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8.4 Relationships Among the Functions
Objective
1. To simplify trigonometric expressions and to prove trigonometric identities.
2. To use the fundamental identities to find the values of other trigonometric functions from the value of a given trigonometric function.
RECIPROCAL IDENTITIES
sin
1csc
cos
1sec
tan
1cot
QUOTIENT IDENTITIES
cos
sintan
sin
coscot
2 21 tan sec
22 csc cot1
PYTHAGOREAN IDENTITIES
1 cossin 22
Note It will be necessary to recognize alternative forms of the identities above, such as sin² = 1 – cos² and cos² = 1 – sin² .
NEGATIVE IDENTITIES
tan)tan(cos)cos(sin)sin(
csc( ) csc sec( ) sec cot( ) cot
All of the identities we learned are found in the back page of your book under the heading Trigonometric Identities and then Fundamental Identities.
You'll need to have these memorized or be able to derive them for this course.
Pythagorean Identities
Note It will be necessary to recognize alternative forms of the identities above, such as sin² = 1 – cos² and cos² = 1 – sin² .
2 2 2 2 2 2sin cos 1 1 tan sec 1 cot csc
Relationships Among the Functions
2 2 22 2
2 2 2
sin cos sin 11 tan 1 seccos cos cos
2 2 2 2 22 2
2 2 2 2sin cos 1y x x y r
r r r r
2 2 22 2
2 2 2
cos sin cos 11 cot 1 cscsin sin sin
Cofunction Relationships
COFUNCTION IDENTITIES
sin cos cos sin
tan cot cot tan
sec csc csc sec
90 90
90 90
90 90
a f a fa f a fa f a f
and
and
and
COFUNCTION IDENTITIES
sin cos cos sin2 2
tan cot cot tan2 2
sec csc csc sec2 2
and
and
and
Each of the trigonometric relationships given is true for all values of the variable for which each side of the equation is defined.
Such relationships are called trigonometric identities.
1 sin sec sin tan sin
cos cos
xx x x x
x x
21 sin
cos cos
x
x x
21 sin
cos
x
x
2cos
cos
x
x cos x
Example 1: Simplify secx – sinx tanx
[Solution]
Example 2: Write tan + cot in terms of sin and cos .
[Solution]
sincos1
sincoscossin
sincoscos
sincossin
sincos
cossin
cottan
22
22
2
2cot 1 tan
Example 3: Prove: csctan
A AA
A
2cot sec
tan
A A
A21
cot sectan
A AA
21 1sec
tan tanA
A A
22
1sec
tanA
A
22
1cot
cosA
A
2
2 2
cos 1
sin cos
A
A A
2
1
sin A 2csc A
[Proof]
An Identity is NOT a Conditional Equation
• Conditional equations are true only for some values of the variable.
• You learned to solve conditional equations in Algebra by “balancing steps,” such as adding the same thing to both sides, or taking the square root of both sides.
• We are not “solving” identities so we must approach identities differently.
22
4
82
912
2
2
2
xorx
x
x
x
Example 4: Prove
[Proof] Manipulate right to look like left. Expand the binomial and express in terms of sin & cos.
2 2 2(csc cot ) csc 2csc cot cotx x x x x x
x
xxx
cos1
cos1)cot(csc 2
2 2
2 2 2
1 2 cos cos 1 2cos cos
sin sin sin sin sin
x x x x
x x x x x
2 2
2 2
(1 cos ) (1 cos ) (1 cos )(1 cos )
sin 1 cos (1 cos )(1 cos )
x x x x
x x x x
1 cos
1 cos
x
x
1. Learn the fundamental identities.
2. Try to rewrite the more complicated side of the equation so that it is identical to the simpler side.
3. It is often helpful to express all functions in terms of sine and cosine and then simplify the result.
4. Usually, any factoring or indicated algebraic operations should be performed. For example,
5. As you select substitutions, keep in mind the side you are not changing, because it represents your goal.
6. If an expression contains 1 + sin x, multiplying both numerator and denominator by 1 – sin x would give 1 – sin² x, which could be replaced with cos² x.
. and ,)1(sin1sin2sin cossinsincos
cos1
sin122
We Verify (or Prove) Identities by Doing the Following:
Suggestions• Start with the more complicated side
• Try substituting basic identities (changing all functions to be in terms of sine and cosine may make things easier)
• Try algebra: factor, multiply, add, simplify, split up fractions
• If you’re really stuck make sure to:Change everything on both sides to sine
and cosine.
• Work with only one side at a time!
Verifying an Identity (Working with One Side)
Example 5: Verify that the following equation is an identity.
cot x + 1 = csc x(cos x + sin x)
Analytic Solution Since the side on the right is more complicated, we work with it.
cot 1 csc (cos sin )1 (cos sin )
sincos sinsin sincot 1
x x x x
x xxx xx xx
Original identity
1cscsin
xx
Distributive property
cos sincot , 1sin sin
x xxx x
Example 6: Verify that the following equation is an identity.
[Proof]
tttttt 22 cscsec
cossincottan
tt
tt
tttt
tttt
ttt
ttt
ttt
ttt
tttt
22
22
cscsec
sin1
cos1
cossin1
sincos
cossin1
cossin
cossin1
cotcossin1
tan
cossincot
cossintan
cossincottan
Verifying an Identity (Working with One Side)
Verifying an Identity (Working with Both Sides)
Example 7: Verify that the following equation is an identity.
