Download Math 30-1 Chapter 6 Lessons

Lesson 1: Reciprocal, Quotient, and Pythagorean Identities
Specific Outcome 6: Prove trigonometric identities, using:
reciprocal
identities, quotient identities, Pythagorean identities
What is the difference between an equation and an identity? Talk through this idea with
the students and see if they can come up with the following ideas.
2𝑥 2 + 3 = 11 is an equation. It is only true for certain values of the variable x. The
solutions to this equation are -2 and 2 which can be verified by substituting these values
into the equation.
(𝑥 + 1)2 = 𝑥 2 + 2𝑥 + 1 is an identity. It is true for all values of the variable x.
Comparing Two Trigonometric Expressions
1. Graph the curves y = sin x and y = cos x tan x over the domain −360° ≤ 𝑥 < 360°.
Graph the curves on separate grids using the same range and scale. What do you
notice?
They are the same graph.
2. Use the table function on your calculator to compare the two graphs. Describe your
findings.
Same values, except for the non-permissible values.
3. Use your knowledge of tan x to simplify the expression cos x tan x
𝑠𝑖𝑛𝑥
cos 𝑥
cos 𝑥
The cos x cancel off to give sin x.
4. What are the non-permissible values of x in the equation sin x = cos x tan x?
Remember that we cannot divide by 0? So use that info to find non-permissible
sin 𝑥
values. Given tan 𝑥 = cos 𝑥, the non-permissible values will occur when cos x = 0.
That occurs at 90˚ and again at 270˚. (denominator cannot = 0)
The equations y = sin x and y = cos x tan x are examples of trigonometric identities.
Trigonometric identities can be verified both numerically and graphically.
Reciprocal Identities:
1
1
csc x 
sec x 
sin x
cos x
Quotient Identities:
sin x
tan x 
cos x
cot x 
cot x 
1
tan x
cos x
sin x
Verifying Identities for a Particular Case: (plugging in values)
When verifying an identity we must treat the left side (LS) and the right side (RS)
separately and work until both sides represent the same value.
This technique does not prove that an identity is true for all values of the variable – only
for the value of the variable being verified.
sin x
.
cos x
a. Determine the non-permissible values, in degrees, for the equation.
Ex. Verify that tan x 
Occurs when cos x = 0 (at 90˚ and 270˚), and can be represented as
𝜃 ≠ 90° + 180°𝑛 (general solution for NPV)
b. Verify that θ= 60˚.
L.S.
R.S.
sin 60
cos 60
tan 60
3
 3 (using unit circle or triangles)

