Download Trigonometric Identities

University of North Georgia
Department of Mathematics
Instructor: Berhanu Kidane
Course: Precalculus Math 1113
Text Books: For this course we use free online resources:
See the folder Educational Resources in Shared class files
1) http://www.stitz-zeager.com/szca07042013.pdf (Book1)
2) Trigonometry by Michael Corral (Book 2)
Other online resources:
E– Book: http://msenux.redwoods.edu/IntAlgText/
Tutorials: http://www.wtamu.edu/academic/anns/mps/math/mathlab/col_algebra/index.htm
o
o
o
o
o
http://archives.math.utk.edu/visual.calculus /
http://www.ltcconline.net/greenl/java/index.html
http://en.wikibooks.org/wiki/Trigonometry
Animation Lessons: http://flashytrig.com/intro/teacherintro.htm
http://www.sosmath.com/trig/trig.html
For more free supportive educational resources consult the syllabus
1
Trigonometric Identities
(Book 2 page 65)
Objectives: By the end of this section student should be able to



Identify Fundamental or Basic Identities
Find trigonometric values using the trig Identities
Evaluate trigonometric functions
Identities
Equations: Three types
1) Conditional equations: These types of equations have finitely number of solutions.
Example: a) 2𝑥𝑥 − 5 = 7𝑥𝑥, b) 3𝑥𝑥 2 − 4𝑥𝑥 − 6 = 0
2) Contradictions: These are equations that do not have solutions
Examples: 2𝑥𝑥 − 1 = 2(𝑥𝑥 − 1) + 6
3) Identities: These types of equations hold true for any value of the variable
Examples: (𝑥𝑥 + 5)(𝑥𝑥 − 5) = 𝑥𝑥² − 25
Trigonometric Identities are identities of the Trigonometric equations. We use an identity to give an
expression a more convenient form. In calculus and all its applications, the trigonometric identities are
of central importance.
Fundamental or Basic Trigonometric Identities
Reciprocal Identities, Quotient Identities and Pythagorean Identities
1)
Reciprocal identities (Page: 65)
𝒔𝒔𝒔𝒔𝒔𝒔 𝜽𝜽 =
𝒄𝒄𝒄𝒄𝒄𝒄 𝜽𝜽 =
𝒕𝒕𝒕𝒕𝒕𝒕 𝜽𝜽 =
𝟏𝟏
𝐜𝐜𝐜𝐜𝐜𝐜𝜽𝜽
𝟏𝟏
𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔
𝟏𝟏
𝒄𝒄𝒄𝒄𝒄𝒄𝒄𝒄
, 𝑎𝑎𝑎𝑎𝑎𝑎 𝒄𝒄𝒄𝒄𝒄𝒄 𝜽𝜽 =
𝟏𝟏
𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔
𝟏𝟏
, and 𝒔𝒔𝒔𝒔𝒔𝒔 𝜽𝜽 = 𝒄𝒄𝒄𝒄𝒄𝒄𝒄𝒄
𝟏𝟏
, and 𝒄𝒄𝒐𝒐𝒐𝒐 𝜽𝜽 = 𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕
Proof: Follows directly from the definition of trig functions.
2)
Quotient Identities
𝒕𝒕𝒕𝒕𝒕𝒕 𝜽𝜽 =
𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔
𝒄𝒄𝒄𝒄𝒄𝒄𝒄𝒄
, and 𝒄𝒄𝒄𝒄𝒄𝒄 𝜽𝜽 =
𝒄𝒄𝒄𝒄𝒄𝒄𝒄𝒄
𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔
Proof: Follows directly from the definition of trig functions.
2
3)
Pythagorean Identities ( page: 66 & 67)
a) 𝒔𝒔𝒔𝒔𝒔𝒔²𝜽𝜽 + 𝒄𝒄𝒄𝒄𝒄𝒄²𝜽𝜽 = 𝟏𝟏
b) 𝟏𝟏 + 𝒕𝒕𝒕𝒕𝒕𝒕²𝜽𝜽 = 𝒔𝒔𝒔𝒔𝒔𝒔²𝜽𝜽
c) 𝟏𝟏 + 𝒄𝒄𝒄𝒄𝒄𝒄²𝜽𝜽 = 𝒄𝒄𝒄𝒄𝒄𝒄 ²𝜽𝜽
Proof: a) Let 𝑷𝑷(𝒙𝒙, 𝒚𝒚) be on the terminal side of the angle 𝜽𝜽.
Then 𝒓𝒓 = �𝒙𝒙𝟐𝟐 + 𝒚𝒚𝟐𝟐 which implies that 𝒓𝒓𝟐𝟐 = 𝒙𝒙𝟐𝟐 + 𝒚𝒚𝟐𝟐 , 𝐬𝐬𝐬𝐬𝐬𝐬 𝜽𝜽 =
And so, 𝒔𝒔𝒔𝒔𝒔𝒔²𝜽𝜽 + 𝒄𝒄𝒄𝒄𝒄𝒄²𝜽𝜽 =
𝒙𝒙𝟐𝟐
𝒓𝒓𝟐𝟐
+
𝒚𝒚𝟐𝟐
𝒓𝒓𝟐𝟐
=
𝒙𝒙𝟐𝟐 + 𝒚𝒚𝟐𝟐
𝒓𝒓𝟐𝟐
=
𝒓𝒓𝟐𝟐
𝒓𝒓𝟐𝟐
= 𝟏𝟏
𝒚𝒚
, and 𝐜𝐜𝐜𝐜𝐜𝐜 𝜽𝜽 =
𝒓𝒓
𝒙𝒙
𝒓𝒓
Example: Book 2: Example 3.1, 3.2, 3.3, 3.4, 3.5, 3.6 and 3.7 reading (67 – 69)
Note:
a) it follows that: 𝒔𝒔𝒔𝒔𝒔𝒔²𝜽𝜽 = 𝟏𝟏 − 𝒄𝒄𝒄𝒄𝒄𝒄²𝜽𝜽 and 𝒄𝒄𝒄𝒄𝒄𝒄²𝜽𝜽 = 𝟏𝟏 − 𝒔𝒔𝒔𝒔𝒔𝒔²𝜽𝜽,

From

b) and c) are similarly proved.
