Chapter Five
Section 5.1
Fundamental Identities
Page186
Objectives:
1. Review of Basic Identities
2. NegativeNegative-Angle Identities
3. Fundamental Identities
Identity
IDENTITY:
• Recall that an identity is an equation that
is true for every value in the domain of its
variable.
• Example:
2(−3 x − 7) = −6 x − 14
and
a 3 + b3 = ( a + b ) ( a 2 − ab + b 2 )
Trigonometric Functions
Review of Basic Identities
Negative-Angle Identities
sin (–
(–x) = –sin x
cos (–
(–x) = cos x
tan (–
(–x) = –tan x
csc (–
(–x) = –csc x
sec (–
(–x) = sec x
cot (–
(–x) = –cot x
Negative-Angle Identities
• One of the easiest ways to remember the
negative--angle identities is to remember that
negative
only cosine
• and its reciprocal, secant are even functions.
For even functions, f(–
f(– x) = f(x) which means
these
• functions have yy-axis symmetry. The other
four trig functions (sine, cosecant, tangent,
and
• cotangent) are odd functions. For odd
functions, f(–
f(– x) = – f(x) which means these
functions have
• origin symmetry.
• Example: Since tangent is an odd function, if
tan x = 2.6, then tan (–
(–x) = –2.6.
Negative-Angle Identities
•
•
•
•
•
•
sin (–
(–x) = –sin x
cos (–
(–x) = cos x
tan (–
(–x) = –tan x
csc (–
(–x) = –csc x
sec (–
(–x) = sec x
cot (–
(–x) = –cot x
(x, y)
(x,(x,-y)
Finding Trigonometric Function Values
Given One Value and the Quadrant
Using Identities to Rewrite
Functions and Expressions
Using Identities to Rewrite
Functions and Expressions
Using Identities to Rewrite
Functions and Expressions
Writing any of the six trigonometric functions in terms
of any of the others
Example: Write tan θ in terms of cos θ.
sin θ
tan θ =
cos θ
2
2
Now, cos θ + sin θ = 1
2
2
2
And sin θ = 1 − cos θ ⇔ sin θ = ± 1 − cos θ
2
2
± 1 − cos θ
1 − cos θ
=±
so we have tan θ =
cos θ
cos θ
Finding all Trigonometric values given one value and the
quadrant
Example: If cos θ = 1/5 , θ in quadrant I, find the
remaining five trigonometric functions.
sin 2 θ + cos 2 θ = 1
Now you figure out the rest!
2
1
sin 2 θ +   = 1
5
1
=1
sin 2 θ +
25
sin 2 θ = 1 −
sin 2 θ =
1
25
24
25
24
2 6
=±
25
5
quadrant I ⇒ positive
sin θ =
sin θ =
2 6
5
Homework Assignment
on the Internet
Section 5.1:
Pages 191 - 193:
1 - 4 all, 12, 23 – 28 all,
30 - 46 even, 50, 60.