Definitions of the Trigonometric Functions
|sine of θ||=||sin θ||=|| opposite
|cosecant of θ||=||csc θ||=||hypotenuse
|cosine of θ||=||cos θ||=|| adjacent
|secant of θ||=||sec θ||=||hypotenuse
|tangent of θ||=||tan θ||=||opposite
|cotangent of θ||=||cot θ||=|| adjacent
Notice that each ratio in the right-hand column is the inverse, 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.
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."
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a) In a right triangle, the side opposite the right angle is called the
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.
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.
|cos φ||=|| 8
|cot φ||=|| 8
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 x-axis. The value
of which function of θ is equal to its slope?
Problem 5. 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, so that the base will be the side c. Show that the height h drawn to that base is
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,
|=||cot A, or x = h cot A . . . (1)|
In the right triangle containing the acute angle B,
|c − x
|=||cot B, or c − x = h cot B.|
On substituting the expression for x from line (1),
c − h cot A = h cot B,
c = h cot A + h cot B = h(cot A + cot B).
Therefore, on solving for h,
cot A + cot B
Problem 6. 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.
The area of a triangle is equal to one-half the base times the height. In triangle ABC, let the base be c. Then
Area = ½ ch.
sin A = h/b,
h = b sin A.
Therefore in the expression for the Area, replace h with b sin A:
Area = ½ cb sin A.
Copyright © 2018 Lawrence Spector
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