A fielder at the boundary gathers the ball and throws it at \(20\,\text{m/s}\), at \(37°\) above the ground. Part of that speed carries the ball towards the wicket-keeper. Another part lifts the ball against gravity. The forward part decides how far the ball goes. The upward part decides how long the ball stays in the air. How much of the \(20\,\text{m/s}\) belongs to each part?
Three ratios of one triangle
Draw a triangle with one right angle, and mark one of its other two angles as \(\theta\). The longest side, across from the right angle, is the hypotenuse. The side across from \(\theta\) is the opposite side. The remaining side, which touches \(\theta\), is the adjacent side. Now make the triangle larger without changing its angles. Every side grows by the same factor, so the ratio of any two sides does not change. The ratios depend on the angle alone.
Because each ratio depends only on \(\theta\), each ratio gets a name. The three that physics uses are the sine, the cosine and the tangent. They are written \(\sin\theta\), \(\cos\theta\) and \(\tan\theta\).
For an acute angle \(\theta\) of a right triangle, the sine of \(\theta\) is the opposite side divided by the hypotenuse. The cosine of \(\theta\) is the adjacent side divided by the hypotenuse. The tangent of \(\theta\) is the opposite side divided by the adjacent side, which is also \(\sin\theta/\cos\theta\).
Three more ratios are the reciprocals of these. The cosecant is \(\text{cosec}\,\theta = 1/\sin\theta\), the secant is \(\sec\theta = 1/\cos\theta\), and the cotangent is \(\cot\theta = 1/\tan\theta\). Physics uses them rarely, mostly inside identities such as \(1 + \tan^2\theta = \sec^2\theta\).
Splitting a vector into two parts
Now put the triangle into physics. Let an arrow of length \(u\) stand for a velocity, and let it make an angle \(\theta\) with the ground. Drop a line from the tip of the arrow straight down to the ground. A right triangle appears, and the arrow is its hypotenuse. The side along the ground is adjacent to \(\theta\), so its length is \(u\cos\theta\). The vertical side is opposite to \(\theta\), so its length is \(u\sin\theta\). These two sides are the components of the velocity.
A short rule prevents most mistakes. The component that touches the angle takes the cosine. The component across from the angle takes the sine. So the rule depends on where the angle is marked, not on which part is horizontal. If the angle is measured from the vertical, the vertical component takes the cosine.
A porter pulls a luggage trolley with a force of \(50\,\text{N}\) along a handle at \(60°\) above the floor. What part of the force pulls the trolley forward along the floor?
The angles physics uses most
A few angles come up in problems again and again. Their ratios follow from two simple shapes. First, cut a square of side 1 along its diagonal. Each half is a right triangle with two angles of \(45°\). By Pythagoras, the hypotenuse is \(\sqrt{1^2 + 1^2} = \sqrt2\). So \(\sin 45° = \cos 45° = 1/\sqrt2 = 0.707\), and \(\tan 45° = 1\).
Second, cut an equilateral triangle of side 2 down its middle. Each half has angles of \(30°\), \(60°\) and \(90°\). Its sides are 1, the half base, and 2, the hypotenuse. The third side is the height, \(\sqrt{2^2 - 1^2} = \sqrt3\). Read the ratios off the triangle. The side of length 1 is opposite \(30°\), and the side \(\sqrt3\) is opposite \(60°\).
So \(\sin 30° = 1/2\), \(\cos 30° = \sqrt3/2 = 0.866\) and \(\tan 30° = 1/\sqrt3 = 0.577\). For \(60°\) the two sides swap roles: \(\sin 60° = 0.866\), \(\cos 60° = 0.5\) and \(\tan 60° = \sqrt3 = 1.732\).
Indian textbooks and entrance papers use one more triangle, with sides 3, 4 and 5. It is a right triangle, because \(3^2 + 4^2 = 25 = 5^2\). Its smaller angle is \(36.87°\), which problems round to \(37°\). Its larger angle is \(53.13°\), rounded to \(53°\). So \(\sin 37° \approx 3/5 = 0.6\), \(\cos 37° \approx 4/5 = 0.8\) and \(\tan 37° \approx 3/4\). For \(53°\) the values swap: \(\sin 53° \approx 0.8\), \(\cos 53° \approx 0.6\) and \(\tan 53° \approx 4/3\).
At the two ends of the range the triangle flattens. At \(0°\) the opposite side shrinks to nothing, so \(\sin 0° = 0\), \(\cos 0° = 1\) and \(\tan 0° = 0\). At \(90°\) the adjacent side shrinks to nothing, so \(\sin 90° = 1\) and \(\cos 90° = 0\). Then \(\tan 90°\) would divide by zero, so it does not exist. Between the two ends, the sine rises from 0 to 1 while the cosine falls from 1 to 0.
The seven angles, with their sine, cosine and tangent:
- \(0°\): \(\sin = 0\), \(\cos = 1\), \(\tan = 0\).
- \(30°\): \(\sin = 0.5\), \(\cos = 0.866\), \(\tan = 0.577\).
- \(37°\): \(\sin \approx 0.6\), \(\cos \approx 0.8\), \(\tan \approx 0.75\).
- \(45°\): \(\sin = \cos = 0.707\), \(\tan = 1\).
- \(53°\): \(\sin \approx 0.8\), \(\cos \approx 0.6\), \(\tan \approx 1.33\).
- \(60°\): \(\sin = 0.866\), \(\cos = 0.5\), \(\tan = 1.732\).
- \(90°\): \(\sin = 1\), \(\cos = 0\), \(\tan\) does not exist.
The two acute angles of a right triangle add to \(90°\). The side opposite one of them is adjacent to the other. So the sine of an angle is the cosine of the other angle: \(\sin\theta = \cos(90° - \theta)\). This is why \(\sin 37° = \cos 53°\) and \(\sin 30° = \cos 60°\).
In the 3-4-5 triangle, what is \(\tan 53°\)?
A block of mass \(2\,\text{kg}\) rests on a smooth incline of \(30°\), held by a string along the slope. Take \(g = 10\,\text{m/s}^2\). The normal force balances the component of the weight at right angles to the slope. Find the normal force, in newtons.
Beyond \(90°\): the unit circle
Physics needs angles larger than \(90°\). A wheel turns through \(120°\), a rope pulls backward at \(150°\), and a phase runs on past \(360°\). A right triangle cannot hold such an angle. So the ratios are defined again, on a circle of radius 1 centred at the origin, the unit circle. Turn a radius anticlockwise from the \(+x\) axis through the angle \(\theta\). Call the tip of the radius P, with coordinates \((x, y)\).
The new definitions are \(\cos\theta = x\), \(\sin\theta = y\) and \(\tan\theta = y/x\). For an angle between \(0°\) and \(90°\), the right triangle under the radius has hypotenuse 1. Its adjacent side is \(x\) and its opposite side is \(y\). So the new definitions agree with the old ratios, and they now work for every angle.
In the first quadrant, from \(0°\) to \(90°\), both \(x\) and \(y\) are positive, so all three ratios are positive. In the second quadrant, from \(90°\) to \(180°\), \(x\) is negative and \(y\) is positive. There only the sine is positive. In the third quadrant both are negative, so only the tangent, \(y/x\), is positive. In the fourth quadrant \(x\) is positive and \(y\) is negative, so only the cosine is positive. The words "All, Sin, Tan, Cos", one per quadrant, keep the order.
