2 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 inverse, or 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.
Each ratio moreover is a function of the acute angle. That is, one quantity is a "function" of another if its value depends on the value of the other. 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 when the radius changes, the circumference also will change. As we will see in the next Topic, the value of each ratio depends only on the value of the acute angle. That is why we say that those ratios are functions of the acute angle. We call them 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.
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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:
8² + 15²  =  17² 
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?
Problem 5. The height of a triangle. Every triangle, rightangled or not, will have at least two acute angles.
Let them be the base angles at A and B, so that the base will be the side c. Show that the height h drawn to that base is
h =  _____c_____ cot A + cot B 
. 
Hint: The height h 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.
In the right triangle containing the acute angle A,
x h 
=  cot A, or x = h cot A . . . (1) 
In the right triangle containing the acute angle B,
c − x h 
=  cot B, or c − x = h cot B. 
On substituting the expression for x from line (1),
c − h cot A = h cot B,
which implies
c = h cot A + h cot B = h(cot A + cot B).
Therefore, on solving for h,
h =  _____c_____ cot A + cot B 
. 
Problem 6. The area of a triangle. Prove: 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, prove that
Area of triangle ABC = ½ sin A bc = ½ bc sin A.
The area of a triangle is equal to onehalf the base times the height. In triangle ABC, let the base be c. 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.
Next Topic: Trigonometry of right triangles
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