 𝒔𝒔𝒔𝒔𝒔𝒔²𝜽𝜽, "sine squared theta", means(𝒔𝒔𝒔𝒔𝒔𝒔 𝜽𝜽)²
Example 1:
a) Express sin 𝜃𝜃 in terms of cos 𝜃𝜃
b) Express cos 𝜃𝜃 in terms of sin 𝜃𝜃
c) Express tan 𝜃𝜃 in terms of cos 𝜃𝜃, where θ in Quadrant II
3
d) If tan 𝜃𝜃 = 2 and θ is in Quadrant III, find sin 𝜃𝜃 and cos 𝜃𝜃
1
e) If cos 𝜃𝜃 = 2 and θ is in Quadrant IV, find all other trig values of θ
f) Use the basic trigonometric identities to determine the other five values of the trigonometric
functions given that sin 𝛼𝛼 = 7/8 and cos 𝛼𝛼 > 0.
g) 𝑥𝑥 is in quadrant II and 𝑠𝑠𝑠𝑠𝑠𝑠 𝑥𝑥 = 1/5. Find 𝑐𝑐𝑐𝑐𝑐𝑐 𝑥𝑥 and 𝑡𝑡𝑡𝑡𝑡𝑡 𝑥𝑥.
Example 2: Prove the Pythagorean Identities b) and c)
Example 3: Homework Reading page 67 – 69 Examples 3.1, 3.2, 3.3, 3.4, 3.5, 3.6 and 3.7
(Book 2) Homework Exercises 3.1 page 70: # 1 – 21 odd numbers
Examples YouTube Videos Trigonometric Identities
1) https://www.youtube.com/watch?v=iKWGv0xcuCA
2) https://www.youtube.com/watch?v=QGk8sYck_ZI
3) https://www.youtube.com/watch?v=raVGSdfBVBg
3
4) Pythagorean Identity
5) Trigonometric Identities
6) Verifying more difficult Trig. Identities
4
7) Review Trig identities (1)
Understanding Trig I
5
More on Trigonometry Identities
Further on Trigonometric Identities
a) Co-Function Identities
𝝅𝝅
𝒔𝒔𝒔𝒔𝒔𝒔 � − 𝜽𝜽� = 𝒄𝒄𝒄𝒄𝒄𝒄 𝜽𝜽,
𝟐𝟐
𝝅𝝅
𝒕𝒕𝒕𝒕𝒕𝒕 � − 𝜽𝜽� = 𝒄𝒄𝒄𝒄𝒄𝒄 𝜽𝜽,
𝟐𝟐
𝝅𝝅
𝒔𝒔𝒔𝒔𝒔𝒔 � − 𝜽𝜽� = 𝒄𝒄𝒄𝒄𝒄𝒄 𝜽𝜽 ,
𝟐𝟐
𝝅𝝅
𝒄𝒄𝒄𝒄𝒄𝒄 � − 𝜽𝜽� = 𝒔𝒔𝒔𝒔𝒔𝒔 𝜽𝜽,
𝟐𝟐
𝝅𝝅
𝒄𝒄𝒄𝒄𝒄𝒄 � − 𝜽𝜽� = 𝒔𝒔𝒔𝒔𝒔𝒔 𝜽𝜽,
𝟐𝟐
𝝅𝝅
𝒄𝒄𝒄𝒄𝒄𝒄 � − 𝜽𝜽� = 𝒕𝒕𝒕𝒕𝐧𝐧 𝜽𝜽
𝟐𝟐
Proof: By definition, referring to the figure above
𝝅𝝅
𝒙𝒙
𝝅𝝅
𝒚𝒚
𝐬𝐬𝐬𝐬𝐬𝐬 � − 𝜽𝜽� = = 𝐜𝐜𝐜𝐜𝐜𝐜 𝜽𝜽
𝟐𝟐
𝒓𝒓
𝐜𝐜𝐜𝐜𝐜𝐜 � − 𝜽𝜽� = = 𝐬𝐬𝐬𝐬𝐬𝐬 𝜽𝜽
𝟐𝟐
𝒓𝒓
Similarly all the remaining cofunction identities follow
6
b) Even-Odd Identities
𝒔𝒔𝒔𝒔𝒔𝒔(−𝜽𝜽) = − 𝒔𝒔𝒔𝒔𝒔𝒔 𝜽𝜽 ,
𝒄𝒄𝒄𝒄𝒄𝒄(−𝜽𝜽) = − 𝒄𝒄𝒄𝒄𝒄𝒄 𝜽𝜽,
𝒄𝒄𝒄𝒄𝒄𝒄(−𝜽𝜽) = 𝒄𝒄𝒄𝒄𝒄𝒄 𝜽𝜽,
𝒕𝒕𝒕𝒕𝒕𝒕(−𝜽𝜽) = − 𝒕𝒕𝒕𝒕𝒕𝒕 𝜽𝜽,
𝒔𝒔𝒔𝒔𝒔𝒔(−𝜽𝜽) = 𝒔𝒔𝒔𝒔𝒔𝒔 𝜽𝜽 ,
𝒄𝒄𝒄𝒄𝒄𝒄(−𝜽𝜽) = − 𝒄𝒄𝒄𝒄𝒄𝒄 𝜽𝜽
st
Proof: Let 𝜽𝜽 be an angle of the 1
Quadrant,
th
then −𝜽𝜽 is angle of 4 Quadrant, see
figure.
𝒓𝒓 = 𝑶𝑶𝑶𝑶 = 𝑶𝑶𝑶𝑶′ . Now,
𝒔𝒔𝒔𝒔𝒔𝒔(−𝜽𝜽) =
−𝒚𝒚
𝒓𝒓
𝒚𝒚
= − = − 𝒔𝒔𝒔𝒔𝒔𝒔 𝜽𝜽
𝒓𝒓
The rest of the Even – Odd Identities
st
for an angle of the 1 Quadrant can be
justified the same way.
