Trig values of the special right triangles
The sine, cosine and tangent of most angles come out of a calculator as decimals that have to be cut off somewhere. Are there angles whose values can be written down exactly, by hand?
Where you are 7 in Right triangle trigonometry
When you reach for it
Ask a calculator for \sin 40^\circ and it answers 0.6427876\ldots, a decimal it has to cut off somewhere. For three angles there is nothing to cut off. The sine, cosine and tangent of 30^\circ, 45^\circ and 60^\circ are exact numbers built from 1, \displaystyle \sqrt{2} and \displaystyle \sqrt{3}, and they can be read straight off two triangles you already know from the special right triangles: half a square, and half an equilateral triangle.
Reach for this whenever one of these three angles turns up and the answer should be exact: an exact length, slope or area, an expression to simplify, or a value to recognize and turn back into an angle. The same three angles come back all the way round the unit circle, so this small table ends up carrying most of the exact values in trigonometry.
Many people learn the nine values by heart. You never need to. Anyone who can draw the two triangles can rebuild the whole table in a minute, and the drawing also tells you why each value is what it is.
The procedure
Why one triangle is enough
The sine, cosine and tangent of an acute angle are ratios of two sides of a right triangle that contains it. Every right triangle with a 30^\circ angle has angles 30^\circ, 60^\circ and 90^\circ, since the angles of a triangle add to 180^\circ. So any two such triangles have two angles equal and are similar by AA similarity. Similar triangles have their sides in proportion, which means the ratio of two sides is the same in both: a triangle twice as big has both sides doubled, and the doubling cancels.
That is the whole license for this lesson. The value of \sin 30^\circ does not depend on which 30-60-90 triangle you measure, so measure the one with the simplest numbers. The same goes for the 45-45-90 triangle.
The two triangles
The 45-45-90 triangle. Cut a square of side 1 along a diagonal. Each half has two legs of length 1 and two angles of 45^\circ. The hypotenuse comes from the Pythagorean theorem:
The 30-60-90 triangle. Take an equilateral triangle of side 2 and draw the segment from one vertex to the middle of the opposite side. Each half has a hypotenuse of 2, a whole side of the equilateral triangle, and a short leg of 1, half the base. The top angle of 60^\circ was split in two, so the short leg faces an angle of 30^\circ, and the angle at the base is still 60^\circ. The long leg \ell comes from the Pythagorean theorem:
Side 2 is chosen so that the short leg is 1 rather than \displaystyle \frac{1}{2}. Both triangles, and the proofs that their sides are always in these ratios, are in the special right triangles.
One fact settles which leg is which if you redraw the 30-60-90 triangle from memory: in any triangle the larger angle faces the longer side. Since \displaystyle 1 < \sqrt{3} < 2, the short leg 1 faces the 30^\circ angle, the long leg \displaystyle \sqrt{3} faces the 60^\circ angle, and the hypotenuse 2 faces the right angle.
Reading the values
Fix the angle you want. Name the three sides from it as opposite, adjacent and hypotenuse: the opposite side is the one that does not touch the angle, the hypotenuse is the one facing the right angle, and the adjacent side is the other one touching the angle. Then form the ratios.
At 30^\circ the opposite side is the short leg 1, the adjacent side is the long leg \displaystyle \sqrt{3}, and the hypotenuse is 2:
The tangent has a root in its denominator. Multiplying top and bottom by \displaystyle \sqrt{3} multiplies by \displaystyle \frac{\sqrt{3}}{\sqrt{3}} = 1, which leaves the value alone and rationalizes the denominator:
The two forms are the same number, and both are correct. The rationalized form is the usual way to give a final answer, and it is the form in the table below.
Now stand at the 60^\circ angle of the same triangle. Nothing in the triangle moves. Only the viewpoint changes: the side that was opposite is now adjacent, the side that was adjacent is now opposite, and the hypotenuse is the hypotenuse from either angle.
With the 45^\circ row worked out in Example 1, the table is complete:
| \theta | \sin\theta | \cos\theta | \tan\theta |
|---|---|---|---|
| 30^\circ | \displaystyle \frac{1}{2} | \displaystyle \frac{\sqrt{3}}{2} | \displaystyle \frac{\sqrt{3}}{3} |
| 45^\circ | \displaystyle \frac{\sqrt{2}}{2} | \displaystyle \frac{\sqrt{2}}{2} | 1 |
| 60^\circ | \displaystyle \frac{\sqrt{3}}{2} | \displaystyle \frac{1}{2} | \displaystyle \sqrt{3} |
Three patterns that check the table
Each of these has a reason behind it, so each one catches a slip.
