sine of θ	=	sin θ	=	opposite   hypotenuse		cosecant of θ	=	csc θ	=	hypotenuse   opposite
cosine of θ	=	cos θ	=	adjacent   hypotenuse		secant of θ	=	sec θ	=	hypotenuse   adjacent
tangent of θ	=	tan θ	=	opposite  adjacent		cotangent of θ	=	cot θ	=	adjacent opposite

Notice that each ratio (function) in the right-hand column is the inverse ratio, or the reciprocal, of the ratio in the left-hand column.

The reciprocal of sin θ is csc θ ; and vice-versa.

The reciprocal of cos θ is sec θ.

And the reciprocal of tan θ is cot θ.

As for why these are called functions, we say that one quantity is a "function" of another if the first depends on the other. For example, the circumference of a circle is a function of the radius, because the size of the circumference depends on the size of the radius. And so we call these trigonometric functions rather than just trigonometric ratios, because each ratio, as we will see, depends on the acute angle.

Problem 1.   Complete the following with either "opposite,"  "adjacent to," or "hypotenuse."

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a) In a right triangle, the side opposite the right angle is called the

a) hypotenuse.

b) CA is called the side opposite angle θ.

c) BC is called the side adjacent to angle θ.

d) AC is called the side adjacent to angle φ.

e) BC is called the side opposite angle φ.

Problem 2.    The sides of a right triangle are in the ratio 3:4:5, as shown.  Name and evaluate the six trigonometric functions of angle θ.

sin θ

=

4 5

csc θ

=

5 4

cos θ

=

3 5

sec θ

=

5 3

tan θ

=

4 3

cot θ

=

3 4

Problem 3.    The sides of a right triangle are in the ratio 8 : 15 : 17, as shown.  Name and evaluate the six trigonometric functions of angle φ.

sin φ

=

15 17

csc φ

=

17 15

cos φ

=

8  17

sec φ

=

17  8

tan φ

=

15  8

cot φ

=

8  15

Notice that the sides of this triangle satisfy, as they must, the Pythagorean theorem:

Problem 4.  The height of a triangle.   Every triangle, right-angled or not, will have at least two acute angles.

Let them be the base angles at A and B.  And call the base c.  Then the height h drawn to that base will cut the entire triangle into two right triangles.

Let x be the segment of the base containing the angle A.  Then the remaining segment is the difference between the whole c and that segment:  c − x.

Show that the height h is

Problem 5.  The area of a triangle.   Prove:   The area of a triangle is equal to one-half the sine of any angle times the product of the two sides that make the angle.

Specifically, prove that

Area of triangle ABC = ½ sin A bc = ½ bc sin A.