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1 GRADE 12 TRIGONOMETRY TRIGONOMETRIC IDENTITIES: Grade 11 Revision: Worksheet handed out on 18/04 COMPOUND ANGLE IDENTITIES: cos(A – B) = cosA.cosB + sinA.sinB cos(A + B) = cosA.cosB - sinA.sinB sin(A – B) = sinA.cosB – cosA.sinB sin(A + B) = sinA.cosB + cosA.sinB Examples: Use the appropriate compound angle formulae to verify that: Example 1: Example 2: sin(90° + ) = cosθ cos3θ.sinθ – cosθ.sin3θ = - sin2θ Solution: Solution: LHS = sin(90° + ) = sin90°.cosθ + cos90°.sinθ = 1.cosθ + 0 = cosθ LHS = cos3θ.sinθ – cosθ.sin3θ = sinθ.cos3θ – cosθ.sin3θ = sin(θ - 3θ) = sin(-2θ) = - sin2θ Examples: Evaluate without using a calculator: Example 3: Example 4: sin 60°.cos30° - sin30°.cos60° Solution: cos79°.cos 311° + sin101°.sin49° = sin(60° - 30°) = sin 30° = cos79°.cos 49° + sin79°.sin49°…reduction formula = cos(79° - 49°) = = cos30° = Example 5: cos 75° Example 6: sin 285° = cos(30° + 45°) = sin (225o + 60o) = cos 30o.cos 45o - sin 30o.sin 45o = sin 225o.cos 60o + sin 60o.cos 225o 3 2 1 2 2 2 2 2 = (– sin 45o)(cos 60o) + (sin 60o)( – cos 45o) 2 6 2 4 4 2 1 3 2 2 2 2 2 6 2 4 2 ( 3 1) 4 DOUBLE-ANGLE IDENTITIES sin2A = 2sinA.cosA cos2A = cos2A– sin2A cos2A = 1 – 2 sin2A cos2A = 2cos2A – 1 Note * 1 2 * sin 4A = 2sin2A.cos2A sin 2A = sinA.cosA Additional Examples: Prove the following identities: 1. tan A 1. 1 cos 2A sin 2A Solution: R.H.S. (sin 2 A cos 2 A) (cos 2 A sin 2 A) 2 sin Acos A 2 sin A cos 2 A cos 2 A sin 2 A 2 sin Acos A 2 2 sin A 2 sin Acos A sin A cos A = tan A = L.H.S. 2 6 4 4 2 (1 3 ) 4 3 cos 2 x sin 2 x cos x sin x (cos x sin x)3 1 sin 2 x 2. Solution:L.H.S. 3. (cos x sin x)(cos x sin x) (cos x sin x)(cos x sin x) 2 cos 2 P cos P sin P cos P sin P L.H.S. cos x sin x cos x 2 sin x cos x sin 2 x cos x sin x 1 2 sin x cos x cos x sin x 1 sin 2 x 2 cos 2 P sin 2 P cos P sin P (cos P sin P)(cos P sin P) cos P sin P cos P sin P = R.H.S. = R.H.S. RESTRICTIONS ON IDENTITIES: AN IDENTITY IS NOT VALID WHEN ANY DENOMINATOR IS ZERO. AN IDENTITY IS NOT VALID WHEN THERE IS A NEGATIVE VALUE UNDERNEATH THE Tanθ is undefined for θ = 90° + 180°.k (remember the graph of tan?) SOLVING TRIGONOMETRIC EQUATIONS: GRADE 11 REVISION Summary of general solutions: When the interval is not restricted, trigonometric equations have infinite many solutions due to the cyclic nature of Trigonometric Functions (Graphs). The general solution of a trigonometric equation is the solution that holds for any interval given. The solution: = a + k.360o OR = a + k.180o sin & cos graphs tan graph where ‘a’ represents an angle and k Z (remember the period of the graph) A ratio (angle) = fraction Example 1 sin = 1 Solution = 45 + k.360o or = 135o + k.360o ; kZ Example 2 Solution cos = 0,5 = 60o + k.360o or = 300o + k.360o ; kZ Example 3 Solution tan = – 1 = 135o + k.180o or = 315o + k.180o ; kZ 2 o 4 Example 4 Solution tan( θ2 15o ) 32 PAA : 56,309932 …ْ Q2 o θ 15 = 180 - 56,3..