Sine, cosine and tangent
Two right triangles share the same sharp angle but are different sizes. What do their sides have in common, and how can that be turned into a number that belongs to the angle?
Where you are 7 in Right triangle trigonometry
Why this exists
A right triangle has three sides and three angles, and before this lesson they are linked only among themselves. The Pythagorean theorem ties the three sides to each other, and the triangle angle sum ties the three angles to each other. But nothing yet joins a side to an angle. Knowing that a ladder leans at 70^\circ to the ground says nothing, so far, about how high up the wall it reaches.
The join comes from similar triangles. Every right triangle with a 70^\circ angle has the same shape, whatever its size, so the ratio of any two of its sides is the same number in all of them. That number belongs to the angle. Three of these ratios get names, sine, cosine and tangent, and with them a question about an angle can be turned into a question about lengths.
The lesson is taught in two parts. The first part defines the three ratios, reads them off triangles, and shows why each one depends only on the angle. The second part shows how the three are tied to each other: the sine of one acute angle is the cosine of the other, the tangent is the sine divided by the cosine, and the squares of the sine and the cosine add to 1.
Definition
Take an acute angle A and any right triangle that has A as one of its two acute angles. Name the sides from A, as in opposite, adjacent and hypotenuse. The three trigonometric ratios of A are
where “opposite” means the length of the side opposite A, and so on. They are read “the sine of A”, “the cosine of A” and “the tangent of A”; \sin, \cos and \tan are abbreviations, and are read aloud as the full words.
In the usual lettering, \triangle ABC has its right angle at C, and each side takes the small letter of the vertex across from it: a = BC, b = AC and c = AB, as in Figure 1. From \angle A, the side a is opposite, b is adjacent and c is the hypotenuse, so
When the angle is given by its measure, the ratio is written with the measure: \sin 35^\circ is the sine of a 35^\circ angle. That this names one number, the same for every right triangle with a 35^\circ angle in it, is the first Result below. Without it, the definition would not make sense, because “any right triangle” could give different answers.
Two points about the notation. \sin A is one number, a length divided by a length; it is not \sin times A, and \sin on its own means nothing. And a length divided by a length has no units: if both sides are measured in centimeters, the centimeters cancel, so a sine, cosine or tangent is a plain number.
Intuition
Three ways to say how steep
Picture a straight ramp rising from level ground at angle A, with the ramp as the hypotenuse, the ground under it as the adjacent side and the height of its top as the opposite side.
- The tangent says how much the ramp rises for each unit it runs forward: rise over run, the same thing as the Slope of the ramp’s line.
- The sine says how much the ramp rises for each unit of ramp you walk along.
- The cosine says how far forward you get for each unit of ramp you walk along.
A steeper ramp rises more and gets less far forward, so it has a larger tangent, a larger sine and a smaller cosine.
Size does not matter
Figure 2 shows one angle at A with two right triangles in it, a small one, \triangle ADE, and a large one, \triangle ACB. The large one is the small one scaled up: every side is longer by the same factor. Scaling changes the sides but not the shape, and when one side is divided by another the factor cancels. That is why \sin A can be a property of the angle alone.
How the ratios move as the angle opens
Since the size does not matter, give the ramp length 1. Then the sine is simply the height of its top and the cosine is how far forward the top is. Now let the angle grow from nearly 0^\circ to nearly 90^\circ. The top of the ramp swings up along a quarter of a circle. Its height, the sine, climbs from nearly 0 to nearly 1. Its forward distance, the cosine, falls from nearly 1 to nearly 0. The tangent, height divided by forward distance, starts near 0 and grows without any limit as the ramp stands up toward vertical.
Results
The ratios belong to the angle
The ratios depend only on the angle. If two right triangles each have an acute angle of the same measure, then the sine of that angle is the same in both triangles, and so are its cosine and its tangent.
Sine and cosine swap
In right \triangle ABC with the right angle at C, the two acute angles are complementary, m\angle A + m\angle B = 90^\circ, and
Written with measures, for every acute angle \theta:
The “co” in cosine is short for complement: the cosine of an angle is the sine of its complement.
