Download The Pythagorean Identity

Lesson #90 –Trig. Equations that require factoring
A2.A.68 Solve trigonometric equations for all values of the variable from 0° to 360°
Trig Function
Is Positive in Quadrants
sin 
Is Negative in Quadrants
cos
tan 
Solve the following equations:
1)
x 2  3 x  2
2)
2x2  x  6  0
3)
3xy  y  0
With any trigonometric equation we are really just following the same steps as if we were solving for a
normal variable. Therefore, when a trigonometric equation is quadratic or has two different functions in
it, we need to factor to solve it. The only new step is taking the inverse trigonometric function to solve
for the angle. You can think about what you would do if there were a normal variable in the place
of the trig function at any time to make the problem easier.
We write
Example: Solve for  in the interval 0° ≤  ≤ 360°: sin 2   3sin   2
1) Solve the quadratic as if the entire trig. function
was a variable.
2) When you are to the point where you have isolated
the trig function, use ASTC to determine the
quadrants for your answers. (You will have to do
this twice since your factoring creates two
equations.)
3) Use the inverse trig function to solve for the
angles.
4) Put your answers in the quadrants you chose.
1
 sin( ) 
2
as
sin 2  . When
you enter it in
your calculator,
you can simply
enter sin   .
2
Do not try to
enter it with the
2
between sine
and ø.
1.
Solve:
cos 2 x  cos x
on the interval 0  x  360
2.
Find to the nearest degree, all values of  in the interval 0    360 that satisfy the
equation 6 cos 2   1  cos  .
3.
If  is an angle in 0    360 and tan 2   4  0 , what is the value of  to the nearest
degree?
2
4.
Solve: 3cos x sin x  sin x  0 on the interval 0  x  360 . Round to the nearest minute.
5.
Challenge Problem: +2 homework points
Solve: 𝑠𝑖𝑛 𝑥 − 6𝑠𝑖𝑛𝑥 + 4 = 0 on the interval 0  x  360 . Round to the nearest degree.
2
3
Lesson #91 – Trigonometric Identities
A2.A.67 Justify the Pythagorean identities
A2.A.58 Know and apply the co-function and reciprocal relationships between trigonometric
ratios
A Trig Identity is a trig equation that is always true, no matter the value of the variable. Identities
that equal each other can be substituted for each other to help solve other trig equations. You already
know the trig. identities below.
Angles add up to 90°
Reciprocal Identities
C. sec =
D. csc =
E. cot =
Cofunction Identities
Quotient Identities
A. tan =
B. cot =
F. sin =
G. cos =
H. cos
I. tan =
J.
M. sec =
K. cot =
L.
N. csc =
O.
Hints for the following problems




It usually helps to write everything in terms of sinø and cosø.
Use the quotient identity for cotø, not the reciprocal identity.
Whenever you see a trig. function with 90-ø, use a cofunction identity.
Be CREATIVE and PERSISTENT. You might not get it the first time, so try something else.
Simplifying, Solving, and Proving with the Basic Identities
1.
3.
Simplify the expression to a single
trigonometric function: (cos  )(tan  ).
2.
Solve for x. (Assume the angle
measures are degrees)
Sin(3x+7)°=Cos(6x+2)°
4.
Simplify the expression to a single trigonometric
function
cot y
csc y .
Simplify the expression to a single trigonometric
function: cos  sin(90   ) 
4
.
5.
Simplify the expression to a single
trigonometric function:
7.
sec 
csc  .
Solve for x. (Assume the angle
measures are degrees)
6.
8.
9.
Simplify the expression. Leave your
answer in terms of sinø and cosø:
sec  csc
11.
13.
Prove that
cot   csc(90   )   csc
Prove that
sin  (csc   1)  tan  (cot   cos  )
If is ø acute, and cos  
following that are true:
tan( x2  4 x)  cot(30)
cofunctions
Prove that tan  csc  sec
4
circle each of the of the
5
4
5
4
4
sin  90    
csc(90   ) 
5
5
5
5
csc  
sec(90   ) 
4
4
5
5
sec  
csc(90   ) 
4
4
2
10.
If A is acute and tan A  , then
3
2
1)
cot A 
3
1
2)
cot A 
3
2
3)
cot(90  A) 
3
1
4)
cot(90  A) 
3
12. If cos72  sin x , find the number of degrees in the
sin  
4
5
sec(90   ) 
measure of acute angle x.
14.
The expression (sec2  )(cot 2  )(sin  ) is equivalent
to
5
5)
6)
7)
8)
sin 
cos
csc
sec
2
The Pythagorean Identity









