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) = a/b
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 a
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 Pythagorean Theorem says that (sin a)^2 + (cos a)^2 = 1.  From this, we may deduce that  1 + (tan a)^2 =(sec a)^2 and 1 + (cot a)^2 = (csc a)^2


.
Recall 5 You will often find that solutions to trigonometric calculations simplify through normal algebraic cancellation laws.  For example, the expression

can be written as



 
Recall 6: In practical problems you may have to solve equations where the unknown is a trigonometric function.  For example, the equation may turn out to be something like

(cos a)^2 = cos a.  How do you solve such an equation when 0 <= a <= 2pi?

Set  (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.


Examples
 
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 identity  1+ (tanx)^2 = (secx)^2 to find sec a and use it to get cos a.  First 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

= 625.

Therefore c = sqrt(625 )= 25.  Now you can find sin a and cos a.


EXERCISES
______________________________________________________________________
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.

________________________________________________________________________

 ANSWERS
 
1 ±sqrt(5)/3 = ±0.7454
 
2 ±3/4
 
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

10 (sinx)^3

11 1

12 2pi/3