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Trigonometry

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

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Key Notation: 𝑃 : represents a point/coordinate 𝑃′: 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: 𝑃 : represents a point/coordinate 𝑃′: represents

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Trig Recap sin πœƒ= π‘œπ‘π‘π‘œπ‘ π‘–π‘‘π‘’ β„Žπ‘¦π‘π‘œπ‘‘β„Žπ‘’π‘›π‘’π‘ π‘’ = 𝑏 𝑐

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Unit Circle – CAST

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Exact Values in First Quadrant
Angle (degrees) 0Β° 30Β° 45Β° 60Β° 90Β° 180Β° 270Β° 360Β° Angle (radians) πœ‹ 6 πœ‹ 4 πœ‹ 3 πœ‹ 2 πœ‹ 3πœ‹ 2 2πœ‹ 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œƒ<90Β° Radians: 0 < πœƒ < πœ‹ 2 π‘₯>0 π‘Žπ‘›π‘‘ 𝑦 >0 sin πœƒ >0 π‘Žπ‘›π‘‘ cos πœƒ >0 => tan πœƒ >0 Note: sin 30Β°+ sin 60Β° β‰  sin 90Β° In general: sin 𝐴+𝐡 β‰  sin 𝐴 +sin(𝐡) cos 𝐴+𝐡 β‰  cos 𝐴 + cos 𝐡 tan 𝐴+𝐡 β‰  tan 𝐴 + tan 𝐡

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Second Quadrant Angle: 180 βˆ’ ΞΈ π‘‘π‘’π‘”π‘Ÿπ‘’π‘’π‘  π‘œπ‘Ÿ Ο€ βˆ’ΞΈ π‘Ÿπ‘Žπ‘‘π‘–π‘Žπ‘›π‘ 
Degrees: 90Β° <πœƒ<180Β° Radians: πœ‹ 2 < πœƒ < πœ‹ π‘₯<0 π‘Žπ‘›π‘‘ 𝑦 >0 sin πœƒ >0 π‘Žπ‘›π‘‘ cos πœƒ <0 => tan πœƒ >0 Point 𝑃(π‘Ž,𝑏) reflected through the y-axis becomes 𝑃 β€² βˆ’π‘Ž,𝑏 Point P makes angle ΞΈ, then P’ makes angle: 180 βˆ’ ΞΈ π‘‘π‘’π‘”π‘Ÿπ‘’π‘’π‘  π‘œπ‘Ÿ Ο€ βˆ’ΞΈ π‘Ÿπ‘Žπ‘‘π‘–π‘Žπ‘›π‘  sin 180Β° βˆ’ πœƒ = sin πœƒ cos 180Β° βˆ’ πœƒ =βˆ’cos(πœƒ) tan 180Β° βˆ’ πœƒ =βˆ’tan(πœƒ) sin πœ‹ βˆ’ πœƒ = sin πœƒ cos πœ‹βˆ’ πœƒ =βˆ’cos(πœƒ) tan πœ‹βˆ’ πœƒ =βˆ’tan(πœƒ)

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Third Quadrant Angle: 180+ ΞΈ π‘‘π‘’π‘”π‘Ÿπ‘’π‘’π‘  π‘œπ‘Ÿ Ο€+ΞΈ π‘Ÿπ‘Žπ‘‘π‘–π‘Žπ‘›π‘ 
Degrees: 180Β° <πœƒ<270Β° Radians: Ο€< πœƒ < 3πœ‹ 2 π‘₯<0 π‘Žπ‘›π‘‘ 𝑦 <0 sin πœƒ <0 π‘Žπ‘›π‘‘ cos πœƒ <0 => tan πœƒ >0 Point 𝑃(π‘Ž,𝑏) reflected through the y-axis becomes 𝑃 β€² βˆ’π‘Ž,βˆ’π‘ Point P makes angle ΞΈ, then P’ makes angle: 180+ ΞΈ π‘‘π‘’π‘”π‘Ÿπ‘’π‘’π‘  π‘œπ‘Ÿ Ο€+ΞΈ π‘Ÿπ‘Žπ‘‘π‘–π‘Žπ‘›π‘  sin 180Β°+ πœƒ = βˆ’sin πœƒ cos 180Β°+ πœƒ =βˆ’cos(πœƒ) tan 180Β°+ πœƒ =tan(πœƒ) sin πœ‹+ πœƒ = βˆ’sin πœƒ cos πœ‹+ πœƒ =βˆ’cos(πœƒ) tan πœ‹+ πœƒ =tan(πœƒ)