1
2
2
3 2

2 1
 3

LS = RS
c. Verify graphically that this is an identity.
Answer: When you graph each side, they are the same graph.
2
Ex. a. Determine the non-permissible values, in radians, of the variable in the
sec 𝑥
expression tan 𝑥
The functions sec x and tan x both have non-permissible values in their domains:
tan 𝑥 =
sin 𝑥
𝜋
, 𝑡ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒 cos 𝑥 ≠ 0, (𝑥 ≠ + 𝜋𝑛)
cos 𝑥
2
1
sec 𝑥 =
= 𝑠𝑎𝑚𝑒 𝑎𝑠 𝑎𝑏𝑜𝑣𝑒.
cos 𝑥
The value of tan x also cannot equal 0. Tan θ =0 at π
𝜋
Therefore, combine the restriction is 𝑥 ≠ 2 𝑛. A restriction will occur every 90˚.
b. Simplify the expression (substitute known identities into the equation)
1
cos 𝑥
sin 𝑥
cos 𝑥
1
cos 𝑥
cos 𝑥
1
×
=
=
= 𝑐𝑠𝑐𝑥
cos 𝑥 sin 𝑥 cos 𝑥 sin 𝑥 sin 𝑥
Pythagorean Identity
Recall that point P on the terminal arm of an angle θ in standard position has
coordinates (cos θ, sin θ). Consider a right triangle with a hypotenuse of 1 and legs of
cos θ and sin θ.
y
P(cosθ, sinθ)
x
The hypotenuse is 1 because it is the radius of the unit circle. Apply the Pythagorean
Theorem in the right triangle to establish the Pythagorean Identity:
𝑥2 + 𝑦2 = 1
𝑐𝑜𝑠 2 𝜃 + 𝑠𝑖𝑛2 𝜃 = 1
Pythagorean Identities:
sin 2 x  cos 2 x  1
1  tan 2 x  sec2 x
1  cot 2 x  csc2 x
3
These identities can be written in several ways:
sin 2 x  1  cos 2 x
tan 2 x  sec2 x  1
cos 2 x  1  sin 2 x
cot 2 x  csc2 x  1
Ex. Verify the identity 1 + 𝑐𝑜𝑡 2 𝑥 = 𝑐𝑠𝑐 2 for the given values:
a. 𝑥 =
𝜋
6
𝜋 2
𝜋 2
L.S. = 1 + (𝑐𝑜𝑡 6 )
R.S. = (𝑐𝑠𝑐 6 )
2
= 22
=4
= 1 + (√3)
=1+3
=4
4𝜋
b. 3
240˚(QIII) will have a ref angle of 60˚, therefore the Tan will be positive and sin
will be negative. tan 60 = √3, sin 60 =
L.S. = 1 + (𝑐𝑜𝑡
4𝜋 2
)
3
2
√3
2
R.S. = (𝑐𝑠𝑐
1
=1+( )
√3
1
=1+
2
(√3)
1
=1+
3
4
=
3
=(
−2
√3
4
=3
4𝜋 2
)
3
2
)
b. Use quotient identities to verify the Pythagorean identity 1 + 𝑐𝑜𝑡 2 𝑥 = 𝑐𝑠𝑐 2 x
