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Trigonometry

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Reciprocal Trigonometry Functions
Connections to the Study Design: AOS 1 β Functions and graphs Graphs of the reciprocal circular functions cosecant, secant and cotangent, and simple transformations of these

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Key Notation: π : represents a point/coordinate πβ²: represents
Key Vocabulary: Functions Unit Circle Derive Relationship Magnitude Radians Compound- Angle Quadrant CAST Equivalent Reciprocal Abbreviate Cosecant Secant Cotangent Inverse Corresponding Rationalise Denominator Simplify Resulting Key Notation: π : represents a point/coordinate πβ²: represents

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Trig Recap sin π= πππππ ππ‘π βπ¦πππ‘βπππ’π π = π π

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Unit Circle – CAST

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Angle (degrees) 0Β° 30Β° 45Β° 60Β° 90Β° 180Β° 270Β° 360Β° Angle (radians) π 6 π 4 π 3 π 2 π 3π 2 2π sin(ΞΈ) 1 2 2 2 3 2 1 -1 cos(ΞΈ) tan(ΞΈ) 3 3 3 Undefined

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First Quadrant Angle: Degrees: 0Β° <π<90Β° Radians: 0 < π < π 2 π₯>0 πππ π¦ >0 sin π >0 πππ cos π >0 => tan π >0 Note: sin 30Β°+ sin 60Β° β  sin 90Β° In general: sin π΄+π΅ β  sin π΄ +sin(π΅) cos π΄+π΅ β  cos π΄ + cos π΅ tan π΄+π΅ β  tan π΄ + tan π΅

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Second Quadrant Angle: 180 β ΞΈ πππππππ  ππ Ο βΞΈ πππππππ
Degrees: 90Β° <π<180Β° Radians: π 2 < π < π π₯<0 πππ π¦ >0 sin π >0 πππ cos π <0 => tan π >0 Point π(π,π) reflected through the y-axis becomes π β² βπ,π Point P makes angle ΞΈ, then Pβ makes angle: 180 β ΞΈ πππππππ  ππ Ο βΞΈ πππππππ  sin 180Β° β π = sin π cos 180Β° β π =βcos(π) tan 180Β° β π =βtan(π) sin π β π = sin π cos πβ π =βcos(π) tan πβ π =βtan(π)

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Third Quadrant Angle: 180+ ΞΈ πππππππ  ππ Ο+ΞΈ πππππππ
Degrees: 180Β° <π<270Β° Radians: Ο< π < 3π 2 π₯<0 πππ π¦ <0 sin π <0 πππ cos π <0 => tan π >0 Point π(π,π) reflected through the y-axis becomes π β² βπ,βπ Point P makes angle ΞΈ, then Pβ makes angle: 180+ ΞΈ πππππππ  ππ Ο+ΞΈ πππππππ  sin 180Β°+ π = βsin π cos 180Β°+ π =βcos(π) tan 180Β°+ π =tan(π) sin π+ π = βsin π cos π+ π =βcos(π) tan π+ π =tan(π)

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Fourth Quadrant Angle: 180+ ΞΈ πππππππ  ππ Ο+ΞΈ πππππππ
Degrees: 270Β° <π<360Β° Radians: 3π 2 < π <π π₯>0 πππ π¦ <0 sin π <0 πππ cos π >0 => tan π >0 Point π(π,π) reflected through the y-axis becomes π β² βπ,βπ Point P makes angle ΞΈ, then Pβ makes angle: 180+ ΞΈ πππππππ  ππ Ο+ΞΈ πππππππ  sin 360Β°β π = βsin π cos 360Β°βπ =cos(π) tan 360Β°βπ =βtan(π) sin 2πβπ = βsin π cos 2πβπ =cos(π) tan 2πβπ =βtan(π)

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Summary sin π =π ππ πβπ =β sin π+π =β sin 2πβπ cos π =β cos πβπ =β cos π+π = cos 2πβπ tan π =β tan πβπ = tan π+π = βtan 2πβπ