The picture to the right is the unit circle.
Label the point on the circle (x,y). Label the angle ø.
Label the base and height of the triangle in terms of x and y.
Write the length of the radius.
Substitute the sides of the triangle you have labeled into the
Pythagorean Theorem.
_______________
-4
-3
-2
From your study of the unit circle, you know that
x= _______ and y= _________.
Substitute these trig functions into your Pythagorean Theorem
equation.
____________________.
This is the Pythagorean Identity.
You can also solve for sin 2  : _______________.
Other Pythagorean Identities
2nd Pythagorean Identity
(Divide by cos 2 x )
1
0.5
-1
1
-0.5
-1
-1.5
Basic Pythagorean Identity
-2
In the same way you can solve for cos 2  : _______________.
Basic Pythagorean
Identity
1.5
-2.5
-3
3rd Pythagorean Identity
(Divide by sin 2 x )
In this course we will not do too much with the 2nd and 3rd Pythagorean Identities, but you do need
to know how to justify them. In other words, you need to know how to get them from the basic
Pythagorean Identity.
6
The flowchart below will help you remember the connections between the three Pythagorean Identities.
It will also help you remember that the identities in the small textboxes are just different ways of
writing the basic, second, and third identities.
Basic Pythagorean Identity
Second Pythagorean Identity
(with tanθ and secθ)
sin2θ =
cos2θ =
7
Third Pythagorean Identity
(with cotθ and cscθ)
Lesson #92 – Trigonometric Sum and Difference Identities
A2.A.76 Apply the angle sum and difference formulas for trigonometric functions
All of the identities from yesterday need to be memorized. Hopefully you already know the reciprocal,
quotient, and cofunction identities from earlier this year. For the next two days we will learn identities
that will be given to you on the A2&T Reference Sheet. You will want to have your purple sheet handy!
a. Sin30°=
b. Sin45°=
c. Sin60°=
Cos30°=
e. Cos 45°=
f.
Cos 60°=
g. Tan30°=
h. Tan 45°=
i.
Tan 60°=
d.
If cos A 
5
,and 270°<A<360°, find sinA.
13
The six trigonometric identities below are given to you on the reference sheet. You do not need to
memorize them. We will just be learning how to work with them in this course.
Functions of the Sum of Two Angles
Functions of the Difference of Two Angles
sin( A  B)  sin A cos B  cos A sin B
sin( A  B)  sin A cos B  cos A sin B
cos( A  B)  cos A cos B  sin A sin B
cos( A  B)  cos A cos B  sin A sin B
tan( A  B) 
tan A  tan B
1  tan A tan B
tan( A  B) 
tan A  tan B
1  tan A tan B
1. Find and expression for the exact value of sin 75 in simplest radical form.
2. Determine the value of cos15 in simplest radical form.
Why does it make sense that sin(75°)=cos(15°)?
8
75= 30 + 45
Therefore you can use
sin(45+30) to find the
value of sin(75).
This does not mean
sin75° = sin45°+ sin30°.
You MUST use the
formula.
When you plug in values for these formulas it is important to figure out if you are plugging in an angle for
just A or B or if you are plugging in a sine or cosine value for the entire sinA, cosA, SinB, or CosB.
For two angles, A and B, it is known that 0  A  90 and 90  B  180 . If cos A 
3
7
and sin B 
5
25
find the following values in simplest form.
3. sin( A  B)
4. cos( A  B)
5. Which of the following is equivalent to cos(100)cos(20)  sin(100)sin  20 ?
1)
cos(120)
2)
sin(120)
3)
cos(80)
4)
sin(80)
6. Which of the following is not equivalent to sin(70) ?
1)
3)
1  cos2 70
sin80 cos10  cos80 sin10
2)