𝑐𝑜𝑡 2 𝑥 + 1 = 𝑐𝑠𝑐 2 𝑥
𝑐𝑜𝑠 2 𝑥
1
+
1
=
𝑠𝑖𝑛2 𝑥
𝑠𝑖𝑛2 𝑥
2
2
𝑐𝑜𝑠 𝑥 𝑠𝑖𝑛 𝑥
1
+
=
2
2
𝑠𝑖𝑛 𝑥 𝑠𝑖𝑛 𝑥 𝑠𝑖𝑛2 𝑥
𝑐𝑜𝑠 2 𝑥 + 𝑠𝑖𝑛2 𝑥
1
=
2
𝑠𝑖𝑛 𝑥
𝑠𝑖𝑛2 𝑥
1
1
=
2
𝑠𝑖𝑛 𝑥 𝑠𝑖𝑛2 𝑥
𝑐𝑠𝑐 2 𝑥 = 𝑐𝑠𝑐 2 𝑥
Assignment: pg. 296-297 #1, 3, 4, 6, 9 – 11, 13 –17, C2
4
Lesson 2: Sum, Difference and Double-Angle Identities
Specific Outcome 6: Prove trigonometric identities, using:
•
•
•
•
•
reciprocal identities
quotient identities
Pythagorean identities
sum or difference identities (restricted to sine, cosine and tangent)
double-angle identities (restricted to sine, cosine and tangent).
Ex. Use exact values to verify which of the following statements are true:
a. sin(60 + 30)º= sin60º + sin30º
b. sin  60  30   sin 60 cos30  sin 30 cos 60
Have the students go through both examples using exact value and see which one is
true. Once they’ve figured it out, go through the different sum and difference identities.
Sum and Difference Identities:
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  tan B
tan( A  B) 
1  tan A tan B
tan A  tan B
tan( A  B) 
1  tan A tan B
Thankfully, these are on the formula sheet so they do not need to be memorized.
However, we can use them in a number of problems.
1. The first type is simplifying expressions into single trigonometric functions:
Ex. Write the expression cos 88° cos 35° + sin 88° sin 35° as a single trigonometric
function.
Look on the formula sheet. It has the same form as the right side of cos(A – B):
Therefore we can substitute 88 for A and 35 for B. cos (88-35) = cos53˚
2. The second type is finding exact values of angles that are not on our special
triangle.
5
Ex. Find the exact value of sin15
The first step is to look at our special triangle angles: 30, 45 and 60. Then see
which combinations break the angle into exact values
sin15  sin  45  30   sin 45 cos30  sin 30 cos 45
2
3 1
2

