What is the exact value of tan pi over six?

Tangent of pi over six equals one half divided by root three over two, which is one over root three, or root three over three.

What is the exact value of tan pi over six?

Tangent of pi over six equals one half divided by root three over two, which is one over root three, or root three over three.

Year 11 Methods (Unit 1 & 2) Trigonometric Functions

Another Trigonometric Function: Tangent

Q: What is the definition of the tangent function?

On the unit circle, tangent t equals sine t divided by cosine t, which is the y coordinate over the x coordinate of the point at angle t.

Q: When is tangent undefined?

Tangent is undefined whenever cosine t is zero, that is at pi over two plus whole multiples of pi. For example at pi over two and three pi over two, where the graph has a vertical asymptote.

Q: What is the period of the tangent function?

The period is pi. The tangent repeats every half turn, so tangent of t plus pi equals tangent of t.

Q: In which quadrants is tangent positive?

Tangent is positive in quadrants one and three, where x and y have the same sign. It is negative in quadrants two and four.

20 practice questions 1 video lesson Theory + worked examples

Learn the tangent function for Queensland Year 11 Mathematical Methods (QCAA). On the unit circle, tangent equals sine divided by cosine — the vertical coordinate over the horizontal one.

You will learn to find exact tangent values at the special angles, see where tangent is undefined because cosine is zero, work with its period of pi, and read its sign by quadrant — leading to the tangent graph in the QCAA course.

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Another Trigonometric Function: Tangent - Video - Finding exact values for tangent Watch

Theory

In Year 11 Mathematical Methods (QCAA, Unit 1), the tangent function is built from sine and cosine: \(\tan t=\dfrac{\sin t}{\cos t}=\dfrac{y}{x}\), the ratio of the coordinates of the point on the unit circle. This page shows how to evaluate \(\tan t\), the exact values at the special angles, where the tangent is undefined, its period of \(\pi\), and its sign by quadrant.

For the point \(P=(x,\,y)=(\cos t,\ \sin t)\) on the unit circle, the tangent of \(t\) is the ratio of the coordinates: \(\tan t=\dfrac{\sin t}{\cos t}=\dfrac{y}{x}\). Geometrically it is the length of the vertical segment cut off where the terminal side (extended) meets the line \(x=1\), at the point \((1,\ \tan t)\).

The tangent is undefined whenever \(\cos t=0\), that is at \(t=\dfrac{\pi}{2}+k\pi\); its graph has a vertical asymptote there. Because the pattern repeats every half turn, the period of tangent is \(\pi\) (not \(2\pi\)).

The sign of \(\tan t\) follows the signs of \(y\) and \(x\): tangent is positive in quadrants 1 and 3 (where \(y\) and \(x\) have the same sign) and negative in quadrants 2 and 4 — the T in ASTC.

Tangent is sine over cosine. Evaluate \(\tan t=\dfrac{\sin t}{\cos t}\); if \(\cos t=0\) the tangent is undefined.

\(\tan t\) is the height where the terminal side meets \(x=1\), the point \((1,\ \tan t)\).

\(y=\tan t\) with asymptotes at \(\dfrac{\pi}{2}\) and \(\dfrac{3\pi}{2}\); the period is \(\pi\).

The definition and the special-angle values:

\[\tan t=\dfrac{\sin t}{\cos t}=\dfrac{y}{x}\]

\tan t = \frac{\sin t}{\cos t}

\[\tan 0=0,\quad \tan\dfrac{\pi}{6}=\dfrac{\sqrt3}{3},\quad \tan\dfrac{\pi}{4}=1,\quad \tan\dfrac{\pi}{3}=\sqrt3\]

\tan \frac{π}{4} = 1

Where it is undefined, and its period:

\[\tan t\ \text{undefined at}\ t=\dfrac{\pi}{2}+k\pi,\qquad \tan(t+\pi)=\tan t\]

\tan (t + π) = \tan t

Sign by quadrant: \(\tan t>0\) in quadrants 1 and 3, \(\tan t<0\) in quadrants 2 and 4.

How to evaluate \(\tan t\)

Check for undefined: if \(\cos t=0\) (at \(\dfrac{\pi}{2}\), \(\dfrac{3\pi}{2}\), ...), the tangent is undefined — stop there.
Reference angle: find the acute angle to the \(x\)-axis and read the size of \(\tan\) (for example \(\tan\dfrac{\pi}{6}=\dfrac{\sqrt3}{3}\)).
Attach the sign: positive in quadrants 1 and 3, negative in quadrants 2 and 4.

Example 1 — Tangent from sine over cosine

Find the exact value of \(\tan\dfrac{\pi}{4}\).

