Trigonometry

After completing this chapter you should know

• the functions of secant ϴ, cosecant ϴ and cotangent ϴ• the graphs of sec ϴ, cosec ϴ and cot ϴ• how to solve equations and prove identities involving sec ϴ,

cosec ϴ and cot ϴ• how to prove and use the identities 1 + tan² ϴ ≡ sec² ϴ 1 + cot² ϴ ≡ cosec² ϴ• how to sketch and use the inverse trigonometric functions

arcsinx, arccosx and arctanx

The functions of secant ϴ, cosecant ϴ and cotangent ϴ

these are more commonly known as sec, cosec and cot.

sec ϴ = undefined for values of cosϴ = 0

cosec ϴ = undefined for values of sinϴ = 0

cot ϴ = undefined for values of tanϴ = 0

Example 2 looks at finding exact values for sec and cot, lets see what they do

Exercise 6A goes over finding these values on your calculator and giving exact values!

the graphs of sec ϴ, cosec ϴ and cot ϴ

These look so pretty in the book that we’ll look at them there ( page 87)

Simplify Expressions

1. sinϴcotϴsecϴapply what you know ( or ½ know)

=

Cancelling gives sinϴcotϴsecϴ = 1

How about sinϴcosϴ(secϴ + cosecϴ)?

Lets look inside the bracket first secϴ + cosecϴ = = This gives us sinϴcosϴ(secϴ + cosecϴ) = sinϴcosϴ X = sinϴ + cosϴ

Prove Identities

Show that

Simplify the numerator cotϴcosecϴ = x = Simplify the denominator sec²ϴ + cosec²ϴ = + = =

Combine these two elements to give you

÷

x

Solve Equations

Solve the equation sec ϴ = -2.5 in the interval 0≤ϴ≤ 360° sec ϴ = -5/4 so cos ϴ = -0.4cos is negative in the second and third quadrant

AS

T C

66.4°

66.4°

ϴ = 113.6°, 246.4° = 114°, 246° (3s.f.)

Solve the equation cot2ϴ = 0.6

So tan2ϴ = 5/3

Let x = 2ϴ then we are solving tanx = 5/3 in the interval 0≤ x ≤ 720°

AS

T C

59.0°

59.0°

x = 59.0°, 239.0°, 419.0, 599.0°

So ϴ = 29.5°, 120°, 210°, 300° (3 s.f)

Exercise 6C page 92 to use and apply this new found knowledge

prove and use these identities 1 + tan² ϴ ≡ sec² ϴ 1 + cot² ϴ ≡ cosec² ϴ

So here goes with the first proof 1 + tan² ϴ ≡ sec² ϴ

You know sin²ϴ + cos²ϴ ≡ 1

Divide through by cos²ϴ gives us

tan²ϴ + 1 ≡

1 + tan²ϴ ≡ sec²ϴ

And for your second proof1 + cot² ϴ ≡ cosec² ϴ

Take sin²ϴ + cos²ϴ ≡ 1 and divide through by sin²ϴ

No I haven’t done this one it’s your turn

Now we have these identities we get to use them in some very exciting (for mathematicians) ways.

Example 12 tells us that tan A = , that A is obtuse, and asks us to find

a) sec A b) sin A

there are two methods for part a

Method 1 uses the identity 1 + tan² A ≡ sec² A sec² A = 1 + sec A =

because A is obtuse sec A = (cos A is negative in the second quadrant)

Method 2this uses Pythagoras to find the hypotenuse, when tanφ =

5

12

φ

13

so secφ =

Part b) asked us to find sin A

from the diagram in method 2 we can see sin A = (sin is positive in the second quadrant so there is no need to adjust this answer)

using identities we have tanA this gives us sinA ≡ tanAcosA ≡ sinA ≡

you may be asked to use these identities to prove other identities

example 13 on page 95 deals with this

or solve equations by substituting in the identities

example 14 on page 96 shows you what may happen

exercise 6D on page 96 lets you practise all this fun stuff

and finally for chapter 6

The inverse trig functions arcsin x (sin-1 x) arccos x (cos-1 x)

and arctan x (tan-1 x)and their graphs.

Lets start with their graphs (because I know how much you like graph work!)

sinx, cosx and tanx only have inverse functions if their domains are restricted so that they are one-to-one functions.

You can draw the graph of arcsin x with the restricted domain of - ≤ x ≤ by drawing the graph of sin x and reflecting it in the

line y = x

Example 15 shows this with picturesExample 16 shows arccos xExample 17 shows arctan x

Exercise 6E page 101 finishes off this chapter