A hypothetical sensor repeats every 6 hours, centered on 10 with deviations up to 2 units. A sinusoid separates midline, amplitude, period and time shift. Use graph transformations and the unit circle.
y(t) = 10 + 2sin((π)/(3)(t − 1)); T = 6, 8 ≤ y ≤ 12.
Time is in hours, readings in abstract units. At t=1 the value is 10 and rising; the maximum 12 occurs at t=2.5. Six hours later the pattern repeats.
y = d + Asin(ω(t − h)), ∣A∣ > 0, ω > 0; T = (2π)/(ω).
Amplitude |A| is maximum deviation from d; peak-to-peak range is 2|A|. h shifts right; ω is angular frequency in radians/hour, making the argument dimensionless. A=0 gives a constant with no least positive period. OpenStax
Signal parameters
y=10+2sin(π(t−1)/3); -- y=10+sin(πt)
Sine and cosine
sin x; -- cos x
Tangent and cotangent
tan x; -- cot x
One circle, different graphs
sin and cos have domain ℝ, range [−1,1], and least positive period 2π. Their shift explains why one harmonic may be written using either.
cosx = sin(x + π/2); tanx = (sinx)/(cosx); cotx = (cosx)/(sinx).
Tangent and cotangent have period π but no finite amplitude: their values are unbounded. Exclude π/2+kπ for tan and kπ for cot, integer k. Never join branches across these vertical asymptotes. On each domain interval tan increases, cot decreases.
An ideal harmonic is a special periodic pattern. A square pulse repeats too but is not one sinusoid. Our sensor is a stipulated model, not measured data; real signal shape and parameters need separate verification.
Optional: the period of a sum
If T is a period of both f and g, then (f+g)(x+T)=f(x+T)+g(x+T)=f(x)+g(x), so T is a period of their sum. For commensurable periods T₁/T₂=p/q with positive integers p,q, a common period is qT₁=pT₂. For example, sin x and sin 2x share 2π. This need not be the sum’s smallest period: cancellation sin x−sin x=0 even produces a constant. Individual periodicity without a shared period alone does not justify that conclusion.