Cosines

Definition of cosine

The cosine of an angle is defined as the sine of the complementary angle. The complementary angle equals the the given angle subtracted from a right angle, 90°. For instance, if the angle is 30°, then its complement is 60°. Generally, for any angle t,

cos t = sin (90° – t).

Written in terms of radian measurement, this identity becomes

cos t = sin (

/2 – t).

Right triangles and cosines

Consider a right triangle ABC with a right angle at C.

As mentioned before, we'll generally use the letter a to denote the side opposite angle A, the letter b to denote the side opposite angle B, and the letter c to denote the side opposite angle C. Since the sum of the angles in a triangle equals 180°, and angle C is 90°, that means angles A and B add up to 90°, that is, they are complementary angles. Therefore the cosine of B equals the sine of A. We saw on the last page that sin A was the opposite side over the hypotenuse, that is, a/c. Hence, cos B equals a/c. In other words, the cosine of an angle in a right triangle equals the adjacent side divided by the hypotenuse:

cos =

adj

hyp

Also, cos A = sin B = b/c.

The Pythagorean identity for sines and cosines

Recall the Pythagorean theorem for right triangles. It says that

a² + b² = c²

where c is the hypotenuse. This translates very easily into a Pythagorean identity for sines and cosines. Divide both sides by c² and you get

a²/c² + b²/c² = 1.

But a²/c² = (sin A)², and b²/c² = (cos A)². In order to reduce the number of parentheses that have to be written, it is a convention that the notation sin² A is an abbreviation for (sin A)², and similarly for powers of the other trig functions. Thus, we have proven that

sin² A + cos² A = 1

when A is an acute angle. We haven't yet seen what sines and cosines of other angles should be, but when we do, we'll have for any angle t one of most important trigonometric identities, the Pythagorean identity for sines and cosines:

sin² t + cos² t = 1.

Sines and cosines for special common angles

We can easily compute the sines and cosines for certain common angles. Consider first the 45° angle. It is found in an isosceles right triangle, that is, a 45°-45°-90° triangle. In any right triangle c² = a² + b², but in this one a = b, so c² = 2a². Hence c = a sqrt

2. Therefore, both the sine and cosine of 45° equal 1/ sqrt

2 which may also be written sqrt

2 / 2.

Next consider 30° and 60° angles. In a 30°-60°-90° right triangle, the ratios of the sides are 1 : sqrt 3 : 2. It follows that sin 30° = cos 60° = 1/2, and sin 60° = cos 30° = sqrt 3 / 2.

These findings are recorded in this table.

Degrees	Radians	cosine	sine
90°	/2	0	1
60°	/3	1/2	3 / 2
45°	/4	2 / 2	2 / 2
30°	/6	3 / 2	1/2
0°	0	1	0

Exercises

These exercises all refer to right triangles with the standard labelling.

30. b = 2.25 meters and cos A = 0.15. Find a and c.

33. b = 12 feet and cos B = 1/3. Find c and a.

35. b = 6.4, c = 7.8. Find A and a.

36. A = 23° 15', c = 12.15. Find a and b.

Hints

30. The cosine of A relates b to the hypotenuse c, so you can first compute c. Once you know b and c, you can find a by the Pythagorean theorem.

33. You know b and cos B. Unfortunately, cos B is the ratio of the two sides you don't know, namely, a/c. Still, this gives you an equation to work with: 1/3 = a/c. Then c = 3a. The Pythagorean theorem then implies that a² + 144 = 9a². You can solve this last equation for a and then find c.

35. b and c give A by cosines and a by the Pythagorean theorem.

36. A and c give a by sines and b by cosines.

Answers

30. c = b/cos A = 2.25/0.15 = 15 meters; a = 14.83 meters.

33. 8a² = 144, so a² = 18. Therefore a is 4.24', or 4'3".
c = 3a which is 12.73', or 12'9".

35. cos A = b/c = 6.4/7.8 = 0.82. Therefore A = 34.86° = 34°52', or about 35°.
a² = 7.8² – 6.4² = 19.9, so a is about 4.5.

36. a = c sin A = 12.15 sin 23°15' = 4.796.
b = c cos A = 12.15 cos 23°15' = 11.17.

Table of contents. | Back to sines. | On to tangents and slopes.

David E. Joyce
Department of Mathematics and Computer Science
Clark University
Worcester, MA 01610

Email: djoyce@clarku.edu

Dave's Short Trig Course is located at http://aleph0.clarku.edu/~djoyce/java/trig