Sine and cosine swap between 30^\circ and 60^\circ. \sin 30^\circ = \cos 60^\circ and \cos 30^\circ = \sin 60^\circ, because the two angles sit in the same triangle and the side opposite one is the side adjacent to the other. The two angles add to 90^\circ, so they are complementary, and this is one case of the cofunction identities: \sin\theta = \cos(90^\circ - \theta) for every acute angle \theta. The angle 45^\circ is its own complement, which is why its sine and cosine are equal.
Sine climbs and cosine falls. Down the sine column the values are \displaystyle \frac{1}{2} < \frac{\sqrt{2}}{2} < \frac{\sqrt{3}}{2}. Picture a right triangle with hypotenuse 1 and let its angle open: the opposite side grows and the adjacent side shrinks. So a larger acute angle always has a larger sine and a smaller cosine. The numerators of the sine column, \displaystyle \sqrt{1}, \displaystyle \sqrt{2} and \displaystyle \sqrt{3}, make a tidy pattern for these three angles, but it is only a pattern of these three; it does not continue to other angles.
Tangent is sine over cosine, and \sin^2\theta + \cos^2\theta = 1. Dividing \displaystyle \frac{\text{opp}}{\text{hyp}} by \displaystyle \frac{\text{adj}}{\text{hyp}} cancels the hypotenuse and leaves \displaystyle \frac{\text{opp}}{\text{adj}}, so
For instance \displaystyle \tan 60^\circ = \frac{\sqrt{3}}{2} \div \frac{1}{2} = \sqrt{3}. The tangent column is never a separate thing to remember. And since \text{opp}^2 + \text{adj}^2 = \text{hyp}^2, dividing by \text{hyp}^2 gives \sin^2\theta + \cos^2\theta = 1, which every row must satisfy: at 30^\circ, \displaystyle \frac{1}{4} + \frac{3}{4} = 1. That is the Pythagorean identity, seen here in three cases.
The same angles in radians
In radian measure a straight angle, 180^\circ, is \pi. So 30^\circ is \displaystyle \frac{\pi}{6}, 45^\circ is \displaystyle \frac{\pi}{4} and 60^\circ is \displaystyle \frac{\pi}{3}. These are new names for the same angles, not new angles, so every value carries over: \displaystyle \sin\frac{\pi}{6} = \sin 30^\circ = \frac{1}{2}.
Worked examples
The three values at 45^\circ
Find \sin 45^\circ, \cos 45^\circ and \tan 45^\circ exactly.
SolutionUse half of a square of side 1: legs 1 and 1, hypotenuse \displaystyle \sqrt{2}. From either 45^\circ angle one leg is opposite and the other is adjacent. Both are 1, so it does not matter which is which.
The adjacent leg is also 1, so \displaystyle \cos 45^\circ = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2} as well. The tangent is leg over leg: \displaystyle \tan 45^\circ = \frac{1}{1} = 1.
Check. \displaystyle \sin^2 45^\circ + \cos^2 45^\circ = \frac{2}{4} + \frac{2}{4} = 1, and the sine divided by the cosine is 1, which is the tangent.
The same values in a larger, turned triangle
Triangle PQR has its right angle at R, the hypotenuse PQ = 10, PR = 5 and \angle P = 60^\circ. Find QR, then \sin P, \cos P and \tan P from the lengths, and then \sin Q.
The third side comes from the Pythagorean theorem, with PQ as the hypotenuse:
Name the sides from P. The side not touching P is QR, so that is the opposite side. Two sides touch P: PQ is the hypotenuse, because it faces the right angle, so the adjacent side is PR. The fact that PQ lies along the bottom of the picture does not make it adjacent in any special sense; only the angle decides.
These are exactly the 60^\circ row. They had to be: this triangle is the triangle with sides 1, \displaystyle \sqrt{3} and 2 scaled by 5, and the 5 cancels in every ratio.
For Q, the angles add to 180^\circ, so \angle Q = 180^\circ - 90^\circ - 60^\circ = 30^\circ. The side opposite Q is PR, so \displaystyle \sin Q = \frac{PR}{PQ} = \frac{5}{10} = \frac{1}{2}, the value of \sin 30^\circ. It is also the value of \cos P, the swap between complementary angles.