Tangent is sine over cosine
For every acute angle A,
The squares add to one
Write \sin^2 A for (\sin A)^2 and \cos^2 A for (\cos A)^2. For every acute angle A,
The sizes of the ratios
For every acute angle A,
The tangent has no upper limit. As the angle grows, its sine grows, its cosine shrinks and its tangent grows. So two different acute angles never have the same sine, never the same cosine and never the same tangent.
Why it is true
The ratios depend only on the angle
Let \triangle ABC and \triangle A'B'C' be right triangles with their right angles at C and C', and with m\angle A = m\angle A'. The right angles are congruent and so are the angles at A and A', so the triangles are similar by AA similarity. The sides of similar triangles are in proportion, so there is one positive number k with
Then the sine of the angle in the second triangle is
The factor k cancels. The cosine and the tangent work the same way, each a side divided by a side. Nothing in the argument depended on the size of either triangle, so every right triangle with that angle gives the same three numbers.
Sine and cosine swap
The three angles of a triangle add to 180^\circ and the right angle takes 90^\circ, so m\angle A + m\angle B = 90^\circ.
The side a = BC is opposite A. Looked at from B, the same side is one of the two sides of \angle B and is not the hypotenuse, so it is the side adjacent to B. In the same way b = AC is adjacent to A and opposite B. The hypotenuse c is the hypotenuse from both. So
and \displaystyle \tan A = \frac{a}{b} while \displaystyle \tan B = \frac{b}{a}, which is its reciprocal. Since m\angle B = 90^\circ - m\angle A, this is the same as \sin\theta = \cos(90^\circ - \theta).
Tangent, and the squares
Dividing the sine by the cosine, the c cancels:
The Pythagorean theorem says a^2 + b^2 = c^2. Divide every term by c^2, which is allowed because c is a length and so is not 0:
and that is \sin^2 A + \cos^2 A = 1.
The sizes
All three ratios are lengths divided by lengths, so all three are positive. The hypotenuse is longer than each leg, since a^2 = c^2 - b^2 < c^2 gives a < c, and in the same way b < c. A fraction with a positive top smaller than its bottom is less than 1, so \sin A < 1 and \cos A < 1. The tangent is one leg divided by the other, and the legs of a right triangle can be any two positive lengths, so the tangent can be any positive number.
For the growth, use the ramp of length 1 from the Intuition: put A at the center of a circle of radius 1, one side of the angle along the ground and the hypotenuse as a radius. The sine is the height of the radius’s far end and the cosine is how far forward that end is. As the radius turns up from the ground toward vertical, its end climbs the quarter circle, so the height grows and the forward distance shrinks. The tangent, a growing number divided by a shrinking positive number, grows. This is a picture argument, not a full proof: it takes from the drawing that a point climbing a quarter circle gets higher and comes back toward the upright. The unit circle sets the same picture on coordinates, where the sine and cosine are read off a circle of radius 1.
Worked examples
Examples 1 to 4 belong to the first part of the lesson, Examples 5 to 9 to the second.
Reading six ratios off one triangle
In Figure 3, \triangle PQR has its right angle at R, with PR = 45, QR = 28 and PQ = 53. Find the sine, cosine and tangent of \angle P and of \angle Q.
Name the sides from P first. PQ is across from the right angle, so it is the hypotenuse, 53. QR does not touch P, so it is the opposite side, 28. PR is the leg that touches P, so it is the adjacent side, 45. By the definitions,
Now stand at Q. The hypotenuse is still PQ = 53, but the other two names trade places: PR does not touch Q, so it is opposite, 45, and QR is adjacent, 28.
Notice that \sin P = \cos Q, \cos P = \sin Q, and \tan Q is \tan P turned upside down. The second part of the lesson shows that this always happens.
Check. 28^2 + 45^2 = 784 + 2025 = 2809 = 53^2, so these sides do make a right triangle with PQ as the hypotenuse. Every sine and cosine found is less than 1, as it must be.
A side to find first
In Figure 4, \triangle KLM has its right angle at L, with hypotenuse KM = 7 and LM = 3. Find \sin M, \cos M and \tan M.
From M, the hypotenuse is KM = 7 and the adjacent side is LM = 3, the leg touching M. The opposite side is KL, which is not given, so find it first with the Pythagorean theorem:
Only the positive square root is taken, because KL is a length, and \displaystyle \sqrt{40} simplifies because 40 = 4 \cdot 10. Now
A trigonometric ratio need not be a fraction of whole numbers; most are irrational.