ْ + k.180 or 2 θ= 2 138,7 + k.180 = 277,4 + k.360 ; θ 2 θ 2 Q4 15 = 360 - 56,3.. + k.180 o = 318,7 + k.180 k Z The general solution for sin and cos is +k.360o as the functions repeat themselves every 360o (period of sin and cos is 360o) The general solution for tan is + k.180 as tan repeats itself every 180 (period of tan is 180) Example 5 sin(3 + 30) = 0,734 Solution PAA : 47,222785 … Q1 Q2 o 3 + 30 = 47,2.. + k.360 3 = 17,2..o + k.360 = 5,7o + k.120 B or 3 + 30 = 180 – 47,2..o + k.360 3 = 102,7..o + k.360 = 34,3o + k.120 ; k Z Use of Identities and factorisation Example 1 3 sin2 + 2 sin – 1 = 0. Determine the general solution and hence solve for , ε[ Solution Let sin = k 3k2 + 2k – 1 = 0 (3k – 1)(k + 1) = 0 k = 13 sin = 13 or k=–1 or sin = – 1 = 19,5o + k.360o or = 160,5o + k.360o or = 270o + k.360o ; k Z SS: ε {19,5o ; 160,5o} Example 2 Solution 6cos2x + cos x – 2 = 0 Find 1. The general solution 2. x for x [-90; 270] (3cos x + 2)(2cos x – 1) = 0 cos x = 23 or cos x = 12 PAA: 48,18 … PAA: 60 x = 131,8o + k.360 or x = 228,2 + k.360 or x = 60 + k.360 or x = 300 + k.360 ; k Z x = 131,8; 228,2; 60 or – 60 SS: x ε {-60 o ; 60 o ;131,8o ; 228,2o} Example 3 Solve for [0; 360] sin2 – sin .cos – 2 cos2 = 0 5 (sin – 2 cos )(sin + cos ) = 0 sin = 2 cos or sin = – cos tan = 2 or tan = – 1 PAA: 63.434949 … or PAA: 45 = 63,4 + k.180 or = 135 + k.180 ε {63,4; 135; 243,4 or 315} Solution C ; kZ ; kZ Use of co- ratios Example 1 sin x = cos 20. Solve for x [-360; 360] Solution sin x = cos(90 – 70) sin x = sin 70 PAA: 70 x = 70 + k.360 or x = 110 + k.360 x = 70 or x = - 290 or x = 110 or x = - 250 SS: x ε {- 290; - 250; 70; 110} Example 2 sin 3x = cos 2x. Solve for x [-180; 180] Solution sin 3x = sin (90 – 2x) PAA: 90 – 2x 3x = 90 – 2x + k.360 or 3x = 180 – (90 – 2x) + k.360 5x = 90 + k.360 or 3x = 90 + 2x + k.360 x = 18 + k.72 or x = 90 + k.360 ; kZ x = 18 or x = 90 or x = 162 or x = - 54 or x = -126 SS: x ε {- 126; - 54; 18; 90; 162} D General Solution of certain special angles sin x = 1 x = 90 + k.360 ; kZ sin x = 0 x = k.180 ; k Z x = 0o + k.360o or x = 180o + k.360o E sin x = – 1 x = – 90 + k.360 ; k Z or x = 270 + k.360 ; kZ cos x = – 1 x = 180 + k.360 ; kZ cos x = 0 x = 90 + k.180 kZ cos x = 1 x = k.360 ; k Z i.e. x = 0o + k.360o tan x = 0 x = k.180 ; k Z General Solution of where some ratios are undefined tan x ; x = 90 + k.180 ; kZ 6 GR.12 TRIGONOMETRY WORKSHEET 1 1. y r sinA = hence y = r sinA y and P(x; y) cosA = hence A O x x r x = r cosA The POLAR CO-ORDINATES of P are: ______________________________ 2. sin2A + cos2A = ______