4)
Sum and difference formulas (page 71-73)
a) 𝒔𝒔𝒔𝒔𝒔𝒔 (𝜶𝜶 + 𝜷𝜷) = 𝒔𝒔𝒔𝒔𝒔𝒔 𝜶𝜶 𝒄𝒄𝒄𝒄𝒄𝒄 𝜷𝜷 + 𝒄𝒄𝒄𝒄𝒄𝒄𝒄𝒄 𝒔𝒔𝒔𝒔𝒔𝒔 𝜷𝜷
b) 𝒔𝒔𝒔𝒔𝒔𝒔 (𝜶𝜶 − 𝜷𝜷) = 𝒔𝒔𝒔𝒔𝒔𝒔 𝜶𝜶 𝒄𝒄𝒄𝒄𝒄𝒄 𝜷𝜷 − 𝒄𝒄𝒄𝒄𝒄𝒄𝒄𝒄 𝒔𝒔𝒔𝒔𝒔𝒔 𝜷𝜷
c) 𝒄𝒄𝒄𝒄𝒄𝒄 ( 𝜶𝜶 + 𝜷𝜷) = 𝒄𝒄𝒄𝒄𝒄𝒄 𝜶𝜶 𝒄𝒄𝒄𝒄𝒄𝒄 𝜷𝜷 − 𝒔𝒔𝒔𝒔𝒔𝒔 𝜶𝜶 𝒔𝒔𝒔𝒔𝒔𝒔 𝜷𝜷
d) 𝒄𝒄𝒄𝒄𝒄𝒄 (𝜶𝜶 − 𝜷𝜷) = 𝒄𝒄𝒄𝒄𝒄𝒄 𝜶𝜶 𝒄𝒄𝒄𝒄𝒄𝒄 𝜷𝜷 + 𝒔𝒔𝒔𝒔𝒔𝒔 𝜶𝜶 𝒔𝒔𝒔𝒔𝒔𝒔 𝜷𝜷
e) 𝒕𝒕𝒕𝒕𝒕𝒕 (𝜶𝜶 + 𝜷𝜷) =
f) 𝒕𝒕𝒕𝒕𝒕𝒕 (𝜶𝜶 − 𝜷𝜷) =
tan 𝛼𝛼 +tan 𝛽𝛽
1−tan 𝛼𝛼 tan 𝛽𝛽
tan 𝛼𝛼 −tan 𝛽𝛽
1+tan 𝛼𝛼 tan 𝛽𝛽
Example 1: Find the exact values of:
a) 𝒔𝒔𝒔𝒔𝒔𝒔 𝟏𝟏𝟏𝟏𝟎𝟎
Example 2: Simplify the following expression. 𝒔𝒔𝒔𝒔𝒔𝒔 (
𝝅𝝅
𝟐𝟐
b) 𝒄𝒄𝒄𝒄𝒄𝒄 𝟕𝟕𝟕𝟕𝟎𝟎
𝝅𝝅
𝝅𝝅
− 𝒙𝒙) − 𝒕𝒕𝒕𝒕𝒕𝒕( − 𝒙𝒙) 𝒔𝒔𝒔𝒔𝒔𝒔( − 𝒙𝒙)
Example: Book 2: Example 3.9, 3.10, 3.11, 3.12 reading (74 – 75)
𝟐𝟐
𝟐𝟐
(Book 2) Homework Exercises 3.2 page 76 –77: # 1 – 16 odd numbers
7
5)
Double-angle formulas (page 78)
a) 𝒔𝒔𝒔𝒔𝒔𝒔 (𝟐𝟐𝟐𝟐) = 𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐(𝜽𝜽)𝒄𝒄𝒄𝒄𝒄𝒄(𝜽𝜽)
b) 𝒄𝒄𝒄𝒄𝒄𝒄 (𝟐𝟐𝟐𝟐) = 𝒄𝒄𝒄𝒄𝒄𝒄𝟐𝟐 𝜽𝜽 − 𝒔𝒔𝒔𝒔𝒔𝒔𝟐𝟐 𝜽𝜽
c) 𝒄𝒄𝒄𝒄𝒄𝒄 (𝟐𝟐𝟐𝟐) = 𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐 𝜽𝜽 − 𝟏𝟏
d) 𝒄𝒄𝒄𝒄𝒄𝒄 (𝟐𝟐𝟐𝟐) = 𝟏𝟏 − 𝟐𝟐𝒔𝒔𝒔𝒔𝒔𝒔𝟐𝟐 𝜽𝜽
e) 𝒕𝒕𝒕𝒕𝒕𝒕 (𝟐𝟐𝟐𝟐) =
𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐
𝟏𝟏−𝒕𝒕𝒕𝒕𝒕𝒕𝟐𝟐 𝜽𝜽
Proof: Proof follows from Sum Difference Formulas
Example: Book 2: Example 3.13, 3.14
Example 2: Given an angle for which 𝒔𝒔𝒔𝒔𝒔𝒔(𝜶𝜶) = −𝟑𝟑/𝟓𝟓 in Quadrant III, determine the values for
𝒔𝒔𝒔𝒔𝒔𝒔(𝟐𝟐𝜶𝜶), 𝒄𝒄𝒄𝒄𝒄𝒄 (𝟐𝟐𝜶𝜶), 𝒕𝒕𝒕𝒕𝒕𝒕(𝟐𝟐𝜶𝜶), 𝒔𝒔𝒔𝒔𝒔𝒔(𝜶𝜶/𝟐𝟐), 𝒄𝒄𝒄𝒄𝒄𝒄(𝜶𝜶/𝟐𝟐), and 𝒕𝒕𝒕𝒕𝒕𝒕(𝜶𝜶/𝟐𝟐).