Check. \displaystyle 2\sqrt{10} = \sqrt{40} lies between 6 and 7, so \sin M is less than 1, and \tan M is more than 1 because the opposite side is longer than the adjacent one. The picture agrees: \angle M is the larger acute angle.
The same angle in two triangles
In Figure 2, \overline{DE} and \overline{CB} are both perpendicular to \overline{AC}, with AD = 8, DE = 6 and AC = 20. Find \sin A, \cos A and \tan A from \triangle ADE. Then use the tangent to find BC, find AB, and work out \sin A again from \triangle ACB.
SolutionIn \triangle ADE the right angle is at D, so AE is the hypotenuse: AE^2 = 8^2 + 6^2 = 100 and AE = 10. From A, the opposite side is DE = 6 and the adjacent side is AD = 8:
The angle at A is also an angle of the right triangle ACB, so by the first Result its tangent there is the same \displaystyle \frac{3}{4}. In \triangle ACB the opposite side is BC and the adjacent side is AC = 20:
Then AB^2 = 20^2 + 15^2 = 625, so AB = 25, and from the large triangle
the same as before. The large triangle is the small one scaled by \displaystyle \frac{20}{8} = \frac{5}{2}: \displaystyle 6 \cdot \frac{5}{2} = 15 and \displaystyle 10 \cdot \frac{5}{2} = 25. That common factor is exactly what cancels in the proof of the first Result.
From one ratio to the other two
An acute angle \theta has \displaystyle \tan\theta = \frac{7}{24}. Find \sin\theta and \cos\theta.
SolutionTake any right triangle with \theta as one of its angles. The tangent is opposite over adjacent, so the two legs are in the ratio 7 : 24: the opposite side is 7k and the adjacent side is 24k for some positive number k. The hypotenuse is
since 49k^2 + 576k^2 = 625k^2. So
The k cancels, which is the first Result again: which triangle was used does not matter. The simplest choice is k = 1, a triangle with sides 7, 24 and 25.
Reading one angle’s ratios from the other’s
The acute angles of a right triangle measure 34^\circ and 56^\circ. A calculator, set to degrees, gives \sin 34^\circ \approx 0.5592 and \cos 34^\circ \approx 0.8290. Without the calculator, find \cos 56^\circ, \sin 56^\circ and \tan 56^\circ.
Solution34 + 56 = 90, so the angles are complementary, and the side opposite one is adjacent to the other. The sine and cosine swap:
The tangent is the sine divided by the cosine:
The two inputs were rounded to four decimal places, so the quotient is only trusted to two. A calculator gives \tan 56^\circ \approx 1.4826, which agrees.
An equation from the swap
Find x if \sin(3x - 5)^\circ = \cos(2x + 15)^\circ and both angles are acute.
SolutionThe cosine of an angle is the sine of its complement, so the right side can be rewritten as a sine:
Now \sin(3x - 5)^\circ = \sin(75 - 2x)^\circ. Two different acute angles never have the same sine, because the sine grows as the angle grows, so the two angles are equal:
Check. 3(16) - 5 = 43 and 2(16) + 15 = 47. Both are acute, and 43 + 47 = 90, so they are complementary and \sin 43^\circ = \cos 47^\circ, as the swap requires.
From the sine to the cosine and tangent
An acute angle \theta has \displaystyle \sin\theta = \frac{2}{5}. Find \cos\theta and \tan\theta without drawing a triangle.
SolutionThe squares of the sine and cosine add to 1, so
The cosine of an acute angle is positive, so take the positive square root: \displaystyle \cos\theta = \frac{\sqrt{21}}{5}. Then the tangent is the sine over the cosine:
The last step rationalizes the denominator, multiplying top and bottom by \displaystyle \sqrt{21}.
Check. With a triangle after all: opposite 2 and hypotenuse 5 give an adjacent side of \displaystyle \sqrt{25 - 4} = \sqrt{21}, so \displaystyle \cos\theta = \frac{\sqrt{21}}{5} and \displaystyle \tan\theta = \frac{2}{\sqrt{21}}, as found.
Comparing without a calculator
Which is larger, \sin 40^\circ or \cos 40^\circ? Is \tan 40^\circ more or less than 1?
