6 Definitions of the Trigonometric Functions

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 in the righthand column is the reciprocal of the ratio in the lefthand column.
The reciprocal of sin θ is csc θ ; and viceversa.
The reciprocal of cos θ is sec θ.
And the reciprocal of tan θ is cot θ.
We will see in the next Topic that the value of each ratio depends only on the value of the acute angle. Therefore we say that each ratio is a function of the acute angle. Hence we call the six ratios the trigonometric functions of an acute angle. All of trigonometry is based on the definitions of those functions.
Problem 1. Complete the following with either "opposite," "adjacent to," or "hypotenuse."
To see the answer, pass your mouse over the colored area.
To cover the answer again, click "Refresh" ("Reload").
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:
8^{2} + 15^{2}  =  17^{2} 
64 + 225  =  289 
Problem 4. A straight line makes an angle θ with the xaxis. The value
of which function of θ is equal to its slope?
Two theorems.
Theorem 1. The area of a triangle. The area of a triangle is equal to onehalf the sine of any angle times the product of the two sides that make the angle.
Specifically:
Area of triangle ABC = ½ sin A bc = ½ cb sin A.
The area of a triangle is equal to onehalf the base times the height. In triangle ABC, let the base be c and the height h. Then
Area = ½ ch.
Now,
sin A = h/b,
so that
h = b sin A.
Therefore in the expression for the Area, replace h with b sin A:
Area = ½ cb sin A.
Which is what we wanted to prove.
Theorem 2. The ratio of the areas of similar triangles.
Similar triangles are to one another as the squares drawn
on their corresponding sides.
When triangles are similar, then by whatever factor side a changes, sides b and c will change by that same factor. Proportionally,
Therefore, according to Theorem 1, the area of the triangle on the left has the following ratio to the triangle on the right:
But the areas of the squares on the corresponding sides have that same ratio.
Therefore, similar triangles are to one another as the squares drawn on their corresponding sides.
Which is what we wanted to prove.
Next Topic: Trigonometry of right triangles
Please make a donation to keep TheMathPage online.
Even $1 will help.
Copyright © 2022 Lawrence Spector
Questions or comments?
Email: teacher@themathpage.com