6)
Half-angle formulas (page 79 – 80)
𝟏𝟏 + 𝒄𝒄𝒄𝒄𝒄𝒄 𝜽𝜽
a) 𝒄𝒄𝒄𝒄𝒄𝒄 (𝜽𝜽/𝟐𝟐) = ±�
𝟐𝟐
𝟏𝟏 − 𝒄𝒄𝒄𝒄𝒄𝒄 𝜽𝜽
b) 𝒔𝒔𝒔𝒔𝒔𝒔 (𝜽𝜽/𝟐𝟐) = ±�
𝜃𝜃
𝟐𝟐
1−cos 𝜃𝜃
c) tan �2� = ±�1+cos 𝜃𝜃
Example: Book 2: Example 3.15 page 81
P roof: Proof follows from Double-angle Formulas
Example 3: Find the Exact value of
7)
Products as sums
8)
Sums as products
a)
b)
c)
d)
a)
b)
c)
d)
𝒔𝒔𝒔𝒔𝒔𝒔 𝛂𝛂 𝒄𝒄𝒄𝒄𝒄𝒄 𝜷𝜷 =
𝒄𝒄𝒄𝒄𝒄𝒄 𝛂𝛂 𝒔𝒔𝒔𝒔𝒔𝒔 𝜷𝜷 =
𝒄𝒄𝒄𝒄𝒄𝒄 𝛂𝛂 𝒄𝒄𝒄𝒄𝒄𝒄 𝜷𝜷 =
𝒔𝒔𝒔𝒔𝒔𝒔 𝛂𝛂 𝒔𝒔𝒔𝒔𝒔𝒔 𝜷𝜷 =
𝒔𝒔𝒔𝒔𝒔𝒔 𝑨𝑨
𝒔𝒔𝒔𝒔𝒔𝒔 𝑨𝑨
𝒄𝒄𝒄𝒄𝒄𝒄 𝑨𝑨
𝒄𝒄𝒄𝒄𝒄𝒄 𝑨𝑨
+
−
+
−
𝝅𝝅
𝐬𝐬𝐬𝐬𝐬𝐬 � � .
𝒔𝒔𝒔𝒔𝒔𝒔 𝑩𝑩
𝒔𝒔𝒔𝒔𝒔𝒔 𝑩𝑩
𝒄𝒄𝒄𝒄𝒄𝒄 𝑩𝑩
𝒄𝒄𝒄𝒄𝒄𝒄 𝑩𝑩
𝟖𝟖
½[𝒔𝒔𝒔𝒔𝒔𝒔 (𝛂𝛂 + 𝜷𝜷) + 𝒔𝒔𝒔𝒔𝒔𝒔 (𝜶𝜶 − 𝜷𝜷)]
½[𝒔𝒔𝒔𝒔𝒔𝒔 (𝛂𝛂 + 𝜷𝜷) − 𝒔𝒔𝒔𝒔𝒔𝒔 (𝛂𝛂 − 𝜷𝜷)]
½[𝒄𝒄𝒄𝒄𝒄𝒄 (𝛂𝛂 + 𝜷𝜷) + 𝒄𝒄𝒄𝒄𝒄𝒄 (𝜶𝜶 − 𝜷𝜷)]
−½[𝒄𝒄𝒄𝒄𝒄𝒄 (𝛂𝛂 + 𝜷𝜷) − 𝒄𝒄𝒄𝒄𝒄𝒄 (𝛂𝛂 − 𝜷𝜷)]
=
=
=
=
𝟐𝟐 𝒔𝒔𝒔𝒔𝒔𝒔 ½ (𝑨𝑨 + 𝑩𝑩) 𝒄𝒄𝒄𝒄𝒄𝒄 ½ (𝑨𝑨 − 𝑩𝑩)
𝟐𝟐 𝒔𝒔𝒔𝒔𝒔𝒔 ½ (𝑨𝑨 − 𝑩𝑩) 𝒄𝒄𝒄𝒄𝒄𝒄 ½ (𝑨𝑨 + 𝑩𝑩)
𝟐𝟐 𝒄𝒄𝒄𝒄𝒄𝒄 ½ (𝑨𝑨 + 𝑩𝑩) 𝒄𝒄𝒄𝒄𝒄𝒄 ½ (𝑨𝑨 − 𝑩𝑩)
−𝟐𝟐 𝒔𝒔𝒔𝒔𝒔𝒔 ½ (𝑨𝑨 + 𝑩𝑩) 𝒔𝒔𝒔𝒔𝒔𝒔 ½ (𝑨𝑨 − 𝑩𝑩)
(Book 2) Homework Exercises 3.3 page 81: # 1 – 18 odd numbers
8
Example 4: Show
that 𝒔𝒔𝒔𝒔𝒔𝒔𝟐𝟐 𝒙𝒙 + 𝒄𝒄𝒄𝒄𝒄𝒄𝟐𝟐 𝒙𝒙 = 𝒔𝒔𝒔𝒔𝒔𝒔𝟐𝟐 𝒙𝒙 𝒄𝒄𝒄𝒄𝒄𝒄𝟐𝟐 𝒙𝒙
Example 5: Prove that
𝐭𝐭𝐭𝐭𝐭𝐭 𝒚𝒚
a)
𝐬𝐬𝐬𝐬𝐬𝐬 𝒚𝒚
= 𝐬𝐬𝐬𝐬𝐬𝐬 𝒚𝒚
𝐬𝐬𝐬𝐬𝐬𝐬 𝒚𝒚 + 𝐬𝐬𝐬𝐬𝐬𝐬 𝒚𝒚 𝐜𝐜𝐜𝐜𝐜𝐜 𝟐𝟐 𝒚𝒚 = 𝐜𝐜𝐜𝐜𝐜𝐜 𝒚𝒚
b)
𝐭𝐭𝐭𝐭𝐭𝐭 𝒙𝒙 + 𝐜𝐜𝐜𝐜𝐜𝐜 𝒙𝒙 = 𝐬𝐬𝐬𝐬𝐬𝐬 𝒙𝒙 𝐜𝐜𝐜𝐜𝐜𝐜 𝒙𝒙
c)
𝟏𝟏+𝐜𝐜𝐜𝐜𝐜𝐜 𝒙𝒙
d)
𝐬𝐬𝐬𝐬𝐬𝐬 𝒙𝒙
=
𝐬𝐬𝐬𝐬𝐬𝐬 𝒙𝒙
𝟏𝟏−𝐜𝐜𝐜𝐜𝐜𝐜 𝒙𝒙
𝐭𝐭𝐭𝐭𝐭𝐭(𝝅𝝅 − 𝒙𝒙) = −𝒕𝒕𝒕𝒕𝒕𝒕 (𝒙𝒙)
e)
𝟑𝟑
𝐭𝐭𝐭𝐭𝐭𝐭 � 𝝅𝝅 + 𝒙𝒙� = −𝒄𝒄𝒄𝒄𝒄𝒄 (𝒙𝒙)
f)
𝟐𝟐
Example 6: Use the basic trigonometric identities to determine the other five values of the
trigonometric functions given that:
a)
𝐬𝐬𝐬𝐬𝐬𝐬 𝜶𝜶 = 𝟕𝟕/𝟖𝟖 and 𝐜𝐜𝐜𝐜𝐜𝐜 𝜶𝜶 > 𝟎𝟎.