Expressions built from the values
Evaluate exactly: (a) \sin^2 30^\circ + \cos^2 30^\circ \qquad (b) 2\sin 30^\circ \cos 30^\circ \qquad (c) \tan 60^\circ - \tan 30^\circ
SolutionReplace each value by its exact form from the table, then do the arithmetic with the roots kept as roots.
(a) \sin^2 30^\circ means (\sin 30^\circ)^2, the square of the value, not the sine of 30^2 degrees.
as the Pythagorean identity says it must be.
(b) Multiply the three factors:
That is \sin 60^\circ, the sine of twice the angle. It is not 2\sin 30^\circ, which is 1: doubling an angle does not double its sine. The match with 2\sin 30^\circ\cos 30^\circ is no accident; it is one case of an identity met later, among the trigonometric identities.
(c) Write \displaystyle \sqrt{3} over the denominator 3 so the two terms can be combined:
The rationalized form of \tan 30^\circ is what made the subtraction easy.
From a value back to the angle
Find the acute angle \theta in each case: (a) \displaystyle \cos\theta = \frac{\sqrt{3}}{2} \qquad (b) \tan\theta = 1 \qquad (c) \displaystyle \sin\theta = \frac{1}{\sqrt{2}} \qquad (d) \displaystyle \tan\theta = \frac{\sqrt{3}}{3}
SolutionRead the table backwards, in the right column. That a match in the table is the answer, and not just one answer, needs a reason. As an acute angle opens, with the hypotenuse held at 1, the opposite side only grows and the adjacent side only shrinks, so the sine only increases, the cosine only decreases, and the tangent, opposite over adjacent, only increases. Each value is therefore taken by at most one acute angle.
(a) \displaystyle \frac{\sqrt{3}}{2} sits in the cosine column at 30^\circ, so \theta = 30^\circ. It also appears in the sine column at 60^\circ, which is why the column matters.
(b) \tan 45^\circ = 1, so \theta = 45^\circ.
(c) \displaystyle \frac{1}{\sqrt{2}} is not in the table as written. Rationalized it is \displaystyle \frac{\sqrt{2}}{2}, which is \sin 45^\circ, so \theta = 45^\circ.
(d) \displaystyle \tan 30^\circ = \frac{\sqrt{3}}{3}, so \theta = 30^\circ.
For a value that is not in the table, such as \sin\theta = 0.3, the inverse ratios take over.
Lines at 30^\circ, 45^\circ and 60^\circ
A line through the origin rises to the right at an angle of 30^\circ to the positive x-axis. Find its slope and its equation. Do the same for 45^\circ and 60^\circ.
SolutionTake any point (x, y) on the line with x > 0 and drop a perpendicular from it to the x-axis. That makes a right triangle with its 30^\circ angle at the origin: the rise y is the side opposite the angle and the run x is the side adjacent to it. The slope is rise over run, which is opposite over adjacent:
The line passes through the origin, so its equation is \displaystyle y = \frac{\sqrt{3}}{3}x. In the same way the line at 45^\circ has slope \tan 45^\circ = 1 and equation y = x, and the line at 60^\circ has slope \displaystyle \tan 60^\circ = \sqrt{3} and equation \displaystyle y = \sqrt{3}\,x. In general, a line rising at an acute angle \theta to the x-axis has slope \tan\theta.
Check the first line with the point \displaystyle (3, \sqrt{3}): its distance from the origin is \displaystyle \sqrt{9 + 3} = \sqrt{12} = 2\sqrt{3}, so the triangle has sides \displaystyle \sqrt{3}, 3 and \displaystyle 2\sqrt{3}, which is \displaystyle 1 : \sqrt{3} : 2 scaled by \displaystyle \sqrt{3}, a 30-60-90 triangle as it should be.
The area of a triangle with a 30^\circ angle
Triangle ABC has AB = 10, AC = 8 and \angle A = 30^\circ. Find its area.
The area of a triangle is half its base times its height. Take AB = 10 as the base. The height is the length h of the perpendicular CD from C to AB, which is not given. But triangle ADC has a right angle at D, the angle 30^\circ at A and the hypotenuse AC = 8, and CD is the side opposite A. So
and the area is
The foot D does lie on the base, as drawn, because \displaystyle AD = 8\cos 30^\circ = 4\sqrt{3}, which is less than 10. With an angle such as 40^\circ the same steps would only give a decimal; the special angle is what makes the area exact. Done with letters instead of numbers, this is the area from two sides and the included angle.
