Download §9.1 Trigonometric Identities

Survey
yes no Was this document useful for you?
   Thank you for your participation!

* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project

Document related concepts

Trigonometric functions wikipedia , lookup

Transcript
§9.1 Trigonometric Identities
sin (θ) = y
and sin (-θ) =
r
so,
sin (-θ) = - sin (θ)
and
cos (θ) = x
and cos (-θ) =
r
so,
cos (-θ) = cos (θ)
r
y
x θ
x
-y -θ
r
-y
r
x
r
And,
Tan (-θ) =
sin (-θ) =
cos (-θ)
- sin (θ) = – sin (θ) = - tan (θ)
cos (θ)
cos (θ)
Note: You may have learned that sine is an odd f(n) [it’s symmetric about the orginin so if (x, y),
then (-x, -y)] and cosine is an even f(n) [it’s symmetric about the y-axis so if (x, y) then
(-x, y)]
Fundamental Identities
Reciprocal Identities
cot θ =
1
tan θ
csc =
1
sec θ =
sin θ
1
cos θ
Quotient Identities
tan θ = sin θ
cos θ
cot θ = cos θ
sin θ
Pythagorean Identities
sin2 θ + cos2 θ = 1
1 + cot2 θ = csc2 θ
tan2 θ + 1 = sec2 θ
sin (-θ) = -sin (θ)
cos (-θ) = cos θ
tan (-θ) = -tan θ
csc (-θ) = -csc θ
sec (-θ) = sec θ
cot (-θ) = -cot θ
Negative Angle Identities
Sometimes it will be necessary to re-express one function in terms of another. This can
be useful in graphing and in differentiation and integration in Calculus. To do this, try
the following:
1) Find an identity that relates the functions
2) Solve for the function of desire
Example:
Express tan θ in terms of cos θ
Now, we’ll do a little more of the same, but try to use only sin and cos. This is an
important skill to possess, hence while we’ll practice this skill by itself in this section.
Here’s a strategy:
1) Use sin & cos to re-write the function(s)
2) Simplify using algebra skills
Example:
Write (sec θ + csc θ) in terms of sine and cosine and simplify
Your Turn
Example:
Write cos x in terms of csc x. (Hint: Think Pythagorean Id. & reciprocate)
Here are some of the hints that Rockswold, Lial & Hornsby have for verifying Trig
Identities in their book, A Graphical Approach to Precalculus with Limits: A Unit Circle
Approach, 5th Ed., p. 633.
We will work on verifying identities first using only one side of the equation. The
purpose in doing this is to prepare for the use of identities in solving Calculus problems.
We need skills of rewriting just one side that we will practice this time and also the skills
of rewriting both sides which we will practice shortly. Don’t be tempted to rewrite both
sides this time, that is not the goal. We’ll practice that in the next group of examples.
Example:
Verify that (cot x)(sin x)(sec x) = 1.
Hint: Put in terms of sin & cos
Example:
Verify that cot2 θ(tan2 + 1) = csc2 θ is an identity.
Hint: Use Pythagorean id & rewrite in terms of sin & cos
Example:
Verify that tan2 s
sec2 s
= (1 + cos s)(1 – cos s)
Hint: Focus on left to restate sec & Pythagorean Id to restate tan.
Example:
Verify that sec s + tan s
sin s
=
csc s
sec s – tan s
is an identity
Hint: Focus on right. Use conjugate with fundamental theorem of fractions. Simplify &
use another Pythagorean ID
Now let’s practice using a different technique. Instead of focusing on just one side we’ll
manipulate both sides. We’ll start with a second look at the third example above.
Example:
Verify that tan2 s
sec2 s
= (1 + cos s)(1 – cos s)
Hint: Rewrite left using sin & cos. Rewrite right by mult. out & using Pythagorean Id.
Example:
Verify that cot θ – csc θ
cot θ + csc θ
=
1 – 2 cos θ + cos2 θ
-sin2 θ
Hint: Right: Factor, Pythagorean Id, Factor
Left: Since csc & cot have sin in common, so mult by sin & simplify
Now, let’s look at an application of identity verification as used in circuit design.
Example:
Energy in a circuit is described by E(t) = C(t) + L(t). For a certain FM
radio station if the energy in the inductor is L(t) = 5 cos2 (62 x 107)t and
the energy stored in the capacitor at time t is C(t) = 5 sin2 (62 x 107)t.
Find the total energy in the circuit at time t.
