Solving Simple Trigonometric Equations

The ratio question run backwards

An equation such as sin x = ½ asks for the angles whose sine is ½. It is the question of finding an angle from a ratio, but now the angle can be any size, not only an acute angle in a right triangle.

One answer is quick to find. The inverse sine gives an angle with sine ½: ${\sin }^{-1}\left(\frac{1}{2}\right)=30^{\circ }$sin⁻¹(½) = 30°, one of the exact values. So x = 30° is a solution. Check: sin 30° = ½.

The same height twice in a turn

On the unit circle, sin x is the height of the point at angle x. A height of ½ is a horizontal line across the circle, halfway up from the center, and that line meets the circle in two places, not one.

One is the point at 30°. The other is its reflection in the y-axis, at 180° − 30° = 150°. The two points are at the same height, so sin 150° = ½ as well. Between 0° and 360° the equation sin x = ½ has two solutions: x = 30° and x = 150°.

On the graph of y = sin x the same thing happens. The line y = ½ cuts the wave twice in each turn: once on the way up to the peak at 90° and once on the way down.

The points at 30° and 150° on the unit circle. The dashed line joining them is level, so their heights are equal: sin 30° = sin 150° = ½.

The level line at height 0.5 cuts the circle at two arms, 30° and 150°, and cuts the wave at the same two angles. Drag the line up or down and both arms move together, the same distance either side of the top of the circle. Turn n to see the same two solutions one full turn further on.

Every sine equation gives a pair

The pair comes from the rule sin (180° − x) = sin x. Whatever angle the inverse sine gives, 180° minus that angle has the same sine.

So for sin x = 0.6, the inverse sine gives $x={\sin }^{-1}\left(0.6\right)=36.87^{\circ }$x = sin⁻¹(0.6) = 36.87°, to 2 decimal places. The second solution is 180° − 36.87° = 143.13°. Check: sin 143.13° = 0.6000.

The pair is always x and 180° − x, never x and 180° + x. Half a turn on from 36.87°, at 216.87°, the point is below the center and the sine is −0.6.

A negative value

Solve sin x = −½ for $0^{\circ }\le x\le 360^{\circ }$0° ≤ x ≤ 360°. The inverse sine gives ${\sin }^{-1}\left(-\frac{1}{2}\right)=-30^{\circ }$sin⁻¹(−½) = −30°. That is a solution of the equation, but it is not between 0° and 360°. Add a full turn: −30° + 360° = 330° is the same position on the circle, so sin 330° = −½.

The pair still comes from 180° minus the angle: 180° − (−30°) = 210°. So the solutions are x = 210° and x = 330°.

Check with the circle: 210° is 30° past a half turn and 330° is 30° short of a full turn. Both points are below the center, at the same height as each other, and sin 30° = ½, so both have sine −½.

Cosine pairs differently

The cosine is the distance across, the x-coordinate of the point. Points with the same x-coordinate are reflections of each other in the x-axis, not the y-axis. The reflection of the point at angle x in the x-axis is the point at 360° − x.

So cos (360° − x) = cos x, and a cosine equation pairs x with 360° − x. For cos x = ½, the inverse cosine gives ${\cos }^{-1}\left(\frac{1}{2}\right)=60^{\circ }$cos⁻¹(½) = 60°, and the second solution is 360° − 60° = 300°.

For cos x = −0.4, the inverse cosine gives ${\cos }^{-1}\left(-0.4\right)=113.58^{\circ }$cos⁻¹(−0.4) = 113.58°, to 2 decimal places. The second solution is 360° − 113.58° = 246.42°. Check: cos 246.42° = −0.4000.

The points at 60° and 300° are reflections of each other in the x-axis. They are the same distance across, so cos 60° = cos 300° = ½.

Tangent repeats every 180°

Tangent pairs x with x + 180°, because its period is 180°. For tan x = 2, the inverse tangent gives ${\tan }^{-1}2=63.43^{\circ }$tan⁻¹ 2 = 63.43°, to 2 decimal places, and the second solution is 63.43° + 180° = 243.43°. Check: tan 243.43° = 2.00, to 2 decimal places.

The method, and two special cases

Rearrange the equation until it reads sin x, cos x or tan x equal to a number. Take the inverse for one solution. Use the pairing for the second: 180° − x for sine, 360° − x for cosine, x + 180° for tangent. Then add or subtract 360° to bring every solution into the range asked for.

For example, solve 2 sin x + 1 = 0 for $0^{\circ }\le x\le 360^{\circ }$0° ≤ x ≤ 360°. Subtract 1 and divide by 2: sin x = −½. That is the equation solved above, so x = 210° or x = 330°.

When the number is 1 or −1, the two solutions in a turn are the same angle. sin x = 1 has only x = 90°, because 180° − 90° = 90°. The level line touches the top of the circle at one point.

When the number is greater than 1 or less than −1, there is no solution at all. The sine and cosine of every angle lie between −1 and 1, so sin x = 1.5 has no answer, and a calculator asked for ${\sin }^{-1}\left(1.5\right)$sin⁻¹(1.5) gives an error.

The usual mistakes

Stopping at the first answer. The inverse gives one angle; between 0° and 360° a sine or cosine equation usually has two.

Using the sine pairing for a cosine. For cos x = ½ the second angle is 360° − 60° = 300°. The angle 180° − 60° = 120° has cosine −½.

Using 180° + x for a sine. Half a turn on, the point is below the center, so sin 210° = −½, not ½.

Leaving a solution outside the range. For sin x = −½ the inverse gives −30°, which must become 330° before it can be an answer between 0° and 360°.

An equation with ${\sin }^2\theta$sin² θ in it

The application below gives a height $H=7.5\sin \theta +5{\sin }^2\theta$H = 7.5 sin θ + 5 sin² θ, where ${\sin }^2\theta$sin² θ means ${\left(\sin \theta \right)}^2$(sin θ)². Setting H = 5 gives an equation with both ${\sin }^2\theta$sin² θ and $\sin \theta$sin θ in it. That is a quadratic equation in $\sin \theta$sin θ.

Write s for $\sin \theta$sin θ. The equation becomes $5s^2+7.5s-5=0$5s² + 7.5s − 5 = 0, an ordinary quadratic, which factors after multiplying every term by $\frac{2}{5}$2/5. Solving it gives two values of s, and each value of s is a sine equation of the kind on this page.

One of the roots is s = −2. No angle has a sine less than −1, so $\sin \theta =-2$sin θ = −2 has no solution, and that root is rejected.