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Fourth Quadrant Angle: 180+ ΞΈ π‘‘π‘’π‘”π‘Ÿπ‘’π‘’π‘  π‘œπ‘Ÿ Ο€+ΞΈ π‘Ÿπ‘Žπ‘‘π‘–π‘Žπ‘›π‘ 
Degrees: 270Β° <πœƒ<360Β° Radians: 3πœ‹ 2 < πœƒ <πœ‹ π‘₯>0 π‘Žπ‘›π‘‘ 𝑦 <0 sin πœƒ <0 π‘Žπ‘›π‘‘ cos πœƒ >0 => tan πœƒ >0 Point 𝑃(π‘Ž,𝑏) reflected through the y-axis becomes 𝑃 β€² βˆ’π‘Ž,βˆ’π‘ Point P makes angle ΞΈ, then P’ makes angle: 180+ ΞΈ π‘‘π‘’π‘”π‘Ÿπ‘’π‘’π‘  π‘œπ‘Ÿ Ο€+ΞΈ π‘Ÿπ‘Žπ‘‘π‘–π‘Žπ‘›π‘  sin 360Β°βˆ’ πœƒ = βˆ’sin πœƒ cos 360Β°βˆ’πœƒ =cos(πœƒ) tan 360Β°βˆ’πœƒ =βˆ’tan(πœƒ) sin 2πœ‹βˆ’πœƒ = βˆ’sin πœƒ cos 2πœ‹βˆ’πœƒ =cos(πœƒ) tan 2πœ‹βˆ’πœƒ =βˆ’tan(πœƒ)

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Summary sin πœƒ =𝑠𝑖𝑛 πœ‹βˆ’πœƒ =βˆ’ sin πœ‹+πœƒ =βˆ’ sin 2πœ‹βˆ’πœƒ cos πœƒ =βˆ’ cos πœ‹βˆ’πœƒ =βˆ’ cos πœ‹+πœƒ = cos 2πœ‹βˆ’πœƒ tan πœƒ =βˆ’ tan πœ‹βˆ’πœƒ = tan πœ‹+πœƒ = βˆ’tan 2πœ‹βˆ’πœƒ

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Negative Angles Point 𝑃(π‘Ž,𝑏) reflected through the x-axis becomes 𝑃 β€² π‘Ž,βˆ’π‘ 𝑃 makes angle ΞΈ with x-axis => 𝑃 β€² makes angle – ΞΈ sin βˆ’πœƒ =βˆ’ sin πœƒ cos βˆ’πœƒ = cos πœƒ tan βˆ’πœƒ =βˆ’tan(πœƒ) A negative angle βˆ’ πœ‹ 2 <πœƒ<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œƒ = sin πœƒ cos 360Β°+πœƒ =cos(πœƒ) tan 360Β°+πœƒ =tan(πœƒ) sin 2πœ‹+πœƒ = sin πœƒ cos 2πœ‹+πœƒ =cos(πœƒ) tan 2πœ‹+πœƒ =tan(πœƒ)

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Reciprocal Trigonometric Functions
Sine Function Cosine Function Tangent Function Reciprocal Trigonometric Function Cosecant Function: cosec(ΞΈ)= 1 sin(πœƒ) , π‘€β„Žπ‘’π‘Ÿπ‘’ sin(πœƒ)β‰ 0 Secant Function: sec ΞΈ = 1 cos(πœƒ) , π‘€β„Žπ‘’π‘Ÿπ‘’ cos(πœƒ)β‰ 0 Cotangent Function: cot ΞΈ = 1 tan(πœƒ) , π‘€β„Žπ‘’π‘Ÿπ‘’ sin πœƒ β‰ 0 Note: these are not inverse trigonometric functions. cosec ΞΈ = β„Žπ‘¦π‘π‘œπ‘‘π‘’π‘›π‘’π‘ π‘’ opposite = 𝑐 𝑏 sec ΞΈ = β„Žπ‘¦π‘π‘œπ‘‘π‘’π‘›π‘’π‘ π‘’ adjacent = 𝑐 π‘Ž cot ΞΈ = π‘Žπ‘‘π‘—π‘Žπ‘π‘’π‘›π‘‘ π‘œπ‘π‘π‘œπ‘ π‘–π‘‘π‘’ = π‘Ž 𝑏

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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‘œπ‘ π‘’π‘( 5πœ‹ 4 )

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Example 2: Using triangles to find values
If π‘π‘œπ‘ π‘’π‘ πœƒ = 7 4 and πœ‹ 2 <πœƒ<πœ‹, find the exact value of cot(πœƒ).