§9.2 Sum and Difference Identities
I debated whether or not to cover the derivation of these formulas in class. I have decided
against it due to time constraints. Suffice it to say that the proof comes from
1)
2 points on a circle with terminal points described by the sine and cosine
2)
The distance formula
3)
Some algebra
I will leave it to a diligent student to investigate further on p. 641-642 of Ed. 5 &
p. 643-644.
Sine of a Sum/Difference
Cosine of a Sum/Difference
sin (A + B) = sinA cosB + cos A sin B
cos (A + B) = cosA cosB – sin A sin B
sin (A – B) = sin A cosB – cos A sin B
cos (A – B) = cosA cosB + sin A sin B
Tangent of a Sum/Difference
tan (A + B) =
tan A + tan B
1 – tan A tan B
tan (A – B) =
tan A – tan B
1 + tan A tan B
There are multiple tasks that we will need to perform using these sum/difference
identities. Some of the tasks may have characteristics from our first section.
1)
We may be required to rewrite an angle as a sum of 2 angles for which we
know exact trig values.
13π
Ex.
15º = 45º – 30º
&
/12 = 9π/12 + 4π/12 = 3π/4 + π/3
Note: Here are some equivalencies that you might find helpful in doing these problems.
2π
2)
3)
4)
5)
6)
3π
4π
/12 = π/6 = 30º
/12 = π/4 = 45º
/12 = π/3 = 60º
We could be called to remember reference angles
Ex.
See the second example above
We could be required to use even/odd identities
Ex.
sin (-15º) = - sin 15º
We could use the identities in reverse with or without skills in 1-3 above
Ex.
tan 100º – tan 70º = tan (100º – 70º)
1 + tan 100ºtan 70º
We might use the identities in general resulting in the use of exact trig
values simplifying down to a single trig function in general
Ex.
Prove sin (θ – 270º) = cos θ
We may use a special formula that comes from substitution in the last case
to re-express a sum so that it is easier to graph as a translation of sine
*See below for the formula & substitution
Example:
Find the exact value for
a)
cos (-75°)
b)
cos (17π/12)
c)
tan (13π/12)
f)
d)
sin (-15°)
Example:
e)
cos173ºcos83º + sin 173º sin83º
tan100º – tan 30º
1 + tan 100º tan 70º
Prove that cos (180 + θ) = -cos θ using the sum/difference identities.
Some tools your book doesn’t give you that might be useful:
The last example is a part of a group of formulas known as the reduction formulas.
Reduction Formulas
cos (90 + θ) = -sin θ
cos (180 – θ) = -cos θ
cos (180 + θ) = -cos θ
cos (270 – θ) = -sin θ
sin (180 + θ) = -sin θ
tan (270 – θ) = cot θ
Example:
Prove each of the following
a)
sin (θ – 270º) = cos θ
b)
Your book makes a point to give you the following theorem.
Sums of Sines and Cosines
If A & B ∈ Real #’s then
A sin x + B cos x = k sin (x + φ)
Where k = √A2 + B2 and φ satisfies
cos φ = A &
sin φ = B
k
k
tan (θ + 3π) = tan θ
Example:
Prove that √3/2 cos θ + -1/2 sin θ = sin (120º + θ)
Example:
Use what you’ve just learned to rewrite & then graph
f(x) = sin x + cos x
§9.3 Further Identities
Double-Angle, Half-Angle & Product-Sum Formulas
Double Angle Identities
Half Angle Identities
cos 2A = cos2 A – sin2 A
√
√
√
cos 2A = 1 – 2 sin2 A
2
cos 2A = 2 cos A – 1
sin 2A = 2sin A cosA
tan 2A =
2tanA
1 – tan2 A
cos A = ±
2
1 + cos A
2
sin A = ±
2
1 – cos A
2
tan A = ±
2
1 – cos A
1 + cos A
tan A =
2
tan A =
2
sin A
1 + cos A
1 – cos A
sin A
* + or – depends on quadrant of A/2
Sum to Product
Product to Sum
1
cosA cos B = /2 [cos (A + B) + cos (A – B)]
1
sinA sin B = /2 [cos (A – B) – cos (A + B)]
sin A cos B = 1/2 [sin (A + B) + sin (A – B)]
cosA sin B = 1/2 [sin (A + B) – sin (A – B)]
(
(
(
(
)
)
)
)
(
(
(
(
cos A – B
2
sin A – sin B = 2cos A + B
2
sin A – B
2
cos A + cos B = 2cos A + B
2
cos A – B
2
cos A – cos B = -2sin A + B
2
sin A – B
2
Formulas to Lower Powers
sin2 A = 1 – cos 2A
2
)
)
)
)
sin A + sin B = 2sin A + B
2
cos2 A = 1 + cos 2A
2
tan2 A =
1 – cos 2A
1 + cos 2A
*Note: These are just the double angle formulas from above manipulated. Some books don’t even
separate them out. Hornsby, Lial and Rockswold don’t.
