = 2\cos^2 A - 1 = 1 - 2\sin^2 A \). Use a calculator in TABLE mode to plot these values and observe the behavior of the cosine function with double angles.
Solución
To compare the values of \( \cos(A - B) \), \( \cos A - \cos B \), and \( \cos A \cdot \cos B + \sin A \cdot \sin B \), we can start by using the cosine subtraction formula and some algebraic manipulations.
### Step 1: Use the Cosine Subtraction Formula
The cosine subtraction formula states:
\[
\cos(A - B) = \cos A \cdot \cos B + \sin A \cdot \sin B
\]
### Step 2: Compare \( \cos(A - B) \) with \( \cos A - \cos B \)
Now, we need to compare \( \cos(A - B) \) with \( \cos A - \cos B \).
1. From the cosine subtraction formula, we have:
\[
\cos(A - B) = \cos A \cdot \cos B + \sin A \cdot \sin B
\]
2. The expression \( \cos A - \cos B \) can be rewritten using the identity:
\[
\cos A - \cos B = -2 \sin\left(\frac{A + B}{2}\right) \sin\left(\frac{A - B}{2}\right)
\]
### Step 3: Analyze the Comparison
Now we can analyze the relationship between these expressions:
- **Equality**: From the cosine subtraction formula, we see that:
\[
\cos(A - B) = \cos A \cdot \cos B + \sin A \cdot \sin B
\]
This shows that \( \cos(A - B) \) is equal to \( \cos A \cdot \cos B + \sin A \cdot \sin B \).
- **Difference**: The expression \( \cos A - \cos B \) is not equal to \( \cos(A - B) \) in general. Instead, it can be negative or positive depending on the values of \( A \) and \( B \).
### General Conclusion
From the above analysis, we can conclude that:
- \( \cos(A - B) \) is always equal to \( \cos A \cdot \cos B + \sin A \cdot \sin B \).
- \( \cos A - \cos B \) can vary and does not have a consistent relationship with \( \cos(A - B) \) or \( \cos A \cdot \cos B + \sin A \cdot \sin B \).
### Question 2: Investigating the Cosine Double Angle
For the cosine double angle, the formula is:
\[
\cos(2A) = \cos^2 A - \sin^2 A
\]
or equivalently,
\[
\cos(2A) = 2\cos^2 A - 1
\]
or
\[
\cos(2A) = 1 - 2\sin^2 A
\]
To investigate this, you can use a calculator in TABLE mode to generate values for \( A \) and \( \cos(2A) \) based on the above formulas.
### Steps to Use the Calculator:
1. Set your calculator to TABLE mode.
2. Input values for \( A \) (e.g., from 0 to \( 360^\circ \) or \( 0 \) to \( 2\pi \) radians).
3. Calculate \( \cos(2A) \) using the double angle formulas.
4. Sketch the graphs of \( \cos(2A) \) against \( A \) on the provided grid.
This will help visualize how the cosine function behaves with respect to the double angle.
Revisado y aprobado por el equipo de tutoría de UpStudy
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