sin 40 cos30  cos 40 sin30
4)
sin 60 cos10  cos60 sin10
9
7. Find the exact value of tan(75°).
8. If
tan A 
9. If cos A 
1
4 and tan B  5 Find the value of tan(A+B).
3 13
and 270  A  360 and tan B  2.5 Find the value of tan(A-B).
13
10
1. Match each of the trigonometric values in column A with an equivalent expression in column B.
Column A
Column B
1) sin 35
a) cos50 cos15  sin50 sin15
2) cos35
b) sin50 cos15  cos50 sin15
3) sin 65
c) sin50 cos15  cos50 sin15
4) cos 65
d) cos50 cos15  sin50 sin15
2. Which of the following is equivalent to cos75 cos 25  sin 75 sin 25 ?
1)
sin100
2)
cos100
3)
sin 50
4)
cos50
3. Which of the following gives the exact value of cos 75 ?
1)
3)
6 2
4
1 3
4
2)
3 2
2
4)
2 6
2
4. The value of sin80 can be expressed in terms of sin 40 and cos 40 as
5)
sin 2 40  cos 2 40
6)
2sin 40 cos 40
7)
1  cos 2 40
8)
sin 2 40  cos 2 40
11
5. For two angles, A and B, it is known that 90  A  180 and 270  B  360 . If cos A  
cos B 
4
find each of the following values in exact, simplest form.
5
(a) sin(A+B)
(b) sin (A-B)
(c) cos(A+B)
(d) cos (A-B)
6. For two angles, A and B, it is known that 0  A  90 and 90  B  180 . If sin A 
sin B 
12
and
13
1
, find sin(A+B) in simplest radical form.
2
12
5
and
3
Lesson #93 – Trigonometric Double and Half Angle Identities
A2.A.77 Apply the double-angle and half-angle formulas for trigonometric functions
Functions of the Double Angle
sin 2 A  2sin A cos A
2 tan A
tan 2 A 
1  tan 2 A
cos 2 A  cos 2 A  sin 2 A
c os 2 A  2cos 2 A  1
cos 2 A  1  2sin 2 A
Options are a good thing. You can use any of the three cos(2A) identities. I always use the 2 nd or 3rd
option depending on if the problem gives me a sine value or a cosine value. Just remember, it does not
really matter – JUST PICK ONE!
1. Verify the identity for sin2A using A  30 and 2 A  60
2. For an angle, A, it is known that 90  A  180 . If cos A  
(a) sin(2A)
(b) cos(2A)
3. For an angle A it is known that 180  A  270 . If sin A  
(a) sin(2A)
(b) cos(2A)
13
3
then find:
5
(c) tan(2A)
11
, find the following:
6
4. Which of the following is equivalent to 2sin50 cos50 ?
1)
2)
sin 25
sin100
1)
2)
cos100
cos 25
5. Which of the following is not equal to cos80 ?
1)
2)
1  sin 2 80
1  2sin 2 40
3)
2 cos 2 60  1
4)
cos 2 40  sin 2 40
6. If cos B = .28, then cos2B is closest to
1)
0.56
3)
-0.68
2)
0.14
4)
-0.84
7. The value of cos130 is equal to
1)
3)
1  sin 2 130
2cos 2 260  1
2)
4)
2cos 65
1  2sin 2 65
There is a + or – in the
front of each half angle
identity because you must
first determine what
quadrant the half angle
will be in. From there you
can decide if the trig
function (sine, cosine, or
tangent) will be positive or
negative. For example, if
270°<A<360°,
Functions of the Half Angle
1  cos A
1 
sin  A   
2
2 
1  cos A
1 
cos  A   
2
2 
1  cos A
1 
tan  A   
1  cos A
2 
then ____< ½ A <____.
8. For an angle, A, it is known that 90  x  180 . If cos A  
1 
A 
2 
a) sin 
1 
A 
2 
b) cos 
14
This is in Q__, so only ____
will be positive.
3
then find:
5
1 
A 
2 
c) tan 
Exercise 4: For the angle A it is known that 180  A  270 and sin A  
1 
A 
2 
1 
A 
2 
a) sin 
b) cos 
5
. Find the following values:
13
2
1. Which of the following is equivalent to the expression 2 cos 60  1 ?
1)
2)
2.
1)
2)
cos30
sin 30
3)
4)
cos120
sin120
Which of the following is equal to sin10 ?
3)
4)
1  2sin 2 20
2sin 20 cos 20
2sin5 cos5
cos 2 5  sin 2 5
3. Which of the following is not equal to cos50 ?
1)
2)
1  sin 2 50
2 cos 2 100  1
4. If cos A 
1)
2)
7
9
5