b) 𝒙𝒙 is an angle in quadrant III and 𝒔𝒔𝒔𝒔𝒔𝒔 𝒙𝒙 = −𝟏𝟏 / 𝟑𝟑.
c) 𝒙𝒙 is an angle in quadrant IV and tan x = -5.
d) 𝒙𝒙 is in quadrant II and 𝒔𝒔𝒔𝒔𝒔𝒔 𝒙𝒙 = 𝟏𝟏/𝟓𝟓.
e) 𝒙𝒙 is in quadrant I and 𝒄𝒄𝒄𝒄𝒄𝒄 𝒙𝒙 = 𝟏𝟏/𝟓𝟓.
Example 7: Write 𝑨𝑨 𝐬𝐬𝐬𝐬𝐬𝐬 𝒃𝒃𝒃𝒃 + 𝑩𝑩 𝐜𝐜𝐜𝐜𝐜𝐜 𝒃𝒃𝒃𝒃 = 𝒂𝒂 𝐬𝐬𝐬𝐬𝐬𝐬(𝒃𝒃𝒃𝒃 + 𝒄𝒄)
Solution:
Solved Examples:
Simplify the following trigonometric expression. 𝒄𝒄𝒄𝒄𝒄𝒄 (𝒙𝒙) 𝒔𝒔𝒔𝒔𝒔𝒔 (𝝅𝝅/𝟐𝟐 − 𝒙𝒙)
1)
Solution:
•
2)
Use the identity 𝒔𝒔𝒔𝒔𝒔𝒔 (𝝅𝝅/𝟐𝟐 − 𝒙𝒙) = 𝒄𝒄𝒄𝒄𝒄𝒄(𝒙𝒙) and simplify
𝒄𝒄𝒄𝒄𝒄𝒄 (𝒙𝒙) 𝒔𝒔𝒔𝒔𝒔𝒔 (𝝅𝝅 /𝟐𝟐 − 𝒙𝒙) = 𝒄𝒄𝒄𝒄𝒄𝒄 (𝒙𝒙) 𝒄𝒄𝒄𝒄𝒄𝒄 (𝒙𝒙) = 𝒄𝒄𝒄𝒄𝒄𝒄 (𝒙𝒙)
Simplify the following trigonometric expression. [𝑠𝑠𝑠𝑠𝑠𝑠 4 𝑥𝑥 − 𝑐𝑐𝑐𝑐𝑐𝑐 4 𝑥𝑥] / [𝑠𝑠𝑠𝑠𝑠𝑠 2 𝑥𝑥 − 𝑐𝑐𝑐𝑐𝑐𝑐 2 𝑥𝑥]
Solution:
• Factor the denominator [𝑠𝑠𝑠𝑠𝑠𝑠 4 𝑥𝑥 − 𝑐𝑐𝑐𝑐𝑐𝑐 4 𝑥𝑥] / [𝑠𝑠𝑠𝑠𝑠𝑠 2 𝑥𝑥 − 𝑐𝑐𝑐𝑐𝑐𝑐 2 𝑥𝑥]
= [𝑠𝑠𝑠𝑠𝑠𝑠2 𝑥𝑥 − 𝑐𝑐𝑐𝑐𝑐𝑐 2 𝑥𝑥][𝑠𝑠𝑠𝑠𝑠𝑠2 𝑥𝑥 + 𝑐𝑐𝑐𝑐𝑐𝑐 2 𝑥𝑥] / [𝑠𝑠𝑠𝑠𝑠𝑠 2 𝑥𝑥 − 𝑐𝑐𝑐𝑐𝑐𝑐 2 𝑥𝑥]
= [𝑠𝑠𝑠𝑠𝑠𝑠2 𝑥𝑥 + 𝑐𝑐𝑐𝑐𝑐𝑐 2 𝑥𝑥] = 1
9
Simplify the following trigonometric expression. [𝑠𝑠𝑠𝑠𝑠𝑠(𝑥𝑥) 𝑠𝑠𝑠𝑠𝑠𝑠2 𝑥𝑥] / [1 + 𝑠𝑠𝑠𝑠𝑠𝑠(𝑥𝑥)]
3)
Solution:
•
•
4)
Substitute sec (x) that is in the numerator by 𝟏𝟏 / 𝒄𝒄𝒄𝒄𝒄𝒄 (𝒙𝒙) and simplify.