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Negative Angles Point π(π,π) reflected through the x-axis becomes π β² π,βπ π makes angle ΞΈ with x-axis => π β² makes angle – ΞΈ sin βπ =β sin π cos βπ = cos π tan βπ =βtan(π) A negative angle β π 2 <π<0 is just the equivalent angle in the fourth quadrant. For positive angles greater than 360Β° or 2Ο, we can just subtract multiples of 360Β° or 2Ο. sin 360Β°+ π = sin π cos 360Β°+π =cos(π) tan 360Β°+π =tan(π) sin 2π+π = sin π cos 2π+π =cos(π) tan 2π+π =tan(π)

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Reciprocal Trigonometric Functions
Sine Function Cosine Function Tangent Function Reciprocal Trigonometric Function Cosecant Function: cosec(ΞΈ)= 1 sin(π) , π€βπππ sin(π)β 0 Secant Function: sec ΞΈ = 1 cos(π) , π€βπππ cos(π)β 0 Cotangent Function: cot ΞΈ = 1 tan(π) , π€βπππ sin π β 0 Note: these are not inverse trigonometric functions. cosec ΞΈ = βπ¦πππ‘πππ’π π opposite = π π sec ΞΈ = βπ¦πππ‘πππ’π π adjacent = π π cot ΞΈ = ππππππππ‘ πππππ ππ‘π = π π

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Exact Values Angles multiples of 30Β° and 45Β°: Exact values for reciprocal trigonometric functions can be found from corresponding trigonometric values.

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Example 1: Exact Values Find the exact value of πππ ππ( 5π 4 )

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Example 2: Using triangles to find values
If πππ ππ π = 7 4 and π 2 <π<π, find the exact value of cot(π).

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Trigonometric Identities using Reciprocal Trigonometric Functions
Connections to the Study Design AOS 1 Functions and graphs Compound and double angle formulas for sine, cosine and tangent and the identities: π ππ 2 π₯ =1+ π‘ππ 2 π₯ πππ ππ 2 π₯ =1+ πππ‘ 2 (π₯)

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Key Vocabulary: Identity Equation Mathematical Convention Transform Simplifying Factorising Cancelling Common Factors Denominators Numerators Prove Fundamental Relations Quotient Key Notation: holds for all values πβπ π π LHS and RHS

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Identities vs Equations
[ sin π ] 2 = π ππ 2 (π) similary [ πππ  π ] 2 = πππ  2 (π) π ππ 2 π + πππ  2 π =1 β Identity not equation, since it holds true for all values of π tan π = sin(π) cos(π) holds for all values of π for which tan(π) is defined, that is for all values where cos(π)β 0, or πβ (2π+1) π 2 where πβπ or odd multiples of π 2

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Identities and Relations
Trigonometric Identities Recap: tan π = sin(π) cos(π) sec π = 1 cos(π) πππ ππ π = 1 sin(π) cot π = 1 tan(π) Fundamental Relations: 1+ πππ‘ 2 π = πππ ππ 2 (π) π‘ππ 2 π +1= π ππ 2 (π)

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Work Example 3: Identity Proof
Prove the identity tan π + cot π = sec π πππ ππ(π)

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Work Example 4: Identity Proof
Prove the identity 1+ π‘ππ 2 (π) 1+ πππ‘ 2 (π) = π‘ππ 2 (π)

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Compound-angle Formulas
Connections to the Study Design AOS 1 Compound and double angle formulas for sine, cosine and tangent and the identities: π ππ 2 π₯ =1+ π‘ππ 2 π₯ πππ ππ 2 π₯ =1+ πππ‘ 2 (π₯)

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Key Vocabulary: Compound Perpendicular Properties Supplementary Substituting Derive Ratio Corresponding Theorems Expanding Expressions Complementary Multiples Argument Key Notation: β π΄π΅πΆ sin ππ 2 Β±π πππ πβπ 0<π΄< π 2