One of the ways that we will see the DOUBLE ANGLE & HALF ANGLE formulas
used is to use our basic knowledge of the 45/45/90 and 30/60/90 triangles to angles that
are half or twice those values.
Example:
a)
Practice your skills with the double angle formula by rewriting the angle
to see it as twice a known angle. You probably already know all the exact
values anyway, but that is not the point.
cos 90º
b)
cos 120º
c)
cos π/3
d)
sin π/2
e)
tan -60º
Example:
Find the exact value of sin 22.5º by using your skill with the 45/45/90
triangle.
Example:
Given that sin θ = 8/17 and cos θ < 0, find sin 2θ, cos 2θ, and tan 2θ
Example:
Given cos x = -3/7 with π < x < 3π/2 find sin (x/2), cos (x/2), tan (x/2)
Example:
Use the double angle formula to find all the values of the 6 trig f(n) of θ
given that cos 2θ = -12/13 and 180º < θ < 270º
Example:
Using a half-angle formula rewrite:
√
1 – cos 8x
2
Note: This is very simple. A = argument of the cosine so see the 1/2 angle formula and use it with A/2.
Example:
Verify that the following is an identity:
cos4 x – sin4 x = cos 2x
Hint: You’ll have to use difference of squares & an identity as well as a double angle formula.
Next we can do some problems that come from the product to sum or sum to product.
The product to sum identities are simple to find by adding the sum and difference
formulas from our last section.
Example:
Write
6 sin 40º sin 15º
as a sum/difference of functions.
Example:
Write
cos 3θ + cos 7θ
as a product of two functions.
Here’s one that uses a few of the formulas together.
Example:
Write cos 3x in terms of cos x.
Hint: Use a sum to rewrite 3x and then manipulate that using a formula, finally wrapping up with some
double angle formulas and some simplification.
§9.4 The Inverse Circular F(n)
First let’s review relations, functions and one-to-one functions.
A relation is any set of ordered pairs. A relation can be shown as a set, a graph or a
function (equation).
a)
b)
c)
{(2,5), (2,6), (2,7)}
y = √x, {x| x ≥ 0, x∈ℜ}
5
-15
x
15
-5
A function is a relation which satisfies the condition that for each of its independent
values (x-values; domain values) there is only 1 dependent value (y-value; f(x) value;
range value).
From above, only
satisfies this requirement
b)
y = √x, {x| x ≥ 0, x∈ℜ}
Note: We can check to see if a relation is a f(n) by seeing if any of the x’s are repeated & go to different
y’s; on a graph a relation must pass vertical line test and if it’s a f(n)/equation then you can think about it
in terms of the domain and range or in terms of its graph.
A one-to-one function is a function that satisfies the condition that each element in its
range is used only once (has a unique x-value; domain value).
From above, only
satisfies this requirement
b)
y = √x, {x| x ≥ 0, x∈ℜ}
Note: We can check to see if a function is 1:1 by seeing if any of the y’s are repeated & go to different x’s;
on a graph a 1:1 f(n) must pass horizontal line test and if it’s a f(n)/equation then you can think about it in
terms of the domain and range or in terms of its graph.
If you remember from your study of Algebra, we care about one-to-one functions because
they have an inverse. The inverse of a function, written f-1(x), is the function for which
the domain and range of the original function f(x) have been reversed. The composite of
an inverse and the original function is always equal to x.
Inverse of f(x)
f -1(x) = {(y, x) | (x, y) ∈ f(x)}
Note: f -1(x) is not the reciprocal of f(x) but the notation used for an inverse function!! The same will be true
with our trigonometric functions.
Facts About Inverse Functions
1)
Function must be 1:1 for an inverse to exist; we will sometimes restrict the
domain of the original function, so this is true, but only if the range is not
effected. Note: You will see this with the trig functions because they are not 1:1 without
the restriction on the domain.
2)
3)
4)
5)
(x, y) for f(x) is (y, x) for f -1(x)
f(x) and f -1(x) are reflections across the y = x line
The composite of f(x) with f -1(x) or f -1(x) with f(x) is the same; it is x
The inverse of a function can be found by changing the x and y and solving for y
and then replacing y with f -1(x). Note: This isn’t that important for our needs here.