[𝑠𝑠𝑠𝑠𝑠𝑠(𝑥𝑥) 𝑠𝑠𝑠𝑠𝑠𝑠2 𝑥𝑥] / [1 + 𝑠𝑠𝑠𝑠𝑠𝑠(𝑥𝑥)]
= 𝑠𝑠𝑠𝑠𝑠𝑠2 𝑥𝑥 / [ 𝑐𝑐𝑐𝑐𝑐𝑐 𝑥𝑥 (1 + 𝑠𝑠𝑠𝑠𝑠𝑠 (𝑥𝑥) ]
= 𝑠𝑠𝑠𝑠𝑠𝑠2 𝑥𝑥 / [ 𝑐𝑐𝑐𝑐𝑐𝑐 𝑥𝑥 + 1 ]
Substitute 𝒔𝒔𝒔𝒔𝒔𝒔𝟐𝟐 𝒙𝒙 𝒃𝒃𝒃𝒃 𝟏𝟏 − 𝒄𝒄𝒄𝒄𝒄𝒄𝟐𝟐 𝒙𝒙 , factor and simplify.
= [ 1 − 𝑐𝑐𝑐𝑐𝑐𝑐 2 𝑥𝑥 ] / [ 𝑐𝑐𝑐𝑐𝑐𝑐 𝑥𝑥 + 1 ]
= [ (1 − 𝑐𝑐𝑐𝑐𝑐𝑐 𝑥𝑥)(1 + 𝑐𝑐𝑐𝑐𝑐𝑐 𝑥𝑥) ] / [ 𝑐𝑐𝑐𝑐𝑐𝑐 𝑥𝑥 + 1 ] = 1 − 𝑐𝑐𝑐𝑐𝑐𝑐 𝑥𝑥
Simplify the following trigonometric expression. 𝒔𝒔𝒔𝒔𝒔𝒔 (−𝒙𝒙) 𝒄𝒄𝒄𝒄𝒄𝒄 (𝝅𝝅/𝟐𝟐 − 𝒙𝒙)
Solution:
•
Use the identities 𝒔𝒔𝒔𝒔𝒔𝒔 (−𝒙𝒙) = − 𝒔𝒔𝒔𝒔𝒔𝒔 (𝒙𝒙) and 𝒄𝒄𝒄𝒄𝒄𝒄 (𝝅𝝅 / 𝟐𝟐 − 𝒙𝒙) = 𝒔𝒔𝒔𝒔𝒔𝒔 (𝒙𝒙) and simplify
𝒔𝒔𝒔𝒔𝒔𝒔 (−𝒙𝒙) 𝒄𝒄𝒄𝒄𝒄𝒄 (𝝅𝝅/ 𝟐𝟐 − 𝒙𝒙) = − 𝒔𝒔𝒔𝒔𝒔𝒔 (𝒙𝒙) 𝒔𝒔𝒔𝒔𝒔𝒔 (𝒙𝒙) = − 𝒔𝒔𝒔𝒔𝒔𝒔𝟐𝟐 𝒙𝒙
Simplify the following trigonometric expression. 𝒔𝒔𝒔𝒔𝒔𝒔𝟐𝟐 𝒙𝒙 − 𝒄𝒄𝒄𝒄𝒄𝒄𝟐𝟐 𝒙𝒙 𝒔𝒔𝒔𝒔𝒔𝒔𝟐𝟐 𝒙𝒙
5)
Solution:
• Factor 𝒔𝒔𝒔𝒔𝒔𝒔𝟐𝟐 𝒙𝒙 out, group and simplify 𝒔𝒔𝒔𝒔𝒔𝒔𝟐𝟐 𝒙𝒙 − 𝒄𝒄𝒄𝒄𝒄𝒄𝟐𝟐 𝒙𝒙 𝒔𝒔𝒔𝒔𝒔𝒔𝟐𝟐 𝒙𝒙
= 𝑠𝑠𝑠𝑠𝑠𝑠2 𝑥𝑥 ( 1 − 𝑐𝑐𝑐𝑐𝑐𝑐 2 𝑥𝑥 )
= 𝑠𝑠𝑠𝑠𝑠𝑠4 𝑥𝑥
Simplify the following trigonometric expression. 𝒕𝒕𝒕𝒕𝒕𝒕𝟒𝟒 𝒙𝒙 + 𝟐𝟐 𝒕𝒕𝒕𝒕𝒕𝒕 𝟐𝟐 𝒙𝒙 + 𝟏𝟏
6)
Solution:
•
•
7)
Note that the given trigonometric expression can be written as a square
𝒕𝒕𝒕𝒕𝒕𝒕𝟒𝟒 𝒙𝒙 + 𝟐𝟐 𝒕𝒕𝒕𝒕𝒕𝒕 𝟐𝟐 𝒙𝒙 + 𝟏𝟏 = ( 𝒕𝒕𝒕𝒕𝒕𝒕𝟐𝟐 𝒙𝒙 + 𝟏𝟏) 𝟐𝟐
We now use the identity 1 + tan 2x = sec 2x
𝒕𝒕𝒕𝒕𝒕𝒕𝟒𝟒 𝒙𝒙 + 𝟐𝟐 𝒕𝒕𝒕𝒕𝒕𝒕 𝟐𝟐 𝒙𝒙 + 𝟏𝟏 = ( 𝒕𝒕𝒕𝒕𝒕𝒕𝟐𝟐 𝒙𝒙 + 𝟏𝟏) 𝟐𝟐 = ( sec 2x ) 2 = sec 4x
Add and simplify.
Solution:
•
𝟏𝟏 / [𝟏𝟏 + 𝒄𝒄𝒄𝒄𝒄𝒄 𝒙𝒙] + 𝟏𝟏 / [𝟏𝟏 − 𝒄𝒄𝒄𝒄𝒄𝒄 𝒙𝒙]
In order to add the fractional trigonometric expressions, we need to have a common denominator
𝟏𝟏 / [𝟏𝟏 + 𝒄𝒄𝒄𝒄𝒄𝒄 𝒙𝒙] + 𝟏𝟏 / [𝟏𝟏 − 𝒄𝒄𝒄𝒄𝒄𝒄 𝒙𝒙]
= [ 1 − 𝑐𝑐𝑐𝑐𝑐𝑐 𝑥𝑥 + 1 + 𝑐𝑐𝑐𝑐𝑐𝑐 𝑥𝑥 ] / [ [1 + 𝑐𝑐𝑐𝑐𝑐𝑐 𝑥𝑥] [1 − 𝑐𝑐𝑐𝑐𝑐𝑐 𝑥𝑥] ]
= 2 / [1 − 𝑐𝑐𝑐𝑐𝑐𝑐 2 𝑥𝑥]
= 2 / 𝑠𝑠𝑠𝑠𝑠𝑠2 𝑥𝑥 = 2 𝑐𝑐𝑐𝑐𝑐𝑐 2 𝑥𝑥
10
Write �( 4 − 4 𝑠𝑠𝑠𝑠𝑠𝑠2 𝑥𝑥 ) without square root for (𝝅𝝅/ 𝟐𝟐) < 𝒙𝒙 < 𝝅𝝅.
