MORE ABOUT TRIGONOMETRIC FUNCTIONS
Objectives: To learn about tangent, cotangent, secant and cosecant functions.
Recall 1: There are four other trigonometric functions that naturally arise when angles become part of a mathematical problem. If we consider the right triangle below to have sides of length a, b and c, then we make the following definitions.
Tangent = tan (alfa) =
Cotangent = cot (alfa) = b/a
Secant = sec (alfa) = c/b
Cosecant = csc (alfa) = c/a
Recall 2: A geometric measure of the tan a can be represented by the height of a right triangle having base angle a and base 1. This is the height of a triangle with base angle a at the center of a circle of radius 1. See illustration below.
Recall 3 Several important relations are displayed below.
tan a = sin a /cos
cot a = cos a /sin a = 1/tan a
sec a = 1/cos a
csc a = 1/sin a
Recall 4 We recall that the
Theorem says that (sin a)^2 + (cos a)^2 = 1. From this,
that 1 + (tan a)^2 =(sec a)^2 and 1 + (cot a)^2 =
can be written as
(cos a)^2 = cos a. How do you solve such an equation when 0 <= a <= 2pi?
(cos a)^2 - cos a = 0.
Then cos a (cos a - 1) = 0.
This tells us that either cos a = 0 or cos a = 1.
The first possibility leads to a = pi/2 or a = 3pi/2 while the second leads to a = 0 or 2pi. Thus, the solutions are 0, pi/2, 3pi/2, and 2pi.
Example 1. Show that
Solution: Work with the more complicated side. In this case the left side is the complicated side. We have
Example 2. Solve the equation (sec x)^2 = 2tan x for -pi/2 <= x <= pi/2.
Solution: Subtract 2tan x from both sides and use the identity 1+ (tan x)^2 = (sec x)^2 to get
(1 + (tan x)^2) - 2tan x = 0.
Then let y = tan x to get
y^2 - 2y + 1 = 0.
Solve this quadratic equation to get y = 1. Hence tan x = 1. Since -pi/2 <= x <= pi/2. we know that x must be pi/4.
Example 3. Suppose that you know that tan a = 7/24 and that 0 <= a <= pi/2. Find sin a and cos a.
Solution: Use the trigonometric
1+ (tanx)^2 = (secx)^2 to find sec a and use it to get cos
draw a triangle like the one below.
Use the Pythagorean theorem to find the length of the hypotenuse of this triangle.
c^2 = 24^2 + 7^2
= 576 + 49
Therefore c = sqrt(625 )= 25. Now
you can find sin
a and cos a.
1. If cos a = 2/3, what is sin a?
2. If cos a = 4/5, what is tan a?
3 A right triangle has hypotenuse 5, base 4 and the angle between the base and the hypotenuse is a. Find tan a, cot a, sec a and csc a.
4 A right triangle has a base of size 4 and height of size 5. Find the values of all the trigonometric functions of the base angle.
5 A line joining the coordinates (0,0) with (3,-4) makes an angle a with the horizontal line through (0,0). Find tan a, cot a, sec a and csc a.
6 A balloon is at an altitude of 200 feet above sea level and directly above the edge of a cliff. An observer on a beach sites the balloon through a telescope aimed at a 30 degree angle from the horizontal. How far is the balloon from the observer?
7 A surveying team measures an angle of ¼/8 radians from a point A on a beach to a point B at the top of a cliff. If point A is 200 meters from the base of the cliff, how high is the cliff?
8 A crane is 80 feet long. For safety reasons it should never be lowered less than 55° from the horizontal. How high will the crane reach?
9 A tower casts a 120 foot shadow on horizontal ground from the sun when it makes an angle of 60 degrees. How tall is the tower?
10 Simplify sin x - (cos x)^2 sin x.
11 Simplify (tan x)^2 (csc x)^2 (cot x)^2 (sin x)^2.
12 Solve the equation 2(sin x)^2 + 3cos x = 0 for 0 <= x <= pi.
1 ±sqrt(5)/3 = ±0.7454
3 tan a = 3/4, cot a = 4/3
sec a = 5/4, csc a = 5/3
4 sin a = 5/sqrt(41), cos a = 4/sqrt(41)
tan a = 5/4, cot a = 4/5
sec a = sqrt(41)/4, csc a = sqrt(41)/5
5 tan a = -4/3,
cot a = -3/4
sec a = 5/3, csc a = -5/4
6 400 feet
7 = 82.84 meters
8 = 65.53 feet
9 120sqrt(3) feet = 207.85 feet