What you should take away from the above list is:
1)
We make the trig functions one-to-one by restricting their domains
2)
(x, y) is (y, x) for the inverse function means that if we have the value of the trig
function, the y, the inverse will find the angle that gives that value, the x
3)
When we graph the inverse functions we will see their relationship to the graphs
of the corresponding trig function as a reflection across y = x.
4)
sin -1 (sin x) is x
Note: You might use this one in your Calculus class.
The Inverse Sine – Also called the ArcSin
If
f(x) = sin x
on
D: [-π/2, π/2] & R: [-1, 1]
Notice: Restriction to the original domain is to QI & QIV of
unit circle!
then
f -1(x) = sin-1 x or f -1(x) = arcsin x
Note: Scan from p. 551 Stewart ed 5
on
D: [-1, 1] & R: [-π/2, π/2]
Note: Scan from p. 551 Stewart ed 5
Finding y = arcsin x
1)
Think of arcsin as finding the x-value (the radian measure or degree measure of the angle)
of sin that will give the value of the argument.
Rewrite y = arcsin x
to
sin y = x
where y = ?
if it helps.
2)
Think of your triangles! What ∠ are you seeing the opposite over hypotenuse
for?
Example:
Find the exact value for y without a calculator. Don’t forget to check domains!
Use radian measure to give the angle.
a)
y = arcsin √3/2
b)
y = sin -1 (-1/2)
c)
y = sin -1 √2
However, it is not always possible to use our prior experience with special triangles to
find the inverse. Sometimes we will need to use the calculator to find the inverse.
Example:
Find the value of the following in radians rounded to 5 decimals
*Precalculus, 6th ed., Stewart, Redlin & Watson p. 467 #10
y = sin -1 (1/3)
The Inverse Cosine – Also called the ArcCos
If
f(x) = cos x on
D: [0, π] & R: [-1, 1]
Notice: Restriction to the original domain is to QI & QII of
unit circle!
then
f -1(x) = cos-1 x or f -1(x) = arccos x on
Note: Scan from p. 553 Stewart ed 5
D: [-1, 1] & R: [0, π]
Note: Scan from p. 553 Stewart ed 5
Example:
Find the exact value of y for each of the following in radians.
a)
y = arccos 0
b)
y = cos -1 (1/2)
Example:
Find the value of the following in radians rounded to 5 decimals
*Precalculus, 6th ed., Stewart, Redlin & Watson p. 467 #8
y = cos-1 (-0.75)
The Inverse Tangent – Also called the ArcTan
If
f(x) = tan x on
D: [-π/2, π/2] & R: [-∞, ∞]
Notice: Restriction to the original domain is to QI & QIV of
unit circle!
then
f -1(x) = tan-1 x or f -1(x) = arctan x
Note: Scan from p. 555 Stewart ed 5
on
D: [-∞, ∞] & R: [-π/2, π/2]
Note: Scan from p. 555 Stewart ed 5
Notice: The inverse tangent has horizontal asymptotes at ±π/2. It might be interesting to
note that the inverse tangent is also an odd function (recall that tan (-x) = -tan x and also the
arctan (-x) = - arctan (x)) just as the tangent is and that both the x & y intercepts are zero as
well as both functions being increasing functions.
Example:
Find the exact value for y, in radians without a calculator.
y = arctan √3
I am not going to spend a great deal of time talking about the other 3 inverse trig
functions in class or quizzing/testing your graphing skills for these trig functions. I do
expect you to know their domains and ranges and how to find exact and approximate
values for these functions.
Summary of Sec-1, Csc-1 and Cot-1
Inverse Function
Domain
y = cot-1 x
(-∞, ∞)
-1
y = sec x
(-∞,-1]∪[1,∞)
-1
y = csc x
(-∞,-1]∪[1,∞)
Example:
Interval
(0, π)
[0, π], y≠π/2
[-π/2, π/2], y≠0
Quadrant on Unit Circle
I & II
I & II
I & IV
Find the exact value for y, in radians without a calculator.
y = arccsc -√2
Before we do any examples that require the use of our calculator to find the inverse of
these functions, it is important that you recall your reciprocal identities! It is by using
these reciprocal identities first that you can find the inverse values on your calculator. Let
me show you an example first.