The circle is symmetric, so every angle is linked to an acute one. The point at \(180° - \theta\) is the mirror image of the point at \(\theta\) in the \(y\) axis. It has the same height and the opposite \(x\). So \(\sin(180° - \theta) = \sin\theta\) and \(\cos(180° - \theta) = -\cos\theta\). For example, \(\sin 120° = \sin 60° = 0.866\) and \(\cos 120° = -0.5\). The point at \(-\theta\) is the mirror image in the \(x\) axis, so \(\cos(-\theta) = \cos\theta\) and \(\sin(-\theta) = -\sin\theta\).
A rope pulls a crate with a force of \(40\,\text{N}\) at \(150°\) from the \(+x\) axis. What is the \(x\) component of the force?
The tangent breaks at \(90°\) and \(270°\), where \(\cos\theta = 0\) and \(y/x\) would divide by zero. A rope that pulls straight up has no horizontal component; nothing becomes infinite. The values \(0.6\) and \(0.8\) for \(37°\) are rounded as well: the true \(\sin 37°\) is \(0.6018\). Use them only when a problem gives them. Finally, set the calculator to degrees for degree angles. In radian mode, \(\sin 30\) gives \(-0.988\), the sine of 30 radians.
Back to physics
The fielder's throw now splits at once. The forward part is \(20\cos 37° = 20 \times 0.8 = 16\,\text{m/s}\). The upward part is \(20\sin 37° = 20 \times 0.6 = 12\,\text{m/s}\). The \(16\,\text{m/s}\) carries the ball towards the keeper, and the \(12\,\text{m/s}\) fights gravity. The two parts add to more than \(20\) because they are not in one direction. By Pythagoras, \(\sqrt{16^2 + 12^2} = 20\,\text{m/s}\) again.
A block of mass \(2\,\text{kg}\) rests on a smooth incline of \(30°\). A string parallel to the slope holds the block at rest. Take \(g = 10\,\text{m/s}^2\). Find the tension in the string and the normal force on the block.
The weight is \(W = mg = 20\,\text{N}\), straight down. Take one axis along the slope and one at right angles to it. The weight makes \(30°\) with the second axis, because the slope makes \(30°\) with the ground.
The component next to the \(30°\) angle, at right angles to the slope, is \(W\cos 30°\). The component across from it, down the slope, is \(W\sin 30°\). The string balances the first; the normal force balances the second.
\(T = W\sin 30° = 20 \times 0.5 = 10\,\text{N}\). \(N = W\cos 30° = 20 \times 0.866 = 17.3\,\text{N}\).
The string pulls with \(10\,\text{N}\) and the incline pushes with \(17.3\,\text{N}\). Check with Pythagoras: \(10^2 + 17.3^2 = 400 = 20^2\).
A ray of light in air meets the surface of water at \(53°\) to the normal. Snell's law says \(\sin i = \mu\sin r\), with \(\mu = 4/3\) for water. At what angle to the normal does the ray travel in the water?
Motion in Two Dimensions splits the launch velocity of a projectile into \(u\cos\theta\) and \(u\sin\theta\). Objects in Equilibrium and Accelerated Objects split the weight on an incline into \(mg\sin\theta\) along it and \(mg\cos\theta\) across it. The friction chapters read the angle of repose from \(\tan\theta = \mu\). Relative Velocity in Two Dimensions splits a boat's velocity across a river and the rain falling on a moving person. Circular motion tilts a cyclist by \(\tan\theta = v^2/rg\). Work uses \(Fs\cos\theta\), torque uses \(rF\sin\theta\), and Snell's law relates the sines of two angles.
- Using the sine for the component next to the angle.Look where the angle is marked. The side that touches the angle takes the cosine, whatever its direction.
- Taking \(\sin 37° = 0.6\) as exact.It is exact for the 3-4-5 triangle and rounded for \(37°\). Use it when the problem gives it.
- Forgetting the sign of a component beyond \(90°\).Read the quadrant first. In the second quadrant the cosine, and so the \(x\) component, is negative.
- Calculating \(\sin 30\) with the calculator in radian mode.Check the mode before the first number. Degrees for degree angles, radians for radian angles.