A fielder throws a ball at \(20\,\text{m/s}\), three times, at three angles. At \(30°\) the ball lands \(34.6\,\text{m}\) away. At \(45°\) it lands \(40\,\text{m}\) away. At \(60°\) it lands \(34.6\,\text{m}\) away again, after a much higher flight. The chapter Motion in Two Dimensions finds the range as \(2u^2\sin\theta\cos\theta/g\). Why should a flat throw and a steep throw tie, and why does the throw in between win?
One identity from the unit circle
An identity is an equation that holds for every value of the angle, not only for a few. The most used one comes from the unit circle. The point at angle \(\theta\) is \((\cos\theta, \sin\theta)\), and its distance from the centre is 1. By Pythagoras, the square of that distance is \(\cos^2\theta + \sin^2\theta\). So this sum is 1 for every angle.
For every angle \(\theta\), \(\sin^2\theta + \cos^2\theta = 1\). Dividing both sides by \(\cos^2\theta\) gives \(\tan^2\theta + 1 = \sec^2\theta\), wherever \(\cos\theta \ne 0\). Dividing by \(\sin^2\theta\) gives \(1 + \cot^2\theta = \text{cosec}^2\,\theta\), wherever \(\sin\theta \ne 0\).
The identity finds one ratio from another. If \(\sin\theta = 0.6\) and \(\theta\) is acute, then \(\cos\theta = \sqrt{1 - 0.36} = 0.8\). The sign of the square root comes from the quadrant. In physics the identity appears wherever two parts share out a fixed total. In simple harmonic motion the kinetic energy goes as \(\cos^2\) and the potential energy as \(\sin^2\). Their sum is constant because \(\sin^2 + \cos^2 = 1\).
An angle \(\theta\) lies between \(90°\) and \(180°\), and \(\sin\theta = 5/13\). What is \(\cos\theta\)?
The sine of a sum
Physics often needs the sine of a sum of two angles: a phase added to \(\omega t\), a double angle, a tilt added to a slope. The sine of a sum is not the sum of the sines. For example, \(\sin(30° + 60°) = \sin 90° = 1\), while \(\sin 30° + \sin 60° = 1.366\). The correct rule mixes the sines and the cosines of both angles.
The two sum formulas are proved from the geometry of two rotations. That proof belongs to the mathematics class. Here the formulas are stated and used. They pass a quick test. Put \(A = 30°\) and \(B = 60°\): \(\sin 30°\cos 60° + \cos 30°\sin 60° = 0.25 + 0.75 = 1\), which is \(\sin 90°\). Put \(B = 90°\) in the second formula: \(\cos(A + 90°) = -\sin A\), the rule of the unit circle.
The tangent of a sum follows by dividing the two formulas: \(\tan(A + B) = (\tan A + \tan B)/(1 - \tan A\tan B)\). If \(\tan A = 1/2\) and \(\tan B = 1/3\), then \(\tan(A + B) = (5/6)/(5/6) = 1\), so \(A + B = 45°\).
Two equal angles
The double-angle formulas are the sum formulas with the two angles equal. The derivation takes five short steps.
\[ \sin(A + B) = \sin A\cos B + \cos A\sin B \]
Why. the sum formula, true for every pair of angles.
\[ \sin 2\theta = \sin\theta\cos\theta + \cos\theta\sin\theta = 2\sin\theta\cos\theta \]
Why. put \(A = B = \theta\); the two terms are equal.
\[ \cos 2\theta = \cos\theta\cos\theta - \sin\theta\sin\theta = \cos^2\theta - \sin^2\theta \]
Why. put \(A = B = \theta\) in the cosine formula.
\[ \cos 2\theta = (1 - \sin^2\theta) - \sin^2\theta = 1 - 2\sin^2\theta \]
Why. replace \(\cos^2\theta\) by \(1 - \sin^2\theta\), from the Pythagorean identity.
\[ \sin^2\theta = \tfrac12(1 - \cos 2\theta) \]
Read it. rearrange step 4. A square of a sine becomes a cosine of the double angle, which physics uses to average \(\sin^2\) over a cycle.
The sim answers the fielder's question. The range is \(R = u^2 \cdot 2\sin\theta\cos\theta/g = u^2\sin 2\theta/g\). The sine of \(2\theta\) is largest, 1, when \(2\theta = 90°\), so the best angle is \(45°\) and \(R_{\max} = u^2/g = 40\,\text{m}\). For \(30°\) and \(60°\) the doubled angles are \(60°\) and \(120°\), and \(\sin 120° = \sin(180° - 60°) = \sin 60°\). So the two throws land together at \(34.6\,\text{m}\).
A fielder throws the ball at \(20\,\text{m/s}\) at \(37°\) above the horizontal. Take \(g = 10\,\text{m/s}^2\), \(\sin 37° = 0.6\) and \(\cos 37° = 0.8\). Find the range with the double-angle formula, and find the other angle that gives the same range.
\(u = 20\,\text{m/s}\), so \(u^2 = 400\,\text{m}^2/\text{s}^2\), and \(\theta = 37°\).
\(\sin 2\theta = 2\sin\theta\cos\theta\), so \(\sin 74° = 2 \times 0.6 \times 0.8\).
\(\sin 74° = 0.96\), so \(R = 400 \times 0.96/10 = 38.4\,\text{m}\).
The ball lands \(38.4\,\text{m}\) away. The angle \(90° - 37° = 53°\) gives \(\sin 106° = \sin 74°\), so a throw at \(53°\) lands at the same spot after a higher flight.
