Robot Arm Alignment with a Squared Equation
A robotic arm is being calibrated on a vertical board. Its end effector is programmed to move on the unit circle, so its position after a rotation through angle θ is P(cos θ, sin θ), with 0 ≤ θ < 2π. A laser guide is set along the line x + y = 1, and the controller must lock the arm only when P lies exactly on that line. A junior engineer squares both sides of the alignment condition and reports four locking angles: 0, π/2, π and 3π/2. The senior engineer is not convinced, because the arm cannot pass through the same laser line at all four positions without a sign error. Before the next test run, the team must determine the genuine locking angles in the stated interval, explain why squaring changed the solution set, and decide how many of the junior engineer’s four settings should be accepted by the controller. The arm has no clearance problem at any of these angles.
1. Solve cos θ + sin θ = 1 for 0 ≤ θ < 2π without squaring, and list the genuine locking angles.
Read the worked answer
Use the R-form: cos θ + sin θ = √2 cos(θ − π/4). The condition becomes √2 cos(θ − π/4) = 1, so cos(θ − π/4) = 1/√2. Hence θ − π/4 = ±π/4 + 2nπ. Therefore θ = 0 + 2nπ or θ = π/2 + 2nπ. In 0 ≤ θ < 2π the genuine angles are θ = 0 and θ = π/2. Check: at 0, 1 + 0 = 1; at π/2, 0 + 1 = 1.
2. The junior squares both sides. Show the resulting equation and explain exactly which extra angles appear and why they are not valid.
Read the worked answer
Squaring gives (cos θ + sin θ)² = 1², so cos²θ + 2 sinθ cosθ + sin²θ = 1. Since cos²θ + sin²θ = 1, this reduces to 1 + sin 2θ = 1, i.e. sin 2θ = 0. Thus 2θ = nπ, so θ = 0, π/2, π, 3π/2 in the interval. The step is not reversible: the squared equation is satisfied by both cosθ + sinθ = 1 and cosθ + sinθ = −1. At θ = π, cosπ + sinπ = −1; at θ = 3π/2, 0 − 1 = −1. These two are extraneous, so only two of the four reported settings are genuine.
3. If the controller also requires the point P to satisfy y ≥ x, which of the genuine locking angles remains?
Read the worked answer
For θ = 0, P = (1, 0) and y = 0 < x = 1, so it fails y ≥ x. For θ = π/2, P = (0, 1) and y = 1 > x = 0, so it passes. Hence the controller should keep only θ = π/2.