SolutionThe complement of 40^\circ is 50^\circ, so \cos 40^\circ = \sin 50^\circ. The sine grows as the angle grows, and 50^\circ is larger than 40^\circ, so \sin 50^\circ > \sin 40^\circ. That is, \cos 40^\circ > \sin 40^\circ.
The tangent is \displaystyle \frac{\sin 40^\circ}{\cos 40^\circ}, a positive number divided by a larger positive number, so \tan 40^\circ < 1.
Check. A calculator gives \sin 40^\circ \approx 0.6428, \cos 40^\circ \approx 0.7660 and \tan 40^\circ \approx 0.8391. The same argument shows that whenever an acute angle is smaller than its complement, its cosine is larger than its sine and its tangent is less than 1.
The angle a line makes with the x-axis
In Figure 5, the line through O(0, 0) and P(12, 5) makes an angle \theta with the positive x-axis. Find \tan\theta, \sin\theta and \cos\theta, and compare \tan\theta with the slope of the line.
Drop a perpendicular from P to the x-axis; it meets the axis at Q(12, 0). Then \triangle OQP has its right angle at Q. From \theta at O, the adjacent side is OQ = 12, the opposite side is QP = 5, and the hypotenuse is
So
The slope of the line is \displaystyle \frac{5 - 0}{12 - 0} = \frac{5}{12}, the same as \tan\theta. That is no accident: the rise and run of the line are the opposite and adjacent sides of the triangle. For any line with positive slope m, a run of 1 gives a rise of m, so \displaystyle \tan\theta = \frac{m}{1} = m. The slope of a line is the tangent of the angle it makes with the x-axis, which is also why slope triangles drawn anywhere on one line all give the same slope.
Boundaries
Acute angles only, for now. The two angles of a right triangle other than the right angle are both between 0^\circ and 90^\circ, so these definitions give \sin A only when 0^\circ < m\angle A < 90^\circ. No right triangle has a second 90^\circ angle or a 120^\circ angle, so \sin 90^\circ and \sin 120^\circ mean nothing yet. Sine and cosine on the unit circle give them a meaning that agrees with this lesson on acute angles.
There must be a right angle. A triangle with no right angle has no hypotenuse, and one of its sides divided by its longest side is not the sine of any of its angles in general. Triangles without a right angle are handled by the law of sines.
The ratio is not proportional to the angle. Doubling an angle does not double its sine: \sin 20^\circ \approx 0.3420, but \sin 40^\circ \approx 0.6428, not 0.6840. It could not be proportional, since every sine is below 1 while the angles run all the way up to 90^\circ.
The names belong to the angle, not the side. In one triangle the same ratio \displaystyle \frac{a}{c} is \sin A and \cos B. A ratio means nothing until it says which angle it is taken from.
A calculator must be set to degrees. The angles here are measured in degrees. Calculators can also measure angles in radians, and in that setting \sin 35 is the sine of a quite different angle.
Going back from the ratio to the angle. Since the sine grows with the angle, an acute angle is fixed by its sine, and likewise by its cosine or its tangent. Finding the angle from the ratio is the job of the inverse ratios.
Where this goes next
For most angles the three ratios are irrational and come from a calculator, but for 30^\circ, 45^\circ and 60^\circ they can be read exactly off two special triangles: trig values of the special right triangles.
The main use of the ratios is the one the opening asked for. Given an angle and one side of a right triangle, a ratio links the side you know to the side you want, which is solving right triangles, and with a second sighting it measures heights and distances nobody can reach, as in angles of elevation and depression. Run the other way, from a ratio back to the angle, it becomes the inverse ratios.
The three ratios turned upside down get names too, secant, cosecant and cotangent. The swap of sine and cosine between complementary angles grows into the cofunction identities, and \sin^2 A + \cos^2 A = 1 is the first of the Pythagorean identities.
Finally, the ramp of length 1 in the Intuition is the start of the unit circle, where the cosine and sine become the coordinates of a point, and where angles of any size, not just acute ones, get a sine and a cosine.
Exercises
Exercises 1 to 12 go with the first part of the lesson and 13 to 24 with the second; the reasoning group draws on both. Give exact answers unless a decimal is asked for. Every angle named \theta, A or B is acute.
Reading the ratios off a triangle
Triangle XYZ has a right angle at Y, with XY = 36, YZ = 77 and XZ = 85.