The same angles in radians
Evaluate exactly: (a) \displaystyle \cos\frac{\pi}{4} \qquad (b) \displaystyle \tan\frac{\pi}{6} \qquad (c) \displaystyle \sin\frac{\pi}{3}\cos\frac{\pi}{6} - \cos\frac{\pi}{3}\sin\frac{\pi}{6}
SolutionSince \pi radians is 180^\circ, \displaystyle \frac{\pi}{4} = 45^\circ, \displaystyle \frac{\pi}{6} = 30^\circ and \displaystyle \frac{\pi}{3} = 60^\circ. Renaming the angle does not change it, so the table applies.
(a) \displaystyle \cos\frac{\pi}{4} = \cos 45^\circ = \frac{\sqrt{2}}{2}.
(b) \displaystyle \tan\frac{\pi}{6} = \tan 30^\circ = \frac{\sqrt{3}}{3}.
(c) Replace all four values, then multiply before subtracting:
The answer is \displaystyle \sin\frac{\pi}{6}, and \displaystyle \frac{\pi}{3} - \frac{\pi}{6} = \frac{\pi}{6}. That too is no coincidence: the expression is a case of the formula for the sine of a difference of two angles, another of the identities met later.
Boundaries
Only three angles. The two triangles give exact values at 30^\circ, 45^\circ and 60^\circ, and nowhere else. For an angle such as 40^\circ they say nothing, and a calculator’s decimal is an approximation.
Acute angles only, for now. Sine, cosine and tangent as ratios in a right triangle need an angle strictly between 0^\circ and 90^\circ. The values at 0^\circ and 90^\circ, and at angles beyond, come from the unit circle in exact values on the unit circle, where these same two triangles return as the triangles of reference angles.
The degree sign matters. \displaystyle \sin 30^\circ = \frac{1}{2}, but \sin 30 with no degree sign means the sine of 30 radians, a different angle with a different sine. In radians the angle of this lesson is \displaystyle \frac{\pi}{6}.
The angle decides opposite and adjacent, not the picture. Turning or resizing the triangle changes nothing (Example 2). Moving to the other acute angle swaps the two legs.
The value is not proportional to the angle. \displaystyle \sin 60^\circ = \frac{\sqrt{3}}{2}, not 2\sin 30^\circ = 1. And \sin^2 30^\circ is (\sin 30^\circ)^2, never the sine of 30^2 = 900 degrees.
Exact until the end. \displaystyle \frac{\sqrt{3}}{2} is exact; 0.866 is not. Round only the final answer, if at all, so that later steps do not carry the rounding error forward.
Where this goes next
The reciprocal ratios in secant, cosecant and cotangent come from flipping the table: \displaystyle \sec 60^\circ = \frac{1}{\cos 60^\circ} = 2. When a right triangle has a special angle, solving right triangles gives exact sides by using these values instead of decimals, and inverse trigonometric ratios run the table backwards for any value at all.
The largest step is the unit circle. Give either triangle a hypotenuse of 1 and place its angle at the origin: the far corner is the point (\cos\theta, \sin\theta), so at 30^\circ it is \displaystyle \left(\frac{\sqrt{3}}{2}, \frac{1}{2}\right). With the values on the axes and reference angles, this one table gives the exact values at 120^\circ, 135^\circ, 210^\circ and the rest of the twelve special points on the circle. The swap between 30^\circ and 60^\circ grows into the cofunction identities, \sin^2\theta + \cos^2\theta = 1 into the Pythagorean identities, and the coincidences in Examples 3 and 7 into the trigonometric identities.
Exercises
Read from the triangles
Find each exact value, with any denominator rationalized.
- \sin 60^\circ
- \cos 45^\circ
- \tan 30^\circ
- \tan 60^\circ
From a triangle with lengths
- A right triangle has hypotenuse 12 and one leg 6. Let \theta be the angle opposite the leg of length 6. Find the other leg, then \sin\theta, \cos\theta and \tan\theta, and name \theta.
- Each leg of a right isosceles triangle is 7. Find the hypotenuse and the sine of either acute angle.
- In triangle ABC the right angle is at C, \angle A = 60^\circ and AC = 4. Find AB and BC, then \sin B, \cos B and \tan B.