8)
Solution:
• Factor, and substitute 𝟏𝟏 − 𝒔𝒔𝒔𝒔𝒔𝒔𝟐𝟐 𝒙𝒙 𝒃𝒃𝒃𝒃 𝒄𝒄𝒄𝒄𝒄𝒄 𝟐𝟐 𝒙𝒙
•
9)
�( 4 − 4 𝑠𝑠𝑠𝑠𝑠𝑠2 𝑥𝑥 ) = �4(1 − 𝑠𝑠𝑠𝑠𝑠𝑠2 𝑥𝑥 ) = 2 √ 𝑐𝑐𝑐𝑐𝑐𝑐 2 𝑥𝑥 = 2 | 𝑐𝑐𝑐𝑐𝑐𝑐 (𝑥𝑥) |
Since (𝝅𝝅/ 𝟐𝟐) < 𝒙𝒙 < 𝝅𝝅, 𝒄𝒄𝒄𝒄𝒄𝒄 𝒙𝒙 is less than zero and the given trigonometric expression
simplifies to = − 𝟐𝟐 𝒄𝒄𝒄𝒄𝒄𝒄 (𝒙𝒙)
Simplify the following expression.
Solution:
•
[𝟏𝟏 − 𝒔𝒔𝒔𝒔𝒔𝒔𝟒𝟒 𝒙𝒙] / [𝟏𝟏 + 𝒔𝒔𝒔𝒔𝒔𝒔𝟐𝟐 𝒙𝒙]
Factor the denominator, and simplify
[𝟏𝟏 − 𝒔𝒔𝒔𝒔𝒔𝒔𝟒𝟒 𝒙𝒙] / [𝟏𝟏 + 𝒔𝒔𝒔𝒔𝒔𝒔𝟐𝟐 𝒙𝒙] = [1 − 𝑠𝑠𝑠𝑠𝑠𝑠2 𝑥𝑥] [1 + 𝑠𝑠𝑠𝑠𝑠𝑠2 𝑥𝑥] / [1 + 𝑠𝑠𝑠𝑠𝑠𝑠2 𝑥𝑥]
= [𝟏𝟏 − 𝒔𝒔𝒔𝒔𝒔𝒔𝟐𝟐 𝒙𝒙] = 𝒄𝒄𝒄𝒄𝒄𝒄𝟐𝟐 𝒙𝒙
𝟏𝟏 / [𝟏𝟏 + 𝒔𝒔𝒔𝒔𝒔𝒔 𝒙𝒙] + 𝟏𝟏 / [𝟏𝟏 − 𝒔𝒔𝒔𝒔𝒔𝒔 𝒙𝒙]
10) Add and simplify.
Solution :
•
Use a common denominator to add
𝟏𝟏
[𝟏𝟏 + 𝒔𝒔𝒔𝒔𝒔𝒔 𝒙𝒙]
+ [𝟏𝟏
𝟏𝟏
− 𝒔𝒔𝒔𝒔𝒔𝒔 𝒙𝒙]
11) Add and simplify.
[𝟏𝟏 − 𝒔𝒔𝒔𝒔𝒔𝒔 𝒙𝒙 + 𝟏𝟏 + 𝒔𝒔𝒔𝒔𝒔𝒔 𝒙𝒙]
= [ (𝟏𝟏
+ 𝒔𝒔𝒔𝒔𝒔𝒔 𝒙𝒙)(𝟏𝟏 − 𝒔𝒔𝒔𝒔𝒔𝒔 𝒙𝒙)]
= 𝟐𝟐 / [ 𝟏𝟏 − 𝒔𝒔𝒔𝒔𝒔𝒔𝟐𝟐 𝒙𝒙 ]
= 𝟐𝟐 / 𝒄𝒄𝒄𝒄𝒄𝒄𝟐𝟐 𝒙𝒙 = 𝟐𝟐 𝒔𝒔𝒔𝒔𝒔𝒔𝟐𝟐 𝒙𝒙
𝒄𝒄𝒄𝒄𝒄𝒄 𝒙𝒙 − 𝒄𝒄𝒄𝒄𝒄𝒄 𝒙𝒙 𝒔𝒔𝒔𝒔𝒔𝒔𝟐𝟐 𝒙𝒙
Solution:
•
factor 𝒄𝒄𝒄𝒄𝒄𝒄 𝒙𝒙 out ;
𝐜𝐜𝐜𝐜𝐜𝐜 𝒙𝒙 – 𝐜𝐜𝐜𝐜𝐜𝐜 𝒙𝒙 𝒔𝒔𝒔𝒔𝒔𝒔𝟐𝟐 𝒙𝒙 = 𝐜𝐜𝐜𝐜𝐜𝐜 𝒙𝒙 �𝟏𝟏 – 𝒔𝒔𝒔𝒔𝒔𝒔𝟐𝟐 𝒙𝒙�
= 𝒄𝒄𝒄𝒄𝒄𝒄 𝒙𝒙 𝒄𝒄𝒄𝒄𝒄𝒄𝟐𝟐 𝒙𝒙
= 𝒄𝒄𝒄𝒄𝒄𝒄𝟑𝟑 𝒙𝒙
11
𝒕𝒕𝒕𝒕𝒕𝒕𝟐𝟐 𝒙𝒙 𝒄𝒄𝒄𝒄𝒄𝒄𝟐𝟐 𝒙𝒙 + 𝒄𝒄𝒄𝒄𝒄𝒄𝟐𝟐 𝒙𝒙 𝒔𝒔𝒔𝒔𝒔𝒔𝟐𝟐 𝒙𝒙
12) Simplify the following expression.
