Show that using the other trig identities found.

Certainly! Let's revisit the solution with a focus on using the trigonometric identities to break down the vector components and calculate the displacement and average velocity.

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### Step 1: Breaking the motion into vector components

We already have the information for the three segments of motion:

- **Segment 1:** 1.29 km due east → This gives the displacement vector:
  - \( x_1 = 1.29 \, \text{km} \)
  - \( y_1 = 0 \, \text{km} \)

- **Segment 2:** 0.275 km due south → This gives the displacement vector:
  - \( x_2 = 0 \, \text{km} \)
  - \( y_2 = -0.275 \, \text{km} \)

- **Segment 3:** 3.32 km at 64.3° north of west

  We need to break this displacement into \( x \) and \( y \) components using trigonometric identities:

- The angle is 64.3° north of west. Since this is north of west, the x-component will be negative (westward), and the y-component will be positive (northward).

Using the trig identities:

- \( \cos(64.3^\circ) = \cos(90^\circ - 64.3^\circ) = \sin(64.3^\circ) \)
- \( \sin(64.3^\circ) = \sin(90^\circ - 64.3^\circ) = \cos(64.3^\circ) \)

The displacement vector for segment 3 can then be calculated as:

\[
x_3 = 3.32 \times \cos(64.3^\circ)
\]
\[
y_3 = 3.32 \times \sin(64.3^\circ)
\]

Let’s calculate the values using the known values of \( \cos(64.3^\circ) \approx 0.438 \) and \( \sin(64.3^\circ) \approx 0.898 \):

\[
x_3 = 3.32 \times (-0.438) \approx -1.45 \, \text{km}
\]
\[
y_3 = 3.32 \times 0.898 \approx 2.98 \, \text{km}
\]

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### Step 2

Show that using the other trig identities found.

Learn how to use trigonometric identities to break down vectors into their components and calculate displacement and velocity. This problem involves decomposing a vector at a specific angle using the identities cos(90° - θ) = sin(θ) and sin(90° - θ) = cos(θ). Suitable for high school students studying trigonometry and vector analysis.

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