- Find \sin X, \cos X and \tan X.
- Find \sin Z, \cos Z and \tan Z.
A side to find first
- A right triangle has legs of lengths 8 and 15, and \theta is the angle opposite the leg of length 8. Find \sin\theta, \cos\theta and \tan\theta.
- A right triangle has hypotenuse 6 and one leg 2, and A is the angle opposite that leg. Find \sin A, \cos A and \tan A.
- A right triangle has legs of lengths 1 and 3, and B is the angle opposite the leg of length 3. Find \sin B, \cos B and \tan B.
One ratio given
- \displaystyle \cos\theta = \frac{8}{17}. Find \sin\theta and \tan\theta.
- \displaystyle \tan\theta = \frac{20}{21}. Find \sin\theta and \cos\theta.
- \displaystyle \sin A = \frac{3}{7}. Find \cos A and \tan A.
- \tan B = 2. Find \sin B and \cos B.
Same angle, any size
- One right triangle has legs 6 and 8; another has legs 15 and 20. In each, find the sine and the tangent of the angle opposite the shorter leg. Then explain, using similar triangles, why the answers had to match.
- In Figure 2, suppose AD = 12, AE = 15 and AC = 40. Find \sin A, \tan A and BC.
- A right triangle has sides 5, 12 and 13, and \theta is the angle opposite the side of length 5. Every side is multiplied by 4. Find \sin\theta in the new triangle, and say why \theta itself has not changed.
Complementary angles
- \sin 23^\circ \approx 0.3907. Without a calculator, give \cos 67^\circ.
- Find x if \sin(4x + 2)^\circ = \cos(2x - 2)^\circ and both angles are acute.
- \tan 35^\circ \approx 0.7002. Without a calculator, find \tan 55^\circ to three decimal places.
- In right \triangle ABC with the right angle at C, \cos A = 0.28. Find \sin B.
Using the identities
- \displaystyle \sin\theta = \frac{4}{5}. Find \cos\theta and \tan\theta.
- \displaystyle \cos\theta = \frac{1}{4}. Find \sin\theta and \tan\theta.
- \displaystyle \sin\theta = \frac{5}{9}. Find \cos\theta and \tan\theta.
- Use a calculator, set to degrees, to find \sin 35^\circ and \cos 35^\circ to four decimal places. Add their squares. Why is the answer not exactly 1?
- Show that \cos\theta \cdot \tan\theta = \sin\theta for every acute angle \theta.
Comparing, and slope
- Without a calculator, put \sin 25^\circ, \cos 25^\circ and \sin 70^\circ in order from least to greatest.
- The line through (0, 0) and (24, 7) makes an angle \theta with the positive x-axis. Find \tan\theta, \sin\theta and \cos\theta.
- The line through (1, 2) and (5, 5) makes an angle \theta with the horizontal. Find \tan\theta, \sin\theta and \cos\theta.
Reasoning
- Explain why \sin A < 1 and \cos A < 1 for every acute angle A of a right triangle.
- Ivo draws a right triangle with a 40^\circ angle and a hypotenuse of 3 cm. Lena draws one with a 40^\circ angle and a hypotenuse of 30 cm. Explain why they get the same value for \cos 40^\circ.
- Can the tangent of an acute angle be 1000? Can its sine? Explain.
- A and B are the two acute angles of a right triangle. Show that \tan A \cdot \tan B = 1.
- Show that \sin A + \cos A > 1 for every acute angle A.
- Mira claims that \sin 2A = 2\sin A. Use A = 40^\circ, and what you know about the size of a sine, to show that she is wrong.
Answers
- \displaystyle \sin X = \frac{77}{85}, \displaystyle \cos X = \frac{36}{85}, \displaystyle \tan X = \frac{77}{36}. From X, the hypotenuse is XZ = 85, the opposite side is YZ = 77 and the adjacent side is XY = 36.
- \displaystyle \sin Z = \frac{36}{85}, \displaystyle \cos Z = \frac{77}{85}, \displaystyle \tan Z = \frac{36}{77}. From Z, the opposite side is XY = 36 and the adjacent side is YZ = 77.
- \displaystyle \sin\theta = \frac{8}{17}, \displaystyle \cos\theta = \frac{15}{17}, \displaystyle \tan\theta = \frac{8}{15}. The hypotenuse is \displaystyle \sqrt{8^2 + 15^2} = \sqrt{289} = 17.