[Proof]
2
2
cos
sinsin21tansectansec
sincos
sincos
sec tan (sec tan )cossec tan (sec tan )cos
sec cos tan cossec cos tan cos1 tan cos1 tan cos1 cos 1 sin1 cos 1 sin
Now work on the right side of the original equation.
We have shown that
sin1sin1
)sin1)(sin1()sin1(
sin1)sin1(
cos)sin1(
cos
sinsin21
2
2
2
2
2
2
2
.cos
sinsin21sin1sin1
tansectansec
2
2
Verifying an Identity (Working with Both Sides)
How to get proficient at verifying identities:
• Once you have proved an identity go back to it, redo the verification or proof without looking at how you did it before, this will make you more comfortable with the steps you should take.
• Redo the examples done in class using the same approach, this will help you build confidence in your instincts!
Don’t Get Discouraged!
• Every identity is different
• Keep trying different approaches
• The more you practice, the easier it will be to figure out efficient techniques
• If a solution eludes you at first, sleep on it! Try again the next day. Don’t give up!
• You will succeed!
3
If the angle is acute (less than 90o) and you have the value of one of the six trigonometry functions, you can find the other five.
Sine is the ratio of which sides of a right triangle?h
o
Draw a right triangle and label and the sides you know.
1
When you know 2 sides of a right triangle you can always find the 3rd with the Pythagorean theorem.
a
222 31 a
228 a
22
Now find the other trig functions
cosh
a
22
3sec3
22
Reciprocal of sine so "flip" sine over csc 3
tana
o
22
1
"flipped" cos
cot 22"flipped"
tan
3
1sin
There is another method for finding the other 5 trig functions of an acute angle when you know one function. This method is to use fundamental identities.
3
1sin We'd still get csc by taking reciprocal of
sincsc 3
Now use the trig identity1cossin 22 Sub in the value of sine that you know1cos
3
1 22
Solve this for cos
9
8cos2
3
22
9
8cos
This matches the answer we got with the other method
You can easily find sec by taking reciprocal of cos.
We won't worry about because angle is acute.
square root both sides
3sec
2 2
Let's list what we have so far: 3
1sin
csc 3
We need to get tangent using fundamental identities.
cos
sintan
Simplify by inverting and multiplying
322
31
tan
3
22cos
Finally you can find cot by taking the reciprocal of this answer.
22
3sec
22
3
3
1
22
1
22cot
[Solution]a) sec To find the value of
this function, look for an identity that relates tangent and secant.
Tip: Use Pythagorean Identities.
When is in quadrant II, cos, sec, tan, cot, and csc are all negative.
Example 8: If and is in quadrant II, find each function value.
[Solution]b) sinTip: Use Quotient Identities.
c) cot Tip: Use Reciprocal
Example 8: If and is in quadrant II, find each function value. (Cont.)
1 3cot
tan 5
d) csc Tip: Use Reciprocal
1 34csc
sin 5 34
34
5
[Solution]e) cosTip: Use Reciprocal.
Example 8: If and is in quadrant II, find each function value. (Cont.)
1 3cos
sec 34
3 34
34
Example 9: If and is in quadrant VI, find each function value.
3sin
5
[Solution]a) cosTip: Use Pythagorean
Identities.
22 3
cos 1 sin 15
4
5
When is in quadrant VI, csc, tan, and cot, are all negative, only cos and sec are positive.
b) secTip: Use Reciprocal Identities.
1 5sec
cos 4
c) tanTip: Use Quotient Identities.
sin 3 / 5 3tan
cos 4 / 5 4
Example 9: If and is in quadrant VI, find each function value. (Cont.)
3sin
5
[Solution]d) cotTip: Use Reciprocal Identities.
1 4cot
tan 3
e) cscTip: Use Reciprocal.
1 5csc
sin 3
Example 10: If and find each function value.
1cos
3
Challenge!
[Solution] Since cos > 0, then is either in Quadrant I or VI.
Case 1) If is in Quadrant I , then
a) sin
22 1
sin 1 cos 13
2 2
3
b) tan
sin 2 2 / 3tan 2 2
cos 1/ 3
c) cot
1 1 2cot
tan 42 2
Example 10: If and find each function value. (Cont.)
1cos
3
Challenge!
[Solution] Since cos > 0, then is either in Quadrant I or VI.
Case 1) If is in Quadrant I , then
d) sec e) csc1 3 3 2
cscsin 42 2
1sec 3
cos
Example 10: If and find each function value. (Cont.)
1cos
3
Challenge!
[Solution] Since cos > 0, then is either in Quadrant I or VI.
Case 2) If is in Quadrant VI , then
a) sin
22 1
sin 1 cos 13
2 2
3
b) tan
sin 2 2 / 3tan 2 2
cos 1/ 3
c) cot
1 1 2cot
tan 42 2
Example 10: If and find each function value. (Cont.)
1cos
3
Challenge!
[Solution] Since cos > 0, then is either in Quadrant I or VI.
Case 2) If is in Quadrant VI , then
d) sec e) csc1 3 3 2
cscsin 42 2
1sec 3
cos