SOLUTION: Point A = (cosθ,sinθ) is at the intersection of x2 +y2 = 1 and a ray starting at the origin that makes an angle, θ, with the positive x-axis. The ray starting at the origin thro

Question 1163520: Point A = (cosθ,sinθ) is at the intersection of x2 +y2 = 1 and a ray starting at the origin that makes an angle, θ, with the positive x-axis. The ray starting at the origin through point P makes an angle of 2θ with the positive x-axis.
(a) Explain why P = (cos 2θ, sin 2θ).
(b) Reflect B = (1,0) over the line CA to get an equivalent
form of the coordinates of P written in terms of cos θ and sin θ.
Found 2 solutions by CPhill, ikleyn:
Answer by CPhill(2264) (Show Source): You can put this solution on YOUR website!
This problem explores the relationship between trigonometric double-angle identities and geometric transformations.
### (a) Explaining why $P = (\cos 2\theta, \sin 2\theta)$
Any point $P$ in the Cartesian plane can be described by its distance from the origin ($r$) and the angle it makes with the positive x-axis ($\alpha$). The coordinates are given by:
$$P = (r \cos \alpha, r \sin \alpha)$$
For point $P$:
1. **Angle:** The problem states the ray through $P$ makes an angle of **$2\theta$** with the positive x-axis. Thus, $\alpha = 2\theta$.
2. **Distance:** While the prompt doesn't explicitly state $P$ is on the unit circle, in the context of deriving trigonometric identities through geometry, $P$ is understood to be the point on the circle $x^2 + y^2 = 1$. Therefore, $r = 1$.
Substituting these into the coordinate formula gives:
**$P = (\cos 2\theta, \sin 2\theta)$**
---
### (b) Reflecting $B=(1,0)$ over the line through the Origin and $A$
To find the reflection of $B(1,0)$ over the line $OA$ (where $A = (\cos\theta, \sin\theta)$), we observe the geometric symmetry.
**1. Geometric Reasoning:**
* The line $OA$ makes an angle of $\theta$ with the positive x-axis.
* Point $B(1,0)$ is at an angle of $0^\circ$ relative to the x-axis.
* The angular distance from $B$ to the line of reflection is $\theta - 0 = \theta$.
* By the definition of reflection, the image $P$ must be the same angular distance on the *other side* of the line.
* Therefore, the angle of $P$ is $\theta + \theta = 2\theta$.
**2. Coordinate Calculation via Projection:**
To reflect a vector $\mathbf{v}$ across a line defined by unit vector $\mathbf{u} = (\cos\theta, \sin\theta)$, we use the formula:
$$\text{Ref}_{\mathbf{u}}(\mathbf{v}) = 2(\mathbf{v} \cdot \mathbf{u})\mathbf{u} - \mathbf{v}$$
Here, $\mathbf{v} = (1, 0)$ and $\mathbf{u} = (\cos\theta, \sin\theta)$.
* **Step 1 (Dot Product):** $\mathbf{v} \cdot \mathbf{u} = (1)(\cos\theta) + (0)(\sin\theta) = \cos\theta$.
* **Step 2 (Scalar Multiplication):** $2(\mathbf{v} \cdot \mathbf{u})\mathbf{u} = 2\cos\theta(\cos\theta, \sin\theta) = (2\cos^2\theta, 2\sin\theta\cos\theta)$.
* **Step 3 (Subtraction):** Subtract $\mathbf{v} = (1, 0)$:
$$P = (2\cos^2\theta - 1, 2\sin\theta\cos\theta)$$
### Summary of Identities
By comparing the results of part (a) and part (b), we have derived the double-angle identities:
* $\cos 2\theta = 2\cos^2\theta - 1$
* $\sin 2\theta = 2\sin\theta\cos\theta$
Does the reflection approach make the relationship between these coordinates clearer than just looking at the formulas?
Answer by ikleyn(53862) (Show Source): You can put this solution on YOUR website!
.
Point A = (cosθ,sinθ) is at the intersection of x2 +y2 = 1 and a ray starting at the origin that makes an angle, θ,
with the positive x-axis. The ray starting at the origin through point P makes an angle of 2θ with the positive x-axis.
(a) Explain why P = (cos 2θ, sin 2θ).
(b) Reflect B = (1,0) over the line CA to get an equivalent
form of the coordinates of P written in terms of cos θ and sin θ.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

The author of this post mentioned/introduced point 'C' (by saying about line CA),
but forgot to define this point (and this line).

As a result, the posted problem is fatally defective and is only good to throw it to a garbage bin.

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