 
2
2 2 2
6
2


4
4

6 2
4
Ex. Find the exact value of 𝑡𝑎𝑛
11𝜋
12
When faced with radians, it may be easier to switch to degrees (165˚).
Otherwise you will have to find common denominators. First step is to try the
special triangle angles or use unit circle. If those don’t work, then we also need
to look at the angles that are in other quadrants. If you are looking at degrees,
we can use 135 and 30.
Rewrite as a addition of two special angles (135 + 30)
tan 𝐴 + tan 𝐵
1 − tan 𝐴 tan 𝐵
tan 135 + tan 30
tan(135 + 30) =
1 − tan 135 tan 30
tan(𝐴 + 𝐵) =
From special triangles tan 30 =
√3
3
For 135, it has a reference angle of 45 in QII (tan is negative). Tan 45 = 1
1
√3
tan(135 + 30) =
1
1 − (−1 ( ))
√3
1
−1 +
√3
tan(135 + 30) =
1
1+
√3
−1 +
6
1
√3
+
tan(135 + 30) = √3 √3
1
√3
+
√3 √3
(common denominators)
Simplify by combining fractions:
−√3 + 1 √3 + 1
tan(135 + 30) =
÷
√3
√3
−√3 + 1
√3
tan(135 + 30) =
×
√3
√3 + 1
−√3 + 1
tan(135 + 30) =
√3 + 1
−
There are also double angle identities. Have the students try to find the identities for
sin2A and cos2A by using the sum and difference identities. Once we have gone
through this, give the identities to the students.
Double Angle Identities:
sin 2 A  2sin A cos A
cos 2 A  cos 2 A  sin 2 A
2 tan A
tan 2 A 
1  tan 2 A
Note: cos2A has two other forms. Go through these with the students as well.
Ex. Determine an identity for cos 2A that contains only the sine ratio.
cos 2𝐴 = 𝑐𝑜𝑠 2 𝐴 − 𝑠𝑖𝑛2 𝐴
cos 2𝐴 = (1 − 𝑠𝑖𝑛2 𝐴) − 𝑠𝑖𝑛2 𝐴 (Pythagorean identity)
cos 2𝐴 = 1 − 2𝑠𝑖𝑛2 𝐴
Ex. Determine an identity for cos 2A that contains only the cosine ratio.
cos 2𝐴 = 𝑐𝑜𝑠 2 𝐴 − 𝑠𝑖𝑛2 𝐴
cos 2𝐴 = 𝑐𝑜𝑠 2 𝐴 − (1 − 𝑐𝑜𝑠 2 𝐴)
cos 2𝐴 = 2𝑐𝑜𝑠 2 𝐴 − 1
These identities are on the formula sheet
7
Ex. Express each in terms of a single trigonometric function:
a. 2 sin 4x cos 4x
This follows (2 sinA cosA) = sin 2(4x)=sin 8x
1
1
A  sin 2 A
2
2
1 
cos 2  A 
2 
 cos A
b. cos 2
5
5
c. sin x cos x
2
2
1
5
5 
 2sin x cos x 
2
2
2 
1
5 
 sin 2  x 
2
2 
1
 sin 5 x
2
(the 1/2 is added to the outside of the brackets to balance off the 2 added to the inside,
the 2 is added to match the identity sin2A=2 sinAcosA)
One of the most common questions is to simplify expressions using identities.
sin 2𝑥
Ex. Simplify the expression cos 2𝑥+1 to one of the three primary trigonometric ratios
2 sin 𝐴 cos 𝐴
(2𝑐𝑜𝑠 2 𝐴 − 1) + 1
2 sin 𝐴 cos 𝐴
=
2𝑐𝑜𝑠 2 𝐴
sin 𝑥
=
= tan 𝑥
cos 𝑥
=
Assignment: pg. 306-308 #1ae, 2cd, 4bc, 5abd, 7, 8acdf, 10, 11, 15, 16b, 17, 19b,
20ad, 24
8
Lesson 3: Proving Identities
Specific Outcome 6: Prove trigonometric identities, using:
•
•
•
•
•
reciprocal identities
quotient identities
Pythagorean identities
sum or difference identities (restricted to sine, cosine and tangent)
double-angle identities (restricted to sine, cosine and tangent).
To prove that an identity is true for all permissible values, it is necessary to express both
sides of the identity in equivalent forms. One or both sides of the identity must be
algebraically manipulated into an equivalent form to match the other side.
You cannot perform operations across the equal sign when proving a potential identity.
The equal sign is a border! Simplify the expressions on each side of the identity
independently.
Hints in Proving an Identity:
1. Begin with the more complex side
2. If possible, use known identities given on the formula sheet, eg. try to use
the Pythagorean identities when squares of trigonometric functions are
involved.
3. If necessary change all trigonometric ratios to sines and/or cosines, eg.
sin x
1
replace tan x by
, or sec x by
.
cos x
cos x
4. If there is a sum or difference of fractions, write as a single fraction
5. Look for factoring as a step in trying to prove an identity.
6. Occasionally, you may need to multiply the numerator or denominator of
a fraction by its conjugate.
It is usually easier to make a complicated expression simpler than it is to make a simple
expression more complicated. In this section we will be doing verifying again as well as
proving. Make sure that you can do both. Verifying means we simply plug in an angle
and prove that left side = right side. Proving means that we need to do it algebraically.
9
Ex. Consider the statement
1
 cos x  sin x tan x
cos x
a. Verify the statement is true for x 
L.S.
1
cos