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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: 𝑠𝑒𝑐 2 π‘₯ =1+ π‘‘π‘Žπ‘› 2 π‘₯ π‘π‘œπ‘ π‘’π‘ 2 π‘₯ =1+ π‘π‘œπ‘‘ 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‘›βˆˆπ‘ 𝑄 𝑅 LHS and RHS

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Identities vs Equations
[ sin πœƒ ] 2 = 𝑠𝑖𝑛 2 (πœƒ) similary [ π‘π‘œπ‘  πœƒ ] 2 = π‘π‘œπ‘  2 (πœƒ) 𝑠𝑖𝑛 2 πœƒ + π‘π‘œπ‘  2 πœƒ =1 – Identity not equation, since it holds true for all values of πœƒ tan πœƒ = sin(πœƒ) cos(πœƒ) holds for all values of πœƒ for which tan(πœƒ) is defined, that is for all values where cos(πœƒ)β‰ 0, or πœƒβ‰ (2𝑛+1) πœ‹ 2 where π‘›βˆˆπ‘ or odd multiples of πœ‹ 2

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Identities and Relations
Trigonometric Identities Recap: tan πœƒ = sin(πœƒ) cos(πœƒ) sec πœƒ = 1 cos(πœƒ) π‘π‘œπ‘ π‘’π‘ πœƒ = 1 sin(πœƒ) cot πœƒ = 1 tan(πœƒ) Fundamental Relations: 1+ π‘π‘œπ‘‘ 2 πœƒ = π‘π‘œπ‘ π‘’π‘ 2 (πœƒ) π‘‘π‘Žπ‘› 2 πœƒ +1= 𝑠𝑒𝑐 2 (πœƒ)

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Work Example 3: Identity Proof
Prove the identity tan πœƒ + cot πœƒ = sec πœƒ π‘π‘œπ‘ π‘’π‘(πœƒ)

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Work Example 4: Identity Proof
Prove the identity 1+ π‘‘π‘Žπ‘› 2 (πœƒ) 1+ π‘π‘œπ‘‘ 2 (πœƒ) = π‘‘π‘Žπ‘› 2 (πœƒ)

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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: 𝑠𝑒𝑐 2 π‘₯ =1+ π‘‘π‘Žπ‘› 2 π‘₯ π‘π‘œπ‘ π‘’π‘ 2 π‘₯ =1+ π‘π‘œπ‘‘ 2 (π‘₯)

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Key Vocabulary: Compound Perpendicular Properties Supplementary Substituting Derive Ratio Corresponding Theorems Expanding Expressions Complementary Multiples Argument Key Notation: ∠𝐴𝐡𝐢 sin π‘›πœ‹ 2 Β±πœƒ π‘“π‘œπ‘Ÿ π‘›βˆˆπ‘ 0<𝐴< πœ‹ 2

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Summary of the compound-angle formulas
sin 𝐴+𝐡 = sin 𝐴 cos 𝐡 + cos 𝐴 sin 𝐡 sin π΄βˆ’π΅ = sin 𝐴 cos 𝐡 βˆ’ cos 𝐴 sin 𝐡 cos 𝐴+𝐡 = cos 𝐴 cos 𝐡 βˆ’ sin 𝐴 sin 𝐡 cos π΄βˆ’π΅ = cos 𝐴 cos 𝐡 + sin 𝐴 sin 𝐡 tan 𝐴+𝐡 = tan 𝐴 +tan(𝐡) 1βˆ’ tan 𝐴 tan(𝐡) tan π΄βˆ’π΅ = tan 𝐴 βˆ’tan(𝐡) 1+ tan 𝐴 tan(𝐡)

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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 𝐬𝐒𝐧 𝟐𝟐° 𝐜𝐨𝐬 πŸ‘πŸ–Β° + 𝐜𝐨𝐬 𝟐𝟐° 𝐬𝐒𝐧 πŸ‘πŸ–Β° | Jay and Vincent Work Example 6: Expanding trigonometric expressions with phase shifts Expand 𝟐 𝐜𝐨𝐬 𝜽+ 𝝅 πŸ‘ | Alex and Star Work Example 7: Simplification of 𝑠𝑖𝑛 π‘›πœ‹ 2 Β±πœƒ and cos π‘›πœ‹ 2 Β±πœƒ Use compound-angle formulas to simplify 𝐜𝐨𝐬 πŸ‘π… 𝟐 βˆ’πœ½ | Yanira and Ruby Work Example 8: Exact values for multiples of πœ‹ 12 Find the exact value of 𝐬𝐒𝐧 πŸπŸ‘π… 𝟏𝟐 | Christian and Vish Work Example 9: Using triangles to find values If 𝐜𝐨𝐬 𝑨 = 𝟏𝟐 πŸπŸ‘ and 𝐬𝐒𝐧 𝑩 = πŸ• πŸπŸ“ , where 𝟎<𝑨< 𝝅 𝟐 and 𝝅 𝟐 <𝑩<𝝅, find the exact value of 𝐬𝐒𝐧(π‘¨βˆ’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: 𝑠𝑒𝑐 2 π‘₯ =1+ π‘‘π‘Žπ‘› 2 π‘₯ π‘π‘œπ‘ π‘’π‘ 2 π‘₯ =1+ π‘π‘œπ‘‘ 2 (π‘₯)

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Key Vocabulary: Equivalent Common Factor Null Factor Law Reciprocal
Key Notation: 𝐡=𝐴 π‘₯∈ 0,2πœ‹

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Double Angle Formulas sin 2𝐴 =2 sin 𝐴 cos 𝐴 cos 2𝐴 = π‘π‘œπ‘  2 𝐴 βˆ’ 𝑠𝑖𝑛 2 𝐴 cos 2𝐴 =1βˆ’2 𝑠𝑖𝑛 2 𝐴 cos 2𝐴 =2 π‘π‘œπ‘  2 𝐴 βˆ’1

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Half-angle formulas 1βˆ’ cos 𝐴 =2 𝑠𝑖𝑛 2 𝐴 2 1+ cos 𝐴 =2 π‘π‘œπ‘  2 ( 𝐴 2 )

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Multiple-angle formulas
sin 3𝐴 =3 sin 𝐴 βˆ’4 𝑠𝑖𝑛 3 𝐴 cos 3𝐴 =4 π‘π‘œπ‘  3 𝐴 βˆ’3 cos 𝐴 tan 3𝐴 = 3 tan 𝐴 βˆ’ π‘‘π‘Žπ‘› 3 (𝐴) 1βˆ’3 π‘‘π‘Žπ‘› 2 (𝐴) sin 4𝐴 =cos(𝐴)(4 sin 𝐴 βˆ’8 𝑠𝑖𝑛 3 𝐴 ) cos 4𝐴 =8 π‘π‘œπ‘  4 𝐴 βˆ’8 π‘π‘œπ‘  2 𝐴 +1 tan 4𝐴 = 4tan(𝐴)(1βˆ’ π‘‘π‘Žπ‘› 2 𝐴 ) 1βˆ’6 π‘‘π‘Žπ‘› 2 𝐴 + π‘‘π‘Žπ‘› 4 (𝐴)

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Collaborative Work Examples
Worked Example 10: Using double-angle formulas in simplifying expressions Find the exact value of sin 7πœ‹ cos 7πœ‹ 12 Worked Example 11: Finding Trigonometry expressions involving double-angle formulas If cos 𝐴 = 1 4 , determine the exact values of sin(2𝐴), cos 2𝐴 and tan(2𝐴) Worked Example 12: Solving trigonometric equations involving double-angle formulas Solve for x if sin 2π‘₯ cos π‘₯ =0 for π‘₯∈[0,2πœ‹] Worked Example 13: Trigonometric identities using doubleβ€”angle formula Prove the identity cos 2𝐴 cos 𝐴 + sin 2𝐴 sin(𝐴) sin 3𝐴 cos 𝐴 βˆ’ cos 3𝐴 sin(𝐴) = 1 2 π‘π‘œπ‘ π‘’π‘(𝐴) Worked Example 14: Half-angle formulas Prove the identity π‘π‘œπ‘ π‘’π‘ 𝐴 βˆ’ cot 𝐴 = tan 𝐴 2 Worked Example 15: Multiple-angle formulas Prove the identity cos 3𝐴 =4 π‘π‘œπ‘  3 𝐴 βˆ’3cos(𝐴)

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