Conveyor Bracket Angles and a Controller Domain Error
A fabrication unit in Ludhiana produces wedge brackets for a conveyor frame. The drawing specifies that a particular bracket must sit at an obtuse angle θ in the second quadrant, with sin θ = 0.5. A technician keys sin⁻¹(0.5) into a scientific calculator and gets 30°, then sets the cutting jig to 30°. The finished bracket does not fit: the assembly needs the supplementary angle. The supervisor also flags a second job where a controller computes f(x) = sin⁻¹(2x − 1) and stops with a 'domain error' whenever the sensor reading x is 1.4. The controller manual says the input to sin⁻¹ must lie in [−1, 1], but the operator has been bypassing the error by ignoring it. Before the next batch, the unit wants a short written check of both jobs, including the correct angle, the allowed sensor range for f(x), and a warning note for the calculator step.
1. Find the correct angle θ in [0, 2π) for the bracket. Explain why the calculator's 30° is a correct value of sin⁻¹(0.5) yet not the angle the drawing requires.
Read the worked answer
sin 30° = 0.5 and sin 150° = 0.5, so the equation sin θ = 0.5 has two solutions in [0, 2π): θ = 30° and θ = 150° (in general, θ = 30° + 360°n or θ = 150° + 360°n). The calculator returns the principal value, which by definition lies in [−90°, 90°]; that is why it shows 30°. The drawing says the bracket is in the second quadrant, where sine is positive and cosine is negative. Of the two candidates, 150° is in the second quadrant, while 30° is in the first. Hence the jig should be set to 150° = 5π/6 rad. The calculator answer is not wrong as a principal value; it is the wrong branch for this physical part.
2. Determine the largest set of sensor readings x for which f(x)=sin⁻¹(2x−1) is defined, and state the corresponding output range.
Read the worked answer
The inverse sine accepts exactly arguments in [−1, 1]. Therefore −1 ≤ 2x−1 ≤ 1. Add 1: 0 ≤ 2x ≤ 2. Divide by 2: 0 ≤ x ≤ 1. So the largest allowed sensor range is the closed interval [0, 1]. As x runs over [0, 1], the argument 2x−1 runs over [−1, 1]. The principal inverse sine then runs over [−π/2, π/2]. Both endpoints are attained, at x=0 and x=1, so the output range is also [−π/2, π/2]. A reading such as x=1.4 gives 2(1.4)−1=1.8, which is outside [−1, 1]; the controller is right to stop.
3. The operator proposes to 'fix' the domain error by replacing 2x−1 with |2x−1|. Test this proposal at x=1.4 and explain what it does to the angle information even when it does produce a value.
Read the worked answer
At x=1.4, 2x−1=1.8, so |2x−1|=1.8, which is still greater than 1. The replacement does not repair this error at all: sin⁻¹(1.8) is undefined. More generally, absolute value only folds negative arguments to positive ones; it cannot bring arguments outside [−1, 1] back inside. Even where it does produce a value, for example x=0.25 gives 2x−1=−0.5 and |2x−1|=0.5, the outputs are sin⁻¹(−0.5)=−π/6 and sin⁻¹(0.5)=π/6. The sign change carries real information about which side of the centreline the mechanism lies, so deleting it would make two different physical positions report the same angle. The correct fix is to clamp or reject x outside [0, 1], not to take absolute value.