Example:
Find the approximate value in radians rounded to the nearest 5 decimals.
y = sec-1 (-4)
STEP 1:
Rewrite as secant
sec y = -4
STEP 2:
Rewrite both sides by taking the reciprocal 1/sec y = -1/4
STEP 3:
Use reciprocal identity to rewrite
cos y = -1/4
STEP 4:
Find the inverse of both sides to solve for y
cos-1(cos y) = cos-1 (-1/4)
STEP 5:
Use your calculator to find
y = cos-1 (-1/4)
Example:
Find the approximate value in radians rounded to the nearest 5 decimals.
y = arccot (-0.2528)
We can also “take a page” from the skill-set that we have just been using and use it to
solve a little more difficult problems – those that deal with the composite of trig and
inverse trig functions. In these type of problems we will work from the inside out. We
don’t actually ever get a degree measure from these problems, because the final answer
is actually the ratio of sides.
1)
We will use the inside function to put together a triangle
2)
Then we use the Pythagorean Theorem to find the missing side
3)
Finally, we use the definitions of the trig functions based upon opposite, adjacent
and hypotenuse lengths to answer the problem.
Example:
Evaluate each of the following without a calculator
a)
cos (sin-1 2/3)
b)
sec (cot-1 (-15/8))
Note that I have only given examples that deal with radians, but there are also problems
that ask for the answers in degrees. Practice both and remember when it comes to
applications that involve arc length the use of radians is necessary. See #42 p. 468 of
edition 6 for example.
Let’s do an application problem that will remind us of our previous instruction on angles
of elevation.
Example:
Find the angle of elevation of the sun.
*Precalculus, 6th ed., Stewart, Redlin & Watson
p. 468 #38
96 ft.
θ
120 ft.
§9.5 Solving Basic Trig Equations
Recall conditional linear equations in Algebra had 1 solution. We are going to study
conditional trig equations first. This means that some values will solve them and some
will not.
Method 1: Linear Methods
1)
Employ algebra to isolate the trig function
2)
Use trig identities as needed
3)
Use definitions as needed
4)
When using the inverse to solve, make sure that you realize that the answer you
get comes from the range of the inverse function & you may need to use that as a
reference angle(point) to find all possible values for original equation.
Recall:
sin(sin-1 x)
on [-1, 1]
sin-1 (sin x)
on [-π/2, π/2]
-1
cos(cos x)
on [-1, 1]
-1
cos (cos x)
on [0, π]
tan(tan -1 x)
on (-∞, ∞)
-1
tan (tan x)
on (-π/2, π/2)
5)
Also realize that when finding all possible solutions that each of the functions
repeats with a certain period and once the solutions are found in the fundamental
period they can then be expanded to encompass all periods.
Recall:
sine & cosine’s period is
2π
∴
2 values w/in fundamental cycle + 2kπ
where k is an integer
tangent’s period is π
∴
2 values w/in fundamental cycle + kπ
where k is an integer
Example:
Solve
3 tan θ – √3 = 0
Method 2: Factoring & Quadratic Methods
1)
Get f(x) = 0 using algebra
2)
Factor & use zero factor property
3)
Use Method 1’s process if needed
Example:
Solve
tan x sin x + sin x = 0
on [0, 360º)
Example:
Solve
cos θ cot θ = -cos θ
on [0, 360º)
Sometimes we will have extraneous roots that we will need to catch as we did when
solving rational equations in algebra.
Method 3: Using Trig Identities
1)
See a relationship that exists between 2 functions
2)
Square sides in order to bring out an identity – be careful to check solutions for
extraneous roots any time you do this
3)
Use an identity to rewrite in terms of one function
4)
Solve as in the last section
Example:
Solve
cot x – √3 = csc x
on [0, 2π)
Note: csc and cot are related via a Pythagorean Identity. You can create a problem by squaring to get
there, so do check extraneous roots.
Example:
Solve
cos 2x = sin x
Note: Did you get 3 solutions? Don’t forget about reference angles.
on [0, 2π)
Example:
Solve
2 cos2 θ – 2 sin2 θ + 1 = 0
on [0, 360º)
Note: 1st because the period is cut in half, the function will oscillate twice the number of times on [0, 360º),
so we need to find the answers on [0, 720º) and furthermore we end up solving for 2A, we need to make
sure we answer for A by dividing all by 2!
Example:
Solve
cos 2x + cos x = 0
§9.6 Trig Equations and Inequalities
When solving a trig inequality, using a graph is the easiest method. Recall that solving a
trig equation graphically requires looking for the x-intercepts. Let’s take a look at the
graphic solution for the second to the last example:
Example:
1)
2)
3)
2 cos2 θ – 2 sin2 θ + 1 = 0
Rewrite using the double angle formula for cos 2 θ
Investigate the domain and range of the function
a)
Domain [0, 2π)
b)
Range [-1, 3]
c)
Scale π/6
The
on [0, 2π)