Use \(\sin(45° + 30°)\) to find \(\sin 75°\).
An angle has \(\sin\theta = 0.6\). Find \(\cos 2\theta\).
A ball is thrown at \(30\,\text{m/s}\) at \(15°\) above the horizontal, on level ground. Take \(g = 10\,\text{m/s}^2\). Find its range, in metres.
A sum becomes a product
Two sounds reach one ear, and two waves meet on one string. Physics adds two sines again and again. A sum of two sines of different angles shows no clear amplitude. A product does: one factor can be read as a slowly changing amplitude for the other. The sum formulas turn one into the other.
\[ \sin(A + B) = \sin A\cos B + \cos A\sin B \]
Why. the sum formula.
\[ \sin(A - B) = \sin A\cos B - \cos A\sin B \]
Why. the difference formula, the lower signs.
\[ \sin(A + B) + \sin(A - B) = 2\sin A\cos B \]
Why. add the two lines; the terms \(\cos A\sin B\) cancel.
\[ P = A + B,\ Q = A - B \ \Rightarrow\ A = \dfrac{P + Q}{2},\ B = \dfrac{P - Q}{2} \]
Why. rename the two angles; adding and subtracting the two definitions gives \(A\) and \(B\).
\[ \sin P + \sin Q = 2\sin\dfrac{P + Q}{2}\cos\dfrac{P - Q}{2} \]
Read it. put the new names into step 3. The other three identities come the same way from the other pairs of sum formulas.
When the two angles are \(P = 2\pi f_1t\) and \(Q = 2\pi f_2t\) with close frequencies, the half-sum oscillates at the mean frequency \((f_1 + f_2)/2\). The half-difference changes slowly, at \((f_2 - f_1)/2\). So the sum is a fast oscillation whose amplitude swells and fades. The ear hears the mean pitch, getting loud and soft. Each cycle of the slow cosine has two loud moments, at \(+1\) and at \(-1\). So the loudness beats \(|f_1 - f_2|\) times a second.
Two tuning forks of \(256\,\text{Hz}\) and \(260\,\text{Hz}\) sound together, each giving the same amplitude \(A\) at the ear. Write the sum \(A\cos(2\pi \cdot 256t) + A\cos(2\pi \cdot 260t)\) as a product. What pitch is heard, and how many beats are heard each second?
Take \(P = 2\pi \cdot 260t\) and \(Q = 2\pi \cdot 256t\), and use \(\cos P + \cos Q = 2\cos\tfrac{P + Q}{2}\cos\tfrac{P - Q}{2}\).
\(\tfrac{P + Q}{2} = 2\pi \cdot 258t\) and \(\tfrac{P - Q}{2} = 2\pi \cdot 2t\).
The sum is \(\big[2A\cos(2\pi \cdot 2t)\big]\cos(2\pi \cdot 258t)\).
The pitch heard is \(258\,\text{Hz}\). The envelope has a frequency of \(2\,\text{Hz}\), with two loud moments in each of its cycles, so the ear hears \(4\) beats each second, \(260 - 256\).
The same identity builds a standing wave. Two waves \(A\sin(kx - \omega t)\) and \(A\sin(kx + \omega t)\) run in opposite directions along a string. The half-sum of their angles is \(kx\) and the half-difference is \(-\omega t\). So their sum is \(2A\sin kx\cos\omega t\): every point swings in time as \(\cos\omega t\), with an amplitude \(2A\sin kx\) that depends on where it is.
Tones of \(440\,\text{Hz}\) and \(444\,\text{Hz}\) of equal amplitude \(A\) sound together. Which product equals their sum \(A\cos(2\pi \cdot 440t) + A\cos(2\pi \cdot 444t)\)?
The sine of a sum is not the sum of the sines, and \(\sin 2\theta\) is not \(2\sin\theta\): \(\sin 60° = 0.866\), while \(2\sin 30° = 1\). Every identity here holds for every angle, so test a doubtful step with \(30°\) and \(60°\) before trusting it. In the sum-to-product identities, halve both the sum and the difference. The slow factor reads as an amplitude only when the two frequencies are close; for \(440\) and \(660\,\text{Hz}\) the ear hears two notes, not beats.
Motion in Two Dimensions turns \(2\sin\theta\cos\theta\) into \(\sin 2\theta\) to find the range, the best angle of \(45°\) and the equal ranges of complementary angles. Friction, Part 2 finds the angle of least pull. Dynamics of SHM adds \(\sin^2\) and \(\cos^2\) to keep the total energy constant. The alternating-current chapter averages \(\sin^2\) over a cycle with \(\sin^2\theta = \tfrac12(1 - \cos 2\theta)\). Beats and the Doppler Effect writes two close tones as a product, and Standing Waves on a String writes two opposite waves as \(2A\sin kx\cos\omega t\). Polarisation uses \(\cos^2\theta\) through three sheets.
- Writing \(\sin(A + B) = \sin A + \sin B\).Test with \(30°\) and \(60°\): the left side is 1, the right side 1.366. Use the sum formula.
- Writing \(\sin 2\theta = 2\sin\theta\).\(\sin 2\theta = 2\sin\theta\cos\theta\). The extra \(\cos\theta\) is what makes the range peak at \(45°\).
- Using the full sum and difference in a sum-to-product identity.Both angles on the right are halved: \((P + Q)/2\) and \((P - Q)/2\).
- Taking the beat frequency as half the difference.The envelope has half the difference, but it is loud twice per cycle. The ear hears \(|f_1 - f_2|\) beats a second.