- \displaystyle \sin A = \frac{1}{3}, \displaystyle \cos A = \frac{2\sqrt{2}}{3}, \displaystyle \tan A = \frac{\sqrt{2}}{4}. The other leg is \displaystyle \sqrt{36 - 4} = \sqrt{32} = 4\sqrt{2}, so \displaystyle \sin A = \frac{2}{6}, \displaystyle \cos A = \frac{4\sqrt{2}}{6} and \displaystyle \tan A = \frac{2}{4\sqrt{2}} = \frac{1}{2\sqrt{2}}, which is \displaystyle \frac{\sqrt{2}}{4} once the denominator is rationalized.
- \displaystyle \sin B = \frac{3\sqrt{10}}{10}, \displaystyle \cos B = \frac{\sqrt{10}}{10}, \tan B = 3. The hypotenuse is \displaystyle \sqrt{1 + 9} = \sqrt{10}, so \displaystyle \sin B = \frac{3}{\sqrt{10}} and \displaystyle \cos B = \frac{1}{\sqrt{10}}; multiply top and bottom by \displaystyle \sqrt{10}.
- \displaystyle \sin\theta = \frac{15}{17}, \displaystyle \tan\theta = \frac{15}{8}. Take the adjacent side 8k and the hypotenuse 17k; the opposite side is \displaystyle \sqrt{289k^2 - 64k^2} = 15k.
- \displaystyle \sin\theta = \frac{20}{29}, \displaystyle \cos\theta = \frac{21}{29}. Take the opposite side 20k and the adjacent side 21k; the hypotenuse is \displaystyle \sqrt{400k^2 + 441k^2} = 29k.
- \displaystyle \cos A = \frac{2\sqrt{10}}{7}, \displaystyle \tan A = \frac{3\sqrt{10}}{20}. Take the opposite side 3k and the hypotenuse 7k; the adjacent side is \displaystyle \sqrt{49k^2 - 9k^2} = 2\sqrt{10}\,k, and \displaystyle \tan A = \frac{3}{2\sqrt{10}} = \frac{3\sqrt{10}}{20}.
- \displaystyle \sin B = \frac{2\sqrt{5}}{5}, \displaystyle \cos B = \frac{\sqrt{5}}{5}. Since \displaystyle \tan B = \frac{2}{1}, take the opposite side 2k and the adjacent side k; the hypotenuse is \displaystyle \sqrt{5}\,k, so \displaystyle \sin B = \frac{2}{\sqrt{5}} and \displaystyle \cos B = \frac{1}{\sqrt{5}}.
- In both, the sine is \displaystyle \frac{3}{5} and the tangent is \displaystyle \frac{3}{4}. The hypotenuses are 10 and 25, and \displaystyle \frac{6}{10} = \frac{15}{25}, \displaystyle \frac{6}{8} = \frac{15}{20}. They had to match: the sides 6, 8, 10 and 15, 20, 25 are in the ratio 2 : 5, so the triangles are similar by SSS similarity, the two angles are congruent, and congruent angles have the same ratios.
- \displaystyle \sin A = \frac{3}{5}, \displaystyle \tan A = \frac{3}{4}, BC = 30. In \triangle ADE, \displaystyle DE = \sqrt{15^2 - 12^2} = 9, so \displaystyle \sin A = \frac{9}{15} and \displaystyle \tan A = \frac{9}{12}. The tangent is the same in \triangle ACB, so \displaystyle BC = 40 \cdot \frac{3}{4}.
- Still \displaystyle \frac{5}{13}. The new sides are 20, 48 and 52, and \displaystyle \frac{20}{52} = \frac{5}{13}. Multiplying every side by 4 gives a similar triangle, and matching angles of similar triangles are congruent, so \theta is unchanged.
- \cos 67^\circ \approx 0.3907. Since 23 + 67 = 90, \cos 67^\circ = \sin 23^\circ.
- x = 15; the angles are 62^\circ and 28^\circ. The cosine is the sine of the complement, so as in Example 6, 4x + 2 = 90 - (2x - 2), giving 6x = 90.