Solution:
•
Use the trigonometric identities 𝒕𝒕𝒕𝒕𝒕𝒕 𝒙𝒙 = 𝒔𝒔𝒔𝒔𝒔𝒔 𝒙𝒙 / 𝒄𝒄𝒄𝒄𝒄𝒄 𝒙𝒙 and 𝑐𝑐𝑐𝑐𝑐𝑐 𝑥𝑥 = 𝑐𝑐𝑐𝑐𝑐𝑐 𝑥𝑥 / 𝑠𝑠𝑠𝑠𝑠𝑠 𝑥𝑥 to write the
given expression as
•
𝒕𝒕𝒕𝒕𝒕𝒕𝟐𝟐 𝒙𝒙 𝒄𝒄𝒄𝒄𝒄𝒄𝟐𝟐 𝒙𝒙 + 𝒄𝒄𝒄𝒄𝒄𝒄𝟐𝟐 𝒙𝒙 𝒔𝒔𝒔𝒔𝒔𝒔𝟐𝟐 𝒙𝒙 = (𝒔𝒔𝒔𝒔𝒔𝒔 𝒙𝒙 / 𝒄𝒄𝒄𝒄𝒄𝒄 𝒙𝒙)𝟐𝟐 𝒄𝒄𝒄𝒄𝒄𝒄𝟐𝟐 𝒙𝒙 + (𝒄𝒄𝒄𝒄𝒄𝒄 𝒙𝒙 / 𝒔𝒔𝒔𝒔𝒔𝒔 𝒙𝒙)𝟐𝟐 𝒔𝒔𝒔𝒔𝒔𝒔𝟐𝟐 𝒙𝒙
= 𝒔𝒔𝒔𝒔𝒔𝒔𝟐𝟐 𝒙𝒙 + 𝒄𝒄𝒄𝒄𝒄𝒄𝟐𝟐 𝒙𝒙 = 𝟏𝟏
and simplify to get:
𝝅𝝅
𝝅𝝅
𝝅𝝅
13) Simplify the following expression. 𝒔𝒔𝒔𝒔𝒔𝒔 ( − 𝒙𝒙) − 𝒕𝒕𝒕𝒕𝒕𝒕( − 𝒙𝒙) 𝒔𝒔𝒔𝒔𝒔𝒔( − 𝒙𝒙)
𝟐𝟐
𝟐𝟐
𝟐𝟐
Solution:
•
𝝅𝝅
𝝅𝝅
𝝅𝝅
Use the identities 𝒔𝒔𝒔𝒔𝒔𝒔 ( 𝟐𝟐 − 𝒙𝒙) = 𝒄𝒄𝒄𝒄𝒄𝒄 𝒙𝒙, 𝒕𝒕𝒕𝒕𝒕𝒕( 𝟐𝟐 − 𝒙𝒙) = 𝒄𝒄𝒄𝒄𝒄𝒄 𝒙𝒙 and 𝒔𝒔𝒔𝒔𝒔𝒔( 𝟐𝟐 − 𝒙𝒙) = 𝒄𝒄𝒄𝒄𝒄𝒄 𝒙𝒙 to write
the given expression as
𝝅𝝅
𝝅𝝅
𝝅𝝅
𝐬𝐬𝐬𝐬𝐬𝐬 � − 𝒙𝒙� − 𝐭𝐭𝐭𝐭𝐭𝐭 � − 𝒙𝒙� 𝐬𝐬𝐬𝐬𝐬𝐬 � − 𝒙𝒙� = 𝐜𝐜𝐜𝐜𝐜𝐜 𝒙𝒙 − 𝐜𝐜𝐜𝐜𝐜𝐜 𝒙𝒙 𝐜𝐜𝐜𝐜𝐜𝐜 𝒙𝒙
𝟐𝟐
𝟐𝟐
𝟐𝟐
= 𝐜𝐜𝐜𝐜𝐜𝐜 𝒙𝒙 −
𝐜𝐜𝐜𝐜𝐜𝐜 𝒙𝒙
𝐬𝐬𝐬𝐬𝐬𝐬 𝒙𝒙
𝒄𝒄𝒄𝒄𝒄𝒄 𝒙𝒙
= 𝒄𝒄𝒄𝒄𝒄𝒄 𝒙𝒙 − 𝒄𝒄𝒄𝒄𝒄𝒄𝟐𝟐 𝒙𝒙/𝒔𝒔𝒔𝒔𝒔𝒔 𝒙𝒙
= 𝟏𝟏/𝒔𝒔𝒔𝒔𝒔𝒔 𝒙𝒙 − 𝒄𝒄𝒄𝒄𝒄𝒄𝟐𝟐 𝒙𝒙/𝒔𝒔𝒔𝒔𝒔𝒔 𝒙𝒙
= (𝟏𝟏 − 𝒄𝒄𝒄𝒄𝒄𝒄𝟐𝟐 𝒙𝒙) /𝒔𝒔𝒔𝒔𝒔𝒔 𝒙𝒙
= 𝒔𝒔𝒔𝒔𝒔𝒔𝟐𝟐 𝒙𝒙/𝒔𝒔𝒔𝒔𝒔𝒔 𝒙𝒙
14) Show that 𝒄𝒄𝒄𝒄𝒄𝒄𝟒𝟒 𝒙𝒙 − 𝒔𝒔𝒔𝒔𝒔𝒔𝟒𝟒 𝒙𝒙 = 𝐜𝐜𝐜𝐜𝐜𝐜(𝟐𝟐𝟐𝟐)
= 𝒔𝒔𝒔𝒔𝒔𝒔 𝒙𝒙
Practice Problems
12