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Summary of the compound-angle formulas
sin π΄+π΅ = sin π΄ cos π΅ + cos π΄ sin π΅ sin π΄βπ΅ = sin π΄ cos π΅ β cos π΄ sin π΅ cos π΄+π΅ = cos π΄ cos π΅ β sin π΄ sin π΅ cos π΄βπ΅ = cos π΄ cos π΅ + sin π΄ sin π΅ tan π΄+π΅ = tan π΄ +tan(π΅) 1β tan π΄ tan(π΅) tan π΄βπ΅ = tan π΄ βtan(π΅) 1+ tan π΄ tan(π΅)

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Proof of the compound-angle formula: tangent

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Worked Examples: Collaborative
Work Example 5: Using compound-angle formulas Evaluate π¬π’π§ ππΒ° ππ¨π¬ ππΒ° + ππ¨π¬ ππΒ° π¬π’π§ ππΒ° | Jay and Vincent Work Example 6: Expanding trigonometric expressions with phase shifts Expand π ππ¨π¬ π½+ π π | Alex and Star Work Example 7: Simplification of π ππ ππ 2 Β±π and cos ππ 2 Β±π Use compound-angle formulas to simplify ππ¨π¬ ππ π βπ½ | Yanira and Ruby Work Example 8: Exact values for multiples of π 12 Find the exact value of π¬π’π§ πππ ππ | Christian and Vish Work Example 9: Using triangles to find values If ππ¨π¬ π¨ = ππ ππ and π¬π’π§ π© = π ππ , where π<π¨< π π and π π <π©<π, find the exact value of π¬π’π§(π¨βπ©) | James and Emily

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Double-angle formulas
Connections to the Study Design AOS 1 Compound and double angle formulas for sine, cosine and tangent and the identities: π ππ 2 π₯ =1+ π‘ππ 2 π₯ πππ ππ 2 π₯ =1+ πππ‘ 2 (π₯)

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Key Vocabulary: Equivalent Common Factor Null Factor Law Reciprocal
Key Notation: π΅=π΄ π₯β 0,2π

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Double Angle Formulas sin 2π΄ =2 sin π΄ cos π΄ cos 2π΄ = πππ  2 π΄ β π ππ 2 π΄ cos 2π΄ =1β2 π ππ 2 π΄ cos 2π΄ =2 πππ  2 π΄ β1

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Half-angle formulas 1β cos π΄ =2 π ππ 2 π΄ 2 1+ cos π΄ =2 πππ  2 ( π΄ 2 )

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Multiple-angle formulas
sin 3π΄ =3 sin π΄ β4 π ππ 3 π΄ cos 3π΄ =4 πππ  3 π΄ β3 cos π΄ tan 3π΄ = 3 tan π΄ β π‘ππ 3 (π΄) 1β3 π‘ππ 2 (π΄) sin 4π΄ =cos(π΄)(4 sin π΄ β8 π ππ 3 π΄ ) cos 4π΄ =8 πππ  4 π΄ β8 πππ  2 π΄ +1 tan 4π΄ = 4tan(π΄)(1β π‘ππ 2 π΄ ) 1β6 π‘ππ 2 π΄ + π‘ππ 4 (π΄)

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Collaborative Work Examples
Worked Example 10: Using double-angle formulas in simplifying expressions Find the exact value of sin 7π cos 7π 12 Worked Example 11: Finding Trigonometry expressions involving double-angle formulas If cos π΄ = 1 4 , determine the exact values of sin(2π΄), cos 2π΄ and tan(2π΄) Worked Example 12: Solving trigonometric equations involving double-angle formulas Solve for x if sin 2π₯ cos π₯ =0 for π₯β[0,2π] Worked Example 13: Trigonometric identities using doubleβangle formula Prove the identity cos 2π΄ cos π΄ + sin 2π΄ sin(π΄) sin 3π΄ cos π΄ β cos 3π΄ sin(π΄) = 1 2 πππ ππ(π΄) Worked Example 14: Half-angle formulas Prove the identity πππ ππ π΄ β cot π΄ = tan π΄ 2 Worked Example 15: Multiple-angle formulas Prove the identity cos 3π΄ =4 πππ  3 π΄ β3cos(π΄)

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