Evaluate exactly
- \sin 30^\circ + \cos 60^\circ
- \sin^2 45^\circ + \cos^2 45^\circ
- \tan 30^\circ \cdot \tan 60^\circ
- \cos^2 30^\circ - \sin^2 30^\circ
- 4\sin 60^\circ \cos 60^\circ
In radians
- \displaystyle \sin\frac{\pi}{4}
- \displaystyle \tan\frac{\pi}{3}
- \displaystyle \cos\frac{\pi}{6} - \sin\frac{\pi}{3}
Back to the angle
Find the acute angle \theta.
- \displaystyle \sin\theta = \frac{\sqrt{3}}{2}
- \displaystyle \tan\theta = \frac{\sqrt{3}}{3}
- \displaystyle \cos\theta = \frac{1}{\sqrt{2}}
Using the values
- A line through the origin rises to the right at 60^\circ to the positive x-axis. Find its slope and its equation.
- A triangle has sides of length 6 and 9 with an angle of 45^\circ between them. Find its area exactly.
- A point in the first quadrant lies on the line through the origin at 30^\circ to the positive x-axis, at distance 10 from the origin. Find its coordinates.
Reasoning
- Explain why \displaystyle \sin 30^\circ = \frac{1}{2} in every right triangle with a 30^\circ angle, however large it is.
- Explain, without working out either value, why \sin 60^\circ = \cos 30^\circ.
- Use the two triangles to explain why \tan 45^\circ = 1, and why \tan 30^\circ < 1 < \tan 60^\circ.
- Doubling 30^\circ gives 60^\circ. Show that doubling the angle does not double the sine here, and explain why the sine of an acute angle could never double every time its angle does.
- Prove that \displaystyle \tan\theta = \frac{\sin\theta}{\cos\theta} for every acute angle \theta, and check it against all three rows of the table.
Answers
- \displaystyle \frac{\sqrt{3}}{2}. In the 30-60-90 triangle the side opposite 60^\circ is the long leg \displaystyle \sqrt{3}, and the hypotenuse is 2.
- \displaystyle \frac{\sqrt{2}}{2}. The adjacent leg 1 over the hypotenuse \displaystyle \sqrt{2} is \displaystyle \frac{1}{\sqrt{2}}, rationalized.
- \displaystyle \frac{\sqrt{3}}{3}. The opposite leg 1 over the adjacent leg \displaystyle \sqrt{3} is \displaystyle \frac{1}{\sqrt{3}}, rationalized.
- \displaystyle \sqrt{3}. The opposite leg \displaystyle \sqrt{3} over the adjacent leg 1.
- The other leg is \displaystyle 6\sqrt{3}; \displaystyle \sin\theta = \frac{1}{2}, \displaystyle \cos\theta = \frac{\sqrt{3}}{2}, \displaystyle \tan\theta = \frac{\sqrt{3}}{3}, so \theta = 30^\circ. The leg is \displaystyle \sqrt{144 - 36} = \sqrt{108} = 6\sqrt{3}, and the ratios are \displaystyle \frac{6}{12}, \displaystyle \frac{6\sqrt{3}}{12} and \displaystyle \frac{6}{6\sqrt{3}}.
- The hypotenuse is \displaystyle 7\sqrt{2} and the sine is \displaystyle \frac{\sqrt{2}}{2}. The hypotenuse is \displaystyle \sqrt{49 + 49} = 7\sqrt{2}, and \displaystyle \frac{7}{7\sqrt{2}} = \frac{1}{\sqrt{2}}: the 7 cancels, as similarity says it must.
- AB = 8, \displaystyle BC = 4\sqrt{3}; \displaystyle \sin B = \frac{1}{2}, \displaystyle \cos B = \frac{\sqrt{3}}{2}, \displaystyle \tan B = \frac{\sqrt{3}}{3}. Here \angle B = 30^\circ, so AC, opposite B, is the short leg; the hypotenuse is twice it and the long leg \displaystyle \sqrt{3} times it. From B the opposite side is 4, the adjacent side \displaystyle 4\sqrt{3} and the hypotenuse 8.
- 1. \displaystyle \frac{1}{2} + \frac{1}{2} = 1.
- 1. \displaystyle \left(\frac{\sqrt{2}}{2}\right)^2 + \left(\frac{\sqrt{2}}{2}\right)^2 = \frac{2}{4} + \frac{2}{4}.