- \tan 55^\circ \approx 1.428. 55^\circ is the complement of 35^\circ, so \displaystyle \tan 55^\circ = \frac{1}{\tan 35^\circ} \approx \frac{1}{0.7002} \approx 1.428.
- \sin B = 0.28. A and B are the two acute angles, so \sin B = \cos A.
- \displaystyle \cos\theta = \frac{3}{5}, \displaystyle \tan\theta = \frac{4}{3}. \displaystyle \cos^2\theta = 1 - \frac{16}{25} = \frac{9}{25} and the cosine is positive; then \displaystyle \tan\theta = \frac{4/5}{3/5}.
- \displaystyle \sin\theta = \frac{\sqrt{15}}{4}, \displaystyle \tan\theta = \sqrt{15}. \displaystyle \sin^2\theta = 1 - \frac{1}{16} = \frac{15}{16}; then \displaystyle \tan\theta = \frac{\sqrt{15}/4}{1/4}.
- \displaystyle \cos\theta = \frac{2\sqrt{14}}{9}, \displaystyle \tan\theta = \frac{5\sqrt{14}}{28}. \displaystyle \cos^2\theta = 1 - \frac{25}{81} = \frac{56}{81} and \displaystyle \sqrt{56} = 2\sqrt{14}; then \displaystyle \tan\theta = \frac{5}{2\sqrt{14}} = \frac{5\sqrt{14}}{28}.
- \sin 35^\circ \approx 0.5736 and \cos 35^\circ \approx 0.8192, and 0.5736^2 + 0.8192^2 \approx 1.0001. The identity is exact, but each value was rounded to four decimal places, and the squares carry those small rounding errors.
- \displaystyle \tan\theta = \frac{\sin\theta}{\cos\theta}, so \displaystyle \cos\theta \cdot \tan\theta = \cos\theta \cdot \frac{\sin\theta}{\cos\theta} = \sin\theta. Canceling is allowed because \cos\theta is positive, so not 0.
- \sin 25^\circ < \cos 25^\circ < \sin 70^\circ. \cos 25^\circ = \sin 65^\circ, and the sine grows with the angle, so \sin 25^\circ < \sin 65^\circ < \sin 70^\circ.
- \displaystyle \tan\theta = \frac{7}{24}, \displaystyle \sin\theta = \frac{7}{25}, \displaystyle \cos\theta = \frac{24}{25}. The point (24, 7) is 24 across and 7 up from the origin, at distance \displaystyle \sqrt{576 + 49} = 25.
- \displaystyle \tan\theta = \frac{3}{4}, \displaystyle \sin\theta = \frac{3}{5}, \displaystyle \cos\theta = \frac{4}{5}. From (1, 2) to (5, 5) the run is 4 and the rise is 3, and the segment has length \displaystyle \sqrt{16 + 9} = 5.
- The hypotenuse is longer than each leg: a^2 = c^2 - b^2 < c^2, so a < c, and likewise b < c. So \displaystyle \sin A = \frac{a}{c} and \displaystyle \cos A = \frac{b}{c} are positive fractions whose top is smaller than their bottom, and each is less than 1.
- Both triangles have a right angle and a 40^\circ angle, so they are similar by AA similarity, and every side of Lena’s is 10 times the matching side of Ivo’s. Adjacent over hypotenuse is then \displaystyle \frac{10b}{10c} = \frac{b}{c}: the factor 10 cancels.
- The tangent can: a right triangle with legs 1000 and 1 has tangent 1000 at the angle opposite the long leg, a steep, thin triangle. The sine cannot, since every sine is less than 1 (Exercise 25). The two legs can be in any ratio, but the hypotenuse is always longer than the opposite side.
- With legs a opposite A and b opposite B, \displaystyle \tan A = \frac{a}{b} and \displaystyle \tan B = \frac{b}{a}, so \displaystyle \tan A \cdot \tan B = \frac{a}{b} \cdot \frac{b}{a} = 1.
- By the triangle inequality, a + b > c. Dividing by the positive number c gives \displaystyle \frac{a}{c} + \frac{b}{c} > 1, which is \sin A + \cos A > 1.
- 2\sin 40^\circ \approx 2(0.6428) = 1.2856, which is more than 1. No sine is more than 1, so \sin 80^\circ cannot equal 2\sin 40^\circ; in fact \sin 80^\circ \approx 0.9848.