Solution

Use \(\tan t=\dfrac{\sin t}{\cos t}\) with the exact values at \(\dfrac{\pi}{4}\):

\(\tan\dfrac{\pi}{4}\)	\(=\)	\(\dfrac{\sin(\pi/4)}{\cos(\pi/4)}\)
	\(=\)	\(\dfrac{\sqrt2/2}{\sqrt2/2}\)

The equal parts cancel:

\(=\)

\(1\)

\(\tan\dfrac{\pi}{4}=1\).

\tan \frac{π}{4} = 1

Example 2 — Tangent at pi over three

Find the exact value of \(\tan\dfrac{\pi}{3}\).

Solution

Divide sine by cosine:

\(\tan\dfrac{\pi}{3}\)	\(=\)	\(\dfrac{\sin(\pi/3)}{\cos(\pi/3)}\)
	\(=\)	\(\dfrac{\sqrt3/2}{1/2}\)

Dividing by \(\dfrac{1}{2}\) doubles the top:

\(=\)

\(\sqrt3\)

\(\tan\dfrac{\pi}{3}=\sqrt3\).

\tan \frac{π}{3} = \sqrt{3}

Example 3 — Where tangent is undefined

For which values of \(t\) in \([0,\,2\pi]\) is \(\tan t\) undefined?

Solution

Tangent is \(\dfrac{\sin t}{\cos t}\), so it is undefined when the denominator \(\cos t=0\):

\(\tan t\)	\(=\)	\(\dfrac{\sin t}{\cos t}\)
\(\cos t\)	\(=\)	\(0\)

On \([0,\,2\pi]\), \(\cos t=0\) at:

\(t\)

\(=\)

\(\dfrac{\pi}{2},\ \ \dfrac{3\pi}{2}\)

These are the vertical asymptotes of \(y=\tan t\); the pattern repeats every \(\pi\).

Undefined at \(t=\dfrac{\pi}{2}\) and \(t=\dfrac{3\pi}{2}\).

t = \frac{π}{2}, \frac{3 π}{2}

Example 4 — Tangent with a quadrant sign

Find the exact value of \(\tan\dfrac{5\pi}{6}\).

Solution

\(\dfrac{5\pi}{6}\) is in the second quadrant, where tangent is negative. Reference angle:

\(\text{ref}\)	\(=\)	\(\pi-\dfrac{5\pi}{6}\)
	\(=\)	\(\dfrac{\pi}{6}\)

Attach the second-quadrant sign to \(\tan\dfrac{\pi}{6}\):

\(\tan\dfrac{5\pi}{6}\)	\(=\)	\(-\tan\dfrac{\pi}{6}\)
	\(=\)	\(-\dfrac{\sqrt3}{3}\)

\(\tan\dfrac{5\pi}{6}=-\dfrac{\sqrt3}{3}\).

\tan \frac{5 π}{6} = - \frac{\sqrt{3}}{3}

Common pitfalls

Treating the period as \(2\pi\). Tangent repeats every \(\pi\), not \(2\pi\); \(\tan(t+\pi)=\tan t\).

Writing tangent \(=0\) instead of undefined. When \(\cos t=0\) the ratio \(\dfrac{\sin t}{\cos t}\) has a zero denominator, so \(\tan t\) is undefined — not zero.

Losing the sign. Tangent is negative in quadrants 2 and 4; a reference angle gives only the size, so \(\tan\dfrac{5\pi}{6}=-\dfrac{\sqrt3}{3}\), not \(+\dfrac{\sqrt3}{3}\).

Frequently asked questions

What is the definition of the tangent function?

On the unit circle \(\tan t=\dfrac{\sin t}{\cos t}=\dfrac{y}{x}\), the ratio of the \(y\)- and \(x\)-coordinates of the point at angle \(t\).

When is tangent undefined?

Whenever \(\cos t=0\), that is at \(t=\dfrac{\pi}{2}+k\pi\) (for example \(\dfrac{\pi}{2}\) and \(\dfrac{3\pi}{2}\)); the graph has a vertical asymptote there.

What is the period of the tangent function?

It is \(\pi\): the tangent repeats every half turn, so \(\tan(t+\pi)=\tan t\).

In which quadrants is tangent positive?

In quadrants 1 and 3, where \(x\) and \(y\) have the same sign. It is negative in quadrants 2 and 4.

What is the exact value of tan(pi/6)?

\(\tan\dfrac{\pi}{6}=\dfrac{\sin(\pi/6)}{\cos(\pi/6)}=\dfrac{1/2}{\sqrt3/2}=\dfrac{1}{\sqrt3}=\dfrac{\sqrt3}{3}\).

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