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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: Eβ 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 ο§ ο§ ο§ 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) 2π₯π₯ β 5 = 7π₯π₯, b) 3π₯π₯ 2 β 4π₯π₯ β 6 = 0 2) Contradictions: These are equations that do not have solutions Examples: 2π₯π₯ β 1 = 2(π₯π₯ β 1) + 6 3) Identities: These types of equations hold true for any value of the variable Examples: (π₯π₯ + 5)(π₯π₯ β 5) = π₯π₯² β 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) ππππππ π½π½ = ππππππ π½π½ = ππππππ π½π½ = ππ πππππππ½π½ ππ ππππππππ ππ ππππππππ , ππππππ ππππππ π½π½ = ππ ππππππππ ππ , and ππππππ π½π½ = ππππππππ ππ , and ππππππ π½π½ = ππππππππ Proof: Follows directly from the definition of trig functions. 2) Quotient Identities ππππππ π½π½ = ππππππππ ππππππππ , and ππππππ π½π½ = ππππππππ ππππππππ Proof: Follows directly from the definition of trig functions. 2 3) Pythagorean Identities ( page: 66 & 67) a) ππππππ²π½π½ + ππππππ²π½π½ = ππ b) ππ + ππππππ²π½π½ = ππππππ²π½π½ c) ππ + ππππππ²π½π½ = ππππππ ²π½π½ Proof: a) Let π·π·(ππ, ππ) be on the terminal side of the angle π½π½. Then ππ = οΏ½ππππ + ππππ which implies that ππππ = ππππ + ππππ , π¬π¬π¬π¬π¬π¬ π½π½ = And so, ππππππ²π½π½ + ππππππ²π½π½ = ππππ ππππ + ππππ ππππ = ππππ + ππππ ππππ = ππππ ππππ = ππ ππ , and ππππππ π½π½ = ππ ππ ππ Example: Book 2: Example 3.1, 3.2, 3.3, 3.4, 3.5, 3.6 and 3.7 reading (67 β 69) Note: a) it follows that: ππππππ²π½π½ = ππ β ππππππ²π½π½ and ππππππ²π½π½ = ππ β ππππππ²π½π½, ο§ From ο§ b) and c) are similarly proved. ο§ ππππππ²π½π½, "sine squared theta", means(ππππππ π½π½)² Example 1: a) Express sin ππ in terms of cos ππ b) Express cos ππ in terms of sin ππ c) Express tan ππ in terms of cos ππ, where ΞΈ in Quadrant II 3 d) If tan ππ = 2 and ΞΈ is in Quadrant III, find sin ππ and cos ππ 1 e) If cos ππ = 2 and ΞΈ is in Quadrant IV, find all other trig values of ΞΈ f) Use the basic trigonometric identities to determine the other five values of the trigonometric functions given that sin πΌπΌ = 7/8 and cos πΌπΌ > 0. g) π₯π₯ is in quadrant II and π π π π π π π₯π₯ = 1/5. Find ππππππ π₯π₯ and π‘π‘π‘π‘π‘π‘ π₯π₯. Example 2: Prove the Pythagorean Identities b) and c) Example 3: Homework Reading page 67 β 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 β 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 π π ππππππ οΏ½ β π½π½οΏ½ = ππππππ π½π½, ππ π π ππππππ οΏ½ β π½π½οΏ½ = ππππππ π½π½, ππ π π ππππππ οΏ½ β π½π½οΏ½ = ππππππ π½π½ , ππ π π ππππππ οΏ½ β π½π½οΏ½ = ππππππ π½π½, ππ π π ππππππ οΏ½ β π½π½οΏ½ = ππππππ π½π½, ππ π π ππππππ οΏ½ β π½π½οΏ½ = πππππ§π§ π½π½ ππ Proof: By definition, referring to the figure above π π ππ π π ππ π¬π¬π¬π¬π¬π¬ οΏ½ β π½π½οΏ½ = = ππππππ π½π½ ππ ππ ππππππ οΏ½ β π½π½οΏ½ = = π¬π¬π¬π¬π¬π¬ π½π½ ππ ππ Similarly all the remaining cofunction identities follow 6 b) Even-Odd Identities ππππππ(βπ½π½) = β ππππππ π½π½ , ππππππ(βπ½π½) = β ππππππ π½π½, ππππππ(βπ½π½) = ππππππ π½π½, ππππππ(βπ½π½) = β ππππππ π½π½, ππππππ(βπ½π½) = ππππππ π½π½ , ππππππ(βπ½π½) = β ππππππ π½π½ st Proof: Let π½π½ be an angle of the 1 Quadrant, th then βπ½π½ is angle of 4 Quadrant, see figure. ππ = πΆπΆπΆπΆ = πΆπΆπΆπΆβ² . Now, ππππππ(βπ½π½) = βππ ππ ππ = β = β ππππππ π½π½ ππ The rest of the Even β 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) ππππππ (πΆπΆ + π·π·) = ππππππ πΆπΆ ππππππ π·π· + ππππππππ ππππππ π·π· b) ππππππ (πΆπΆ β π·π·) = ππππππ πΆπΆ ππππππ π·π· β ππππππππ ππππππ π·π· c) ππππππ ( πΆπΆ + π·π·) = ππππππ πΆπΆ ππππππ π·π· β ππππππ πΆπΆ ππππππ π·π· d) ππππππ (πΆπΆ β π·π·) = ππππππ πΆπΆ ππππππ π·π· + ππππππ πΆπΆ ππππππ π·π· e) ππππππ (πΆπΆ + π·π·) = f) ππππππ (πΆπΆ β π·π·) = tan πΌπΌ +tan π½π½ 1βtan πΌπΌ tan π½π½ tan πΌπΌ βtan π½π½ 1+tan πΌπΌ tan π½π½ Example 1: Find the exact values of: a) ππππππ ππππππ Example 2: Simplify the following expression. ππππππ ( π π ππ b) ππππππ ππππππ π π π π β ππ) β ππππππ( β ππ) ππππππ( β ππ) Example: Book 2: Example 3.9, 3.10, 3.11, 3.12 reading (74 β 75) ππ ππ (Book 2) Homework Exercises 3.2 page 76 β77: # 1 β 16 odd numbers 7 5) Double-angle formulas (page 78) a) ππππππ (ππππ) = ππππππππ(π½π½)ππππππ(π½π½) b) ππππππ (ππππ) = ππππππππ π½π½ β ππππππππ π½π½ c) ππππππ (ππππ) = ππππππππππ π½π½ β ππ d) ππππππ (ππππ) = ππ β ππππππππππ π½π½ e) ππππππ (ππππ) = ππππππππππ ππβππππππππ π½π½ Proof: Proof follows from Sum Difference Formulas Example: Book 2: Example 3.13, 3.14 Example 2: Given an angle for which ππππππ(πΆπΆ) = βππ/ππ in Quadrant III, determine the values for ππππππ(πππΆπΆ), ππππππ (πππΆπΆ), ππππππ(πππΆπΆ), ππππππ(πΆπΆ/ππ), ππππππ(πΆπΆ/ππ), and ππππππ(πΆπΆ/ππ). 6) Half-angle formulas (page 79 β 80) ππ + ππππππ π½π½ a) ππππππ (π½π½/ππ) = ±οΏ½ ππ ππ β ππππππ π½π½ b) ππππππ (π½π½/ππ) = ±οΏ½ ππ ππ 1βcos ππ c) tan οΏ½2οΏ½ = ±οΏ½1+cos ππ 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) ππππππ ππ ππππππ π·π· = ππππππ ππ ππππππ π·π· = ππππππ ππ ππππππ π·π· = ππππππ ππ ππππππ π·π· = ππππππ π¨π¨ ππππππ π¨π¨ ππππππ π¨π¨ ππππππ π¨π¨ + β + β π π π¬π¬π¬π¬π¬π¬ οΏ½ οΏ½ . ππππππ π©π© ππππππ π©π© ππππππ π©π© ππππππ π©π© ππ ½[ππππππ (ππ + π·π·) + ππππππ (πΆπΆ β π·π·)] ½[ππππππ (ππ + π·π·) β ππππππ (ππ β π·π·)] ½[ππππππ (ππ + π·π·) + ππππππ (πΆπΆ β π·π·)] β½[ππππππ (ππ + π·π·) β ππππππ (ππ β π·π·)] = = = = ππ ππππππ ½ (π¨π¨ + π©π©) ππππππ ½ (π¨π¨ β π©π©) ππ ππππππ ½ (π¨π¨ β π©π©) ππππππ ½ (π¨π¨ + π©π©) ππ ππππππ ½ (π¨π¨ + π©π©) ππππππ ½ (π¨π¨ β π©π©) βππ ππππππ ½ (π¨π¨ + π©π©) ππππππ ½ (π¨π¨ β π©π©) (Book 2) Homework Exercises 3.3 page 81: # 1 β 18 odd numbers 8 Example 4: Show that ππππππππ ππ + ππππππππ ππ = ππππππππ ππ ππππππππ ππ Example 5: Prove that ππππππ ππ a) π¬π¬π¬π¬π¬π¬ ππ = π¬π¬π¬π¬π¬π¬ ππ π¬π¬π¬π¬π¬π¬ ππ + π¬π¬π¬π¬π¬π¬ ππ ππππππ ππ ππ = ππππππ ππ b) ππππππ ππ + ππππππ ππ = π¬π¬π¬π¬π¬π¬ ππ ππππππ ππ c) ππ+ππππππ ππ d) π¬π¬π¬π¬π¬π¬ ππ = π¬π¬π¬π¬π¬π¬ ππ ππβππππππ ππ ππππππ(π π β ππ) = βππππππ (ππ) e) ππ ππππππ οΏ½ π π + πποΏ½ = βππππππ (ππ) f) ππ Example 6: Use the basic trigonometric identities to determine the other five values of the trigonometric functions given that: a) π¬π¬π¬π¬π¬π¬ πΆπΆ = ππ/ππ and ππππππ πΆπΆ > ππ. b) ππ is an angle in quadrant III and ππππππ ππ = βππ / ππ. c) ππ is an angle in quadrant IV and tan x = -5. d) ππ is in quadrant II and ππππππ ππ = ππ/ππ. e) ππ is in quadrant I and ππππππ ππ = ππ/ππ. Example 7: Write π¨π¨ π¬π¬π¬π¬π¬π¬ ππππ + π©π© ππππππ ππππ = ππ π¬π¬π¬π¬π¬π¬(ππππ + ππ) Solution: Solved Examples: Simplify the following trigonometric expression. ππππππ (ππ) ππππππ (π π /ππ β ππ) 1) Solution: β’ 2) Use the identity ππππππ (π π /ππ β ππ) = ππππππ(ππ) and simplify ππππππ (ππ) ππππππ (π π /ππ β ππ) = ππππππ (ππ) ππππππ (ππ) = ππππππ (ππ) Simplify the following trigonometric expression. [π π π π π π 4 π₯π₯ β ππππππ 4 π₯π₯] / [π π π π π π 2 π₯π₯ β ππππππ 2 π₯π₯] Solution: β’ Factor the denominator [π π π π π π 4 π₯π₯ β ππππππ 4 π₯π₯] / [π π π π π π 2 π₯π₯ β ππππππ 2 π₯π₯] = [π π π π π π 2 π₯π₯ β ππππππ 2 π₯π₯][π π π π π π 2 π₯π₯ + ππππππ 2 π₯π₯] / [π π π π π π 2 π₯π₯ β ππππππ 2 π₯π₯] = [π π π π π π 2 π₯π₯ + ππππππ 2 π₯π₯] = 1 9 Simplify the following trigonometric expression. [π π π π π π (π₯π₯) π π π π π π 2 π₯π₯] / [1 + π π π π π π (π₯π₯)] 3) Solution: β’ β’ 4) Substitute sec (x) that is in the numerator by ππ / ππππππ (ππ) and simplify. [π π π π π π (π₯π₯) π π π π π π 2 π₯π₯] / [1 + π π π π π π (π₯π₯)] = π π π π π π 2 π₯π₯ / [ ππππππ π₯π₯ (1 + π π π π π π (π₯π₯) ] = π π π π π π 2 π₯π₯ / [ ππππππ π₯π₯ + 1 ] Substitute ππππππππ ππ ππππ ππ β ππππππππ ππ , factor and simplify. = [ 1 β ππππππ 2 π₯π₯ ] / [ ππππππ π₯π₯ + 1 ] = [ (1 β ππππππ π₯π₯)(1 + ππππππ π₯π₯) ] / [ ππππππ π₯π₯ + 1 ] = 1 β ππππππ π₯π₯ Simplify the following trigonometric expression. ππππππ (βππ) ππππππ (π π /ππ β ππ) Solution: β’ Use the identities ππππππ (βππ) = β ππππππ (ππ) and ππππππ (π π / ππ β ππ) = ππππππ (ππ) and simplify ππππππ (βππ) ππππππ (π π / ππ β ππ) = β ππππππ (ππ) ππππππ (ππ) = β ππππππππ ππ Simplify the following trigonometric expression. ππππππππ ππ β ππππππππ ππ ππππππππ ππ 5) Solution: β’ Factor ππππππππ ππ out, group and simplify ππππππππ ππ β ππππππππ ππ ππππππππ ππ = π π π π π π 2 π₯π₯ ( 1 β ππππππ 2 π₯π₯ ) = π π π π π π 4 π₯π₯ Simplify the following trigonometric expression. ππππππππ ππ + ππ ππππππ ππ ππ + ππ 6) Solution: β’ β’ 7) Note that the given trigonometric expression can be written as a square ππππππππ ππ + ππ ππππππ ππ ππ + ππ = ( ππππππππ ππ + ππ) ππ We now use the identity 1 + tan 2x = sec 2x ππππππππ ππ + ππ ππππππ ππ ππ + ππ = ( ππππππππ ππ + ππ) ππ = ( sec 2x ) 2 = sec 4x Add and simplify. Solution: β’ ππ / [ππ + ππππππ ππ] + ππ / [ππ β ππππππ ππ] In order to add the fractional trigonometric expressions, we need to have a common denominator ππ / [ππ + ππππππ ππ] + ππ / [ππ β ππππππ ππ] = [ 1 β ππππππ π₯π₯ + 1 + ππππππ π₯π₯ ] / [ [1 + ππππππ π₯π₯] [1 β ππππππ π₯π₯] ] = 2 / [1 β ππππππ 2 π₯π₯] = 2 / π π π π π π 2 π₯π₯ = 2 ππππππ 2 π₯π₯ 10 Write οΏ½( 4 β 4 π π π π π π 2 π₯π₯ ) without square root for (π π / ππ) < ππ < π π . 8) Solution: β’ Factor, and substitute ππ β ππππππππ ππ ππππ ππππππ ππ ππ β’ 9) οΏ½( 4 β 4 π π π π π π 2 π₯π₯ ) = οΏ½4(1 β π π π π π π 2 π₯π₯ ) = 2 β ππππππ 2 π₯π₯ = 2 | ππππππ (π₯π₯) | Since (π π / ππ) < ππ < π π , ππππππ ππ is less than zero and the given trigonometric expression simplifies to = β ππ ππππππ (ππ) Simplify the following expression. Solution: β’ [ππ β ππππππππ ππ] / [ππ + ππππππππ ππ] Factor the denominator, and simplify [ππ β ππππππππ ππ] / [ππ + ππππππππ ππ] = [1 β π π π π π π 2 π₯π₯] [1 + π π π π π π 2 π₯π₯] / [1 + π π π π π π 2 π₯π₯] = [ππ β ππππππππ ππ] = ππππππππ ππ ππ / [ππ + ππππππ ππ] + ππ / [ππ β ππππππ ππ] 10) Add and simplify. Solution : β’ Use a common denominator to add ππ [ππ + ππππππ ππ] + [ππ ππ β ππππππ ππ] 11) Add and simplify. [ππ β ππππππ ππ + ππ + ππππππ ππ] = [ (ππ + ππππππ ππ)(ππ β ππππππ ππ)] = ππ / [ ππ β ππππππππ ππ ] = ππ / ππππππππ ππ = ππ ππππππππ ππ ππππππ ππ β ππππππ ππ ππππππππ ππ Solution: β’ factor ππππππ ππ out ; ππππππ ππ β ππππππ ππ ππππππππ ππ = ππππππ ππ οΏ½ππ β ππππππππ πποΏ½ = ππππππ ππ ππππππππ ππ = ππππππππ ππ 11 ππππππππ ππ ππππππππ ππ + ππππππππ ππ ππππππππ ππ 12) Simplify the following expression. Solution: β’ Use the trigonometric identities ππππππ ππ = ππππππ ππ / ππππππ ππ and ππππππ π₯π₯ = ππππππ π₯π₯ / π π π π π π π₯π₯ to write the given expression as β’ ππππππππ ππ ππππππππ ππ + ππππππππ ππ ππππππππ ππ = (ππππππ ππ / ππππππ ππ)ππ ππππππππ ππ + (ππππππ ππ / ππππππ ππ)ππ ππππππππ ππ = ππππππππ ππ + ππππππππ ππ = ππ and simplify to get: π π π π π π 13) Simplify the following expression. ππππππ ( β ππ) β ππππππ( β ππ) ππππππ( β ππ) ππ ππ ππ Solution: β’ π π π π π π Use the identities ππππππ ( ππ β ππ) = ππππππ ππ, ππππππ( ππ β ππ) = ππππππ ππ and ππππππ( ππ β ππ) = ππππππ ππ to write the given expression as π π π π π π π¬π¬π¬π¬π¬π¬ οΏ½ β πποΏ½ β ππππππ οΏ½ β πποΏ½ π¬π¬π¬π¬π¬π¬ οΏ½ β πποΏ½ = ππππππ ππ β ππππππ ππ ππππππ ππ ππ ππ ππ = ππππππ ππ β ππππππ ππ π¬π¬π¬π¬π¬π¬ ππ ππππππ ππ = ππππππ ππ β ππππππππ ππ/ππππππ ππ = ππ/ππππππ ππ β ππππππππ ππ/ππππππ ππ = (ππ β ππππππππ ππ) /ππππππ ππ = ππππππππ ππ/ππππππ ππ 14) Show that ππππππππ ππ β ππππππππ ππ = ππππππ(ππππ) = ππππππ ππ Practice Problems 12