- 1. \displaystyle \frac{\sqrt{3}}{3} \cdot \sqrt{3} = \frac{3}{3}. In one 30-60-90 triangle the two tangents are \displaystyle \frac{1}{\sqrt{3}} and \displaystyle \frac{\sqrt{3}}{1}, the same two legs divided in opposite orders, so they are reciprocals.
- \displaystyle \frac{1}{2}. \displaystyle \frac{3}{4} - \frac{1}{4} = \frac{2}{4}.
- \displaystyle \sqrt{3}. \displaystyle 4 \cdot \frac{\sqrt{3}}{2} \cdot \frac{1}{2} = \frac{4\sqrt{3}}{4}.
- \displaystyle \frac{\sqrt{2}}{2}, since \displaystyle \frac{\pi}{4} is 45^\circ.
- \displaystyle \sqrt{3}, since \displaystyle \frac{\pi}{3} is 60^\circ.
- 0. \displaystyle \frac{\pi}{6} is 30^\circ and \displaystyle \frac{\pi}{3} is 60^\circ, and \displaystyle \cos 30^\circ = \sin 60^\circ = \frac{\sqrt{3}}{2}.
- \theta = 60^\circ. \displaystyle \frac{\sqrt{3}}{2} is in the sine column at 60^\circ.
- \theta = 30^\circ. \displaystyle \frac{\sqrt{3}}{3} is in the tangent column at 30^\circ.
- \theta = 45^\circ. Rationalized, \displaystyle \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}, which is \cos 45^\circ.
- Slope \displaystyle \sqrt{3}, equation \displaystyle y = \sqrt{3}\,x. Rise over run is opposite over adjacent for the 60^\circ angle, which is \tan 60^\circ, and the line passes through the origin.
- \displaystyle \frac{27\sqrt{2}}{2}. Take the side 9 as the base. The height is the side opposite 45^\circ in a right triangle with hypotenuse 6, so it is \displaystyle 6\sin 45^\circ = 6 \cdot \frac{\sqrt{2}}{2} = 3\sqrt{2}, and the area is \displaystyle \frac{1}{2} \cdot 9 \cdot 3\sqrt{2}.
- \displaystyle (5\sqrt{3}, 5). Drop a perpendicular to the x-axis: the right triangle has hypotenuse 10 and a 30^\circ angle at the origin, so \displaystyle x = 10\cos 30^\circ = 5\sqrt{3} and y = 10\sin 30^\circ = 5.
- Every right triangle with a 30^\circ angle has angles 30^\circ, 60^\circ and 90^\circ, so any two of them are similar by AA. Their sides are in proportion, so the side opposite 30^\circ divided by the hypotenuse is the same number in all of them. In half of an equilateral triangle of side 2 that number is \displaystyle \frac{1}{2}.
- In a right triangle with angles 30^\circ and 60^\circ, the side opposite the 60^\circ angle is the side adjacent to the 30^\circ angle, and the hypotenuse is shared. So \sin 60^\circ and \cos 30^\circ are the same side divided by the same side.
- In the 45-45-90 triangle the legs are equal, so opposite over adjacent is 1. In the 30-60-90 triangle, seen from 30^\circ the opposite leg is the shorter one, so the ratio is less than 1; seen from 60^\circ the opposite leg is the longer one, so the ratio is more than 1. The larger angle faces the longer leg.
- \displaystyle \sin 60^\circ = \frac{\sqrt{3}}{2}, about 0.87, while 2\sin 30^\circ = 1. The sine of an acute angle is a leg over the hypotenuse, and a leg is always shorter than the hypotenuse, so every such sine is less than 1. If doubling the angle always doubled the sine, then \sin 60^\circ would be \displaystyle 2 \cdot \frac{1}{2} = 1: a leg as long as the hypotenuse, which no right triangle has.
- \displaystyle \frac{\sin\theta}{\cos\theta} = \frac{\text{opp}/\text{hyp}}{\text{adj}/\text{hyp}} = \frac{\text{opp}}{\text{adj}} = \tan\theta, since the hypotenuse cancels, and this uses nothing about \theta except that it is an acute angle of a right triangle. Rows: \displaystyle \frac{1}{2} \div \frac{\sqrt{3}}{2} = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}, \displaystyle \frac{\sqrt{2}}{2} \div \frac{\sqrt{2}}{2} = 1 and \displaystyle \frac{\sqrt{3}}{2} \div \frac{1}{2} = \sqrt{3}.