 cos

3
R.S.

3
sin
3
1 1


1
2
2
1
 2
2
3

2

tan

3
3
3

 3
2
3

2
b. Prove the statement algebraically:
L.S.
1
 cos x
cos x
1  cos 2 x

cos x
sin 2 x

cos x
R.S.
sin x tan x
sin x
 sin x 
cos x
2
sin x

cos x
c. State the non-permissible values for x.
The only non-permissible values for either side is cos x ≠ 0.
𝑥≠
𝜋
+ 𝜋𝑛
2
2 tan 𝑥
Ex. Consider the identity 1+𝑡𝑎𝑛2 𝑥 = sin 2𝑥
a. Describe how to use a graphing calculator to verify the identity.
2 tan 𝑥
If you graph 𝑦1 = 1+𝑡𝑎𝑛2 𝑥 and graph 𝑦2 = sin 2𝑥, The graphs should be identical
10
b. Prove the identity
L.S.
sin 𝑥
2 cos 𝑥
=
𝑠𝑒𝑐 2 𝑥
2 sin 𝑥
= cos 𝑥
1
𝑐𝑜𝑠 2 𝑥
R.S.
= 2 sin 𝑥 cos 𝑥
2 sin 𝑥 𝑐𝑜𝑠 2 𝑥
×
cos 𝑥
1
= 2 sin 𝑥 cos 𝑥
=
1
Ex. Prove that 1−sin 𝐴 =
1+sin 𝐴
𝑐𝑜𝑠2 𝐴
is an identity for all permissible values of x.
(this is an example of multiplying by the conjugate. Also show them how to do the
identity substituting in for the cos denominator and factoring difference of squares.)
𝐿𝑆 =
2 cos 𝑥−1
1
1 + sin 𝐴
×
1 − sin 𝐴 1 + sin 𝐴
1 + sin 𝐴
=
1 − 𝑠𝑖𝑛2 𝐴
1 + sin 𝐴
=
𝑐𝑜𝑠 2 𝐴
1
Ex. Prove that 2𝑐𝑜𝑠2 𝑥−7 cos 𝑥+3 = cos 𝑥−3 is an identity. For what values of x is this identity
undefined?
2 cos 𝑥 − 1
− 6 cos 𝑥 − cos 𝑥 + 3
2 cos 𝑥 − 1
=
2 cos 𝑥(cos 𝑥 − 3) − 1(cos 𝑥 − 3)
2 cos 𝑥 − 1
=
(cos 𝑥 − 3)(2 cos 𝑥 − 1)
1
=
cos 𝑥 − 3
𝐿𝑆 =
2𝑐𝑜𝑠 2 𝑥
11
For non-permissible values, we need to find the equations cos x – 3 = 0 or 2cos x – 1 =
0 are undefined.
1
2
Quad. I and IV (positive)
Ref angle is 60˚
QI: 60˚
QIV: 360˚-60˚ = 300˚
Undefined for
x = 60˚+360˚n, nεI
x = 300˚+360˚n, nεI
cos 𝑥 = 3
No solution.
cos 𝑥 =
I
Assignment: pg. 314-315 #1ad, 2ac, 3acd, 5, 7b, 10ac, 11bc, 12, 14, 18, C1, C2
12
Lesson 4: Solving Trigonometric Equations Using Identities
Specific Outcome 5: Solve, algebraically and graphically, first and second degree trigonometric equations with the
domain expressed in degrees and radians.
For our last section of trig, we are going to combine identities and solving equations.
Investigation
1. Graph the function 𝑦 = sin 2𝑥 − sin 𝑥 over the domain −720° < 𝑥 ≤ 720°. Describe
the graph.
2. From the graph, determine an expression for the zeros of the function over the
domain of all real numbers. (don’t do all…just find a couple for practice)
-660˚, -540˚, -420˚, -360˚, -300˚,-180˚, -60˚, 0˚, 60˚, 180˚, 300˚, 360˚, 420˚, 540˚, 660˚,
720˚
4 solutions for each cycle (look from max to max or min to min)
3. Algebraically solve sin 2𝑥 − 𝑠𝑖𝑛𝑥 = 0 over the domain. Compare your answers to the
question above.
sin 2𝑥 − 𝑠𝑖𝑛𝑥 = 0
= 2 sin 𝑥 cos 𝑥 − sin 𝑥
= sin 𝑥 (2 cos 𝑥 − 1)
sin 𝑥 = 0
At 0˚, 180˚, 360˚, ….
𝑥 = 0° + 180°𝑛
1
2
Ref angle is 60˚(answers in QI and QIV)
= 60˚, 300˚, 420˚, 660˚…
𝑥 = 60° + 360°𝑛
𝑥 = 300° + 360°𝑛
cos 𝑥 =
4. Which method, graphic or algebraic, do you prefer to solve the above equation?
To solve some trigonometric equations, you need to make substitutions using the
trigonometric identities that you have studied in this chapter. This often involves
ensuring that the equation is expressed in terms of one trigonometric function.
13
Ex. Solve the following equation where 0  x  2 . (easier to use graphing
calculator…know both)
2 cos 2 x  3sin x  0
2(1  sin 2 x)  3sin x  0  Pythagorean identity
2  2sin 2 x  3sin x  0  Expand brackets
0  2sin 2 x  3sin x  2  Need to factor
 2sin x 1sin x  2
sin x 
1
2
sin x  2
Positive in Quad I and II
ref   sin 1 (0.5)  30 
x
 
