Bowstring

Sanskrit jyā, a bowstring — the chord an archer draws. Arabic scribes wrote it jība; a Latin translator misread it as jayb, a fold or bay, and rendered it sinus. Nine centuries later we still call it the sine.

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Angle —

Trigonometry, built rather than recited

Guided investigation — ages 14–18 mathematics

Each investigation asks for a prediction in writing before you touch the controls. Steps I–III need no physics; the bridge at the end is for anyone heading into waves and simple harmonic motion.

IThe ratio

What stays still when everything moves

Step I  ·  15 min

Predict

You double every side of a right triangle. State what happens to the length of the opposite side, and what happens to sin θ. If your two answers differ, explain why.

  1. Set θ to 35°. Sweep the scale slider from 0.35 to 2.4 and watch the six side lengths. Record the largest and smallest value each ratio takes.
  2. Explain why the two triangles must give identical ratios, using the word similar.
  3. Now hold the scale fixed and change θ instead. State what the ratios depend on.
  4. Complete the table. The final column checks whether your two ratios really do satisfy Pythagoras.
  5. Suggest why it is misleading to say “sin θ is the opposite side”.
θsin θcos θtan θ sin θ / cos θsin2+cos2
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IIThe circle

Past the corner

Steps II–III  ·  20 min

Predict

A right triangle cannot contain an angle of 150°. Yet sin 150° has a definite value. Predict that value, and explain what has to change about the definition of sine for the question to make sense at all.

  1. In Step II, drag the angle slowly from 0° to 360°. Identify the quadrants where sine is negative and where cosine is negative, and explain both using coordinates rather than a mnemonic.
  2. Switch on Squares on the legs. Describe how the two areas change as θ varies, and state what their total always is.
  3. Switch on the tangent segment and take θ towards 90°. Explain, geometrically, why tan 90° has no value.
  4. Move to Step III and set the angle to 2π. State how many whole radii fit around the circle and what is left over. Deduce why π appears in the answer.
  5. Using s = rθ, determine the arc length swept by a 2.5 m pendulum through 0.30 rad.
IIIThe wave

Where the wave was hiding

Steps IV–V  ·  25 min

Predict

In y = A sin Bx, you double B. Predict whether the wave becomes wider or narrower, and state what happens to the period. Most people get this the wrong way round — commit to an answer first.

  1. In Step IV, watch one full turn. Describe the relationship between the height of the rotating point and the height of the curve being drawn.
  2. Switch on Plot cosine. Measure the horizontal shift between the two curves, in degrees and in radians.
  3. Move to Step V. Change A alone, then D alone. State which one moves the midline and which stretches the curve.
  4. Set B = 1 and read the period. Set B = 2 and read it again. Compare with your prediction and explain the inverse relationship.
  5. Find values of A, B, C and D giving a wave of amplitude 2, period π, and midline y = −1. Verify against the measured readouts, then state the maximum and minimum without reading them off.
Where this leads — simple harmonic motion

Why your waves topic starts with a circle

Step VI casts a shadow of the rotating point onto a vertical screen. That shadow is doing SHM — and the whole standard treatment falls out of the geometry rather than being asserted.

  1. Run Step VI and watch −a/x in the panel. State what it stays equal to, and show that this is the defining condition a = −ω2x.
  2. Identify where in the cycle the shadow moves fastest and where its acceleration is greatest. Explain why these are never the same instant.
  3. The velocity curve leads the displacement curve by a quarter period. Deduce what mathematical operation relates them, and confirm it against x = A cos ωt.
  4. Change ω and watch the period. Derive T = 2π/ω from the fact that one period is one full turn of the circle.
  5. Outline why a pendulum is only approximately simple harmonic, given that this construction is exact.
Colour codeSine is red, cosine is blue, tangent is teal — in every step, on the circle, on the graph and in the readouts. Nothing changes colour when it changes representation.
What is exactAll values are computed live from the angle, not sampled from a table. The tangent segment is drawn on the true tangent line at x = 1; the wrapped radii in Step III are arcs of true length 1.
The nameSine reached English through a chain of translation errors from the Sanskrit jyā-ardha, half-chord. The bowstring in Step II is the original object: sin θ is literally half the chord subtending 2θ.
UnitsAngles are shown in degrees and radians together throughout. Step VI uses arbitrary units where the circle has radius 1 and ω is in rad s−1.