9

3)
cos310
4)
1  2sin 2 25
1
, then cos(2 A)  ?
3
2
3)

3
4)

3
3
15
1 
A 
2 
c) tan 
6.
For an angle, B, it is known that 90    180 and cos B  
(a) sin(2B)
7. For an angle,  , it is known that cos  
5
. Find, in simplest form:
13
(b) cos(2B)
2
and 270    360 . Find, in simplest form:
4
(a) sin(2  )
(b) cos(2  )
16
Lesson #94 Simplifying with the Identities
A2.A.77 Apply the double-angle and half-angle formulas for trigonometric functions
A2.A.67 Justify the Pythagorean identities
A2.A.58 Know and apply the co-function and reciprocal relationships between trigonometric
ratios
The Sine and Cosine Functions of the Double Angle
(from the reference sheet)
Sin2A=
Cos2A=
Cos2A=
Cos2A=
We already did some simplifying in the second lesson of this unit. Now we will learn how to work
with the double-angle formulas and Pythagorean identities when simplifying. We will still be
using the reciprocal and co-function relationships as well. Remember, the double angle
identities are given to you on the reference sheet, but the others are not.
See how many of the identities you can remember. Use the process we learned to find the 2nd
and 3rd Pythagorean Identities.
Reciprocal Identities
sec =
csc =
cot =
Double Angle Identities
Reminder, but will not be used this lesson
sin  2   2sin  cos
Cofunction Identities
cos  2   cos 2   sin 2 
sin =
cos =
cos  2   2cos 2   1
tan =
cot =
cos  2   1  2sin 2 
sec =
csc =
Quotient Identities
Basic Pythagorean Identity
tan =
cot =
Second Pythagorean Identity
sin2θ =
cos2θ =
(with tanθ and secθ)
17
Third Pythagorean Identity
(with cotθ and cscθ)
General Hints
 Look for Double Angles (sin2A or cos2A). Whenever you see them, make a substitution.
 When you see the cosine of a double angle (cos2A), choose the substitution that will make
other parts of that problem cancel.
 Whenever you see a reciprocal function or quotient function, make a substitution.
 Be on the lookout for squared trigonometric functions ( sin 2 x, cos2 x, etc. ) and be ready to
use a Pythagorean Identity. This one is not automatic though.
1.
Simplify the expression to a single
trigonometric function:
3.
Simplify the expression to a single
trigonometric function : 1  sin 2 x .
4.
Prove the Identity:
2
Simplify the expression to a single
trigonometric function:
2.
sin x  cos x
.
cos x
2
sin  (csc  sin  )  cos2 
2
cos x
.
1  cos 2 x
5.
6.
7.
The expression (1 + cos x)(1 - cos x) is 8.
equivalent to
2
(1) 1
(3) sin x
2
(2) sec x
9.
1  cos x   csc x   sin x
Prove the identity:
2
2
(4) csc x
Simplify: cos 2 x  cos 2 tan 2 x
10.
Prove the identity:
18
sin x cos x
1


cos x sin x cos x sin x
11.
cos(2 A)  sin 2 A is equivalent to which of the
12.
following?
(1) 1
2
(3) sin A
2
(2) cos A
2
(4)  sin A
The expression
2 cos
is equivalent to
sin 2
(1) cscø
(3) cotø
(2) secø
(4) sinø
13.
Simplify this complex fraction into a single
trigonometric function:
14.
2  2 sin 2 x
Express in simplest terms:
cos x .
15.
Simplify the expression to a single
sin(2 x)
trigonometric function:
tan x
16.
24. Prove the identity:
17.
18.
19.
20.
sin x  cos x  tan x  csc x sec x cot x   1
Challenge:
Express
cos 2 A  sin 2 A
as a single
cos A
trigonometric function for all values of A for
which the fraction is defined.
19
Lesson #95 – Solving Trigonometric Equations Graphically
A2.A.68 Solve trigonometric equations for all values of the variable from 0° to 360°
2cos x 1  0
0  x  360
It helps to make a scale that goes by 90°.
Set xscl on your calculator to match.
Since our
equations ask for
the answers in
degrees, this is
the only time we
graph in degree
mode!
You used the variable, y, in this
equation just so that you could
graph. Therefore, just the x
values of the intersections are
your answers.
Hint: Find and label the intersections
first before sketching the graph.
15cos 2 x  7 cos x  2  0
0  x  360
20
Note: Equations do not have to be set equal to zero. Sometimes it is easier to find the solutions when
they are not. For example, set the next equation equal to 6.
3cos 2 x  7 cos x  6  0
0  x  360 .
Note: On the regents, the problem will most
likely not tell you to solve graphically. You
have to REALIZE you can do so.
In addition, even if a Regents problem tells
you to solve algebraically, you can solve
graphically and still receive ½ credit.
On June 18th, I can’t tell you to graph – you
must remember!
3sin(2 )  cos   0
2 cos 2 x  3sin x  3  0
21
0  x  360
Lesson #96 – Using Identities to Solve Trig. Equations
A2.A.68 Solve trigonometric equations for all values of the variable from 0° to 360°
A2.A.77 Apply the double-angle and half-angle formulas for trigonometric functions
A2.A.67 Justify the Pythagorean identities
The following equations are probably the most difficult to solve because they require a substitution.
Remember, any trigonometric equation can be solved graphically. You will even receive half credit for
this if the question specifically says to solve algebraically.
Basic Pythagorean Identity:
Trigonometric Identities for Equation Solving
(2nd option)
cos2A=
(3rd option)
Sin2A =
cos2A=
0  A  360
sin(2 A)  cos A  0
cos(2 A)  5cos A  3  0
22
0  A  360
cos 2 x  4sin x  4  0
0  x  360
0    360
3cos 2  2
cos(2 A)  3sin 2 A  6sin A  7  0
Try graphing
this one too!
23