, 
6 
No solution

6
 5

  or x  ,
6 6
6
But, of course, you can use the graphing calculator.
Ex. Solve for x as an exact value where 0 ≤ 𝑥 ≤ 2𝜋
3 − 3 sin 𝑥 − 2𝑐𝑜𝑠 2 𝑥
3 − 3 sin 𝑥 − 2(1 − 𝑠𝑖𝑛2 𝑥) = 0
3 − 3 sin 𝑥 − 2 + 2𝑠𝑖𝑛2 𝑥 = 0
2𝑠𝑖𝑛2 𝑥 − 3 sin 𝑥 + 1 = 0
2𝑠𝑖𝑛2 𝑥 − 2 sin 𝑥 − sin 𝑥 + 1 = 0
2 sin 𝑥 (sin 𝑥 − 1) − 1(sin 𝑥 − 1) = 0
(sin 𝑥 − 1)(2 sin 𝑥 − 1) = 0
1
2
𝜋
Quad I, II, ref angle is 6
sin 𝑥 = 1
𝜋
Quad I, II, ref angle is
2
𝜋
𝑥=
2
sin 𝑥 =
𝑥=
14
𝜋 5𝜋
,
6 6
1
Ex. Solve the equation 𝑠𝑖𝑛2 𝑥 = 2 tan 𝑥 cos 𝑥 algebraically over the domain 0° ≤ 𝑥 <
360°.
1 sin 𝑥
𝑠𝑖𝑛2 𝑥 = ×
× cos 𝑥
2 cos 𝑥
2𝑠𝑖𝑛2 𝑥 = sin 𝑥
2𝑠𝑖𝑛2 𝑥 − sin 𝑥 = 0
sin 𝑥 (2 sin 𝑥 − 1)
1
2
Ref angle is 30˚, pos in QI and II
𝑥 = 30°, 150°
sin 𝑥 = 0
sin 𝑥 =
𝑥 = 180°
Check for any non-permissible values: occur when cos x = 0, at 90˚, 270˚
The non-permissible values are not part of our solution set, therefore our solutions are x
= 30˚, 150˚, and 180˚
b. Verify your answer graphically.
Ex. Algebraically solve cos 2𝑥 = cos 𝑥. Give general solutions expressed in radians.
𝑐𝑜𝑠 2 𝐴 − 𝑠𝑖𝑛2 𝐴 = cos 𝑥
𝑐𝑜𝑠 2 𝐴 − (1 − 𝑐𝑜𝑠 2 𝐴) = cos 𝑥
2𝑐𝑜𝑠 2 𝐴 − 1 = cos 𝑥
2𝑐𝑜𝑠 2 𝐴 − cos 𝑥 − 1 = 0
2
2𝑐𝑜𝑠 𝐴 − 2 cos 𝐴 + cos 𝐴 − 1 = 0
(2 cos 𝐴 + 1)(cos 𝐴 − 1) = 0
1
2
𝜋
Ref angle is 3 , neg in QII andQIII
𝜋 2𝜋
𝑥 =𝜋− =
3
3
𝜋 4𝜋
𝑥 =𝜋+ =
3
3
𝐶𝑜𝑠 𝐴 = 1
Occurs at 0
cos 𝐴 = −
General Solutions:
2𝜋
3
+ 2𝜋𝑛,
4𝜋
3
+ 2𝜋𝑛, 0 + 2𝜋𝑛
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Ex. Algebraically solve 3 cos 𝑥 + 2 = 5 sec 𝑥. Give general solutions expressed in
radians.
3 cos 𝑥 + 2 = 5 sec 𝑥
1
3 cos 𝑥 + 2 = 5 ×
cos 𝑥
5
3 cos 𝑥 + 2 =
cos 𝑥
cos 𝑥 (3 cos 𝑥 + 2) = 5
3𝑐𝑜𝑠 2 𝑥 + 2 cos 𝑥 − 5 = 0
3𝑐𝑜𝑠 2 𝑥 − 3 cos 𝑥 + 5 cos 𝑥 − 5 = 0
(cos 𝑥 − 1)(3 cos 𝑥 + 5) = 0
5
3
No solution.
cos 𝑥 = 1
Occurs at 0
cos 𝑥 = −
Only non-permissible value occurs when cosx = 0, (90˚, 270˚)
The non-permissible values do not change our original solution of 0 +2πn
Unless the domain is restricted, give general solutions. Check that the solutions for an
equation do not include non-permissible values from the original equation.
Assignment: pg. 320-321 #1ad, 2bc, 3bc, 5, 6, 8, 10 – 12, 15, 18, C2
16