The Graph of Tangent

Tangent is sine divided by cosine

In a right triangle, $\tan \theta =$tan θ = opposite ÷ adjacent. Divide the top and the bottom of that fraction by the hypotenuse. The value does not change, and it becomes (opposite ÷ hypotenuse) ÷ (adjacent ÷ hypotenuse), which is $\sin \theta /\cos \theta$sin θ / cos θ.

So $\tan \theta =\sin \theta /\cos \theta$tan θ = sin θ / cos θ. On the unit circle, where the point at angle $\theta$θ is $\left(\cos \theta ,\sin \theta \right)$(cos θ, sin θ), this is the height of the point divided by its distance across, $y/x$y / x. That is the gradient of the radius. So $\tan \theta$tan θ is the gradient of the radius at angle $\theta$θ, and this is how the tangent of any angle, not only an acute one, is defined.

For example, at 45° the sine and cosine are equal, both $\frac{\sqrt{2}}{2}$√2/2, so tan 45° = 1. The radius at 45° has gradient 1.

Where the cosine is 0

At 90° the point on the unit circle is (0, 1), so sin 90° = 1 and cos 90° = 0. Then tan 90° would be 1 ÷ 0, and dividing by 0 has no answer. The tangent of 90° does not exist. The radius at 90° is vertical, and a vertical line has no gradient.

Look at angles close to 90°. To 2 decimal places, tan 80° = 5.67, tan 89° = 57.29 and tan 89.9° = 572.96. As the angle approaches 90° from below, the cosine gets closer and closer to 0 while the sine stays close to 1, so the tangent grows without limit.

Just past 90° the cosine is a small negative number, and tan 91° = −57.29. So on the right of 90° the graph comes up from very large negative values.

The vertical line $\theta =90^{\circ }$θ = 90° is a vertical asymptote of the graph: the curve gets closer and closer to it and never meets it. This is the same thing that happens to any quotient where the denominator is 0 and the numerator is not. The same happens at 270°, where cos 270° = 0 and sin 270° = −1.

$y=\tan \theta$y = tan θ from 0° to 360°, with one square across for every 45°. The dashed vertical lines at 90° and 270° are asymptotes, where $\cos \theta =0$cos θ = 0 and the tangent has no value. The curve rises through each branch without ever crossing them.

The arm at 40°, extended to meet the vertical line x = 1. That line is 1 across from the center, so the height where the arm meets it is the arm’s gradient, tan 40° = 0.84, to 2 decimal places. Turn the arm toward 90° and the height climbs off the top: at 90° the arm is parallel to the line and never meets it.

The sign of the tangent

The tangent is positive when the sine and cosine have the same sign and negative when they have opposite signs.

From 0° to 90° both are positive, so the tangent is positive. From 90° to 180° the point is to the left of the center and above it: the sine is positive and the cosine negative, so the tangent is negative. For example, $\tan 135^{\circ }=\left(\frac{\sqrt{2}}{2}\right)/\left(-\frac{\sqrt{2}}{2}\right)=-1$tan 135° = (√2/2) / (−√2/2) = −1. From 180° to 270° both are negative, so the tangent is positive again. From 270° to 360° it is negative.

The tangent is 0 where the sine is 0, at 0°, 180° and 360°. Those are where the graph crosses the axis.

The point at 135° is above the center and to the left of it. The height $\frac{\sqrt{2}}{2}$√2/2 divided by the distance across, $-\frac{\sqrt{2}}{2}$−√2/2, gives tan 135° = −1.

A period of 180°

Turn the radius half a turn further, from $\theta$θ to $\theta +180^{\circ }$θ + 180°. The point moves to the opposite side of the circle, so both of its coordinates change sign: $\sin \left(\theta +180^{\circ }\right)=-\sin \theta$sin (θ + 180°) = −sin θ and $\cos \left(\theta +180^{\circ }\right)=-\cos \theta$cos (θ + 180°) = −cos θ.

Divide one by the other and the two minus signs cancel: $\tan \left(\theta +180^{\circ }\right)=-\sin \theta /\left(-\cos \theta \right)=\sin \theta /\cos \theta =\tan \theta$tan (θ + 180°) = −sin θ / (−cos θ) = sin θ / cos θ = tan θ. So the tangent repeats every 180°, not every 360°. Its period is 180°.

The gradient says the same thing. After half a turn the radius points the opposite way along the same straight line, and a line has one gradient. So tan 225° = tan 45° = 1.

Every value, and no maximum

Sine and cosine stay between −1 and 1. The tangent does not: between −90° and 90° it rises through every number, from very large negative values near −90° to very large positive values near 90°. Its range is every real number, so $\tan \theta =1000$tan θ = 1000 has a solution, which $\sin \theta =1000$sin θ = 1000 does not.

Turning the radius clockwise through $\theta$θ, to $-\theta$−θ, reflects the point in the x-axis. The sine changes sign and the cosine does not, so $\tan \left(-\theta \right)=-\tan \theta$tan (−θ) = −tan θ: for example tan (−45°) = −1. On the graph, the branch from −90° to 90° is symmetrical about the origin: half a turn about (0, 0) takes it onto itself.

To sketch $y=\tan \theta$y = tan θ from 0° to 360°: draw the asymptotes at 90° and 270°; mark the zeros at 0°, 180° and 360°; mark tan 45° = 1 and tan 135° = −1; then draw three branches, each rising from left to right and bending toward the asymptotes on either side.

The usual mistakes

Joining the branches across the asymptote. The graph is not a continuous wave: at 90° the tangent has no value, and the curve on the left goes up while the curve on the right comes from below.

Writing tan 90° = 0 or tan 90° = 1. The cosine is 0 at 90°, and division by 0 has no answer, so tan 90° is undefined.

Giving the tangent a period of 360°, like sine and cosine. Half a turn changes the sign of both the sine and the cosine, so their quotient repeats every 180°.

Taking the tangent of an obtuse angle as positive. Between 90° and 180° the cosine is negative, so $\tan 120^{\circ }=-\sqrt{3}$tan 120° = −√3.

An exact value with $\sqrt{3}$√3 underneath

The application below writes an exact value in its simplest form. The exact value of tan 30° is $\frac{1}{\sqrt{3}}$1/√3, so $8\tan 30^{\circ }=\frac{8}{\sqrt{3}}$8 tan 30° = 8/√3. It is usual to write such a value with no surd in the denominator, which is called rationalizing the denominator.

Multiply the numerator and the denominator by $\sqrt{3}$√3. The value is unchanged, because this multiplies by $\frac{\sqrt{3}}{\sqrt{3}}$√3/√3, which is 1, and $\sqrt{3}\times \sqrt{3}=3$√3 × √3 = 3. So $\frac{8}{\sqrt{3}}=\frac{8\sqrt{3}}{3}$8/√3 = 8√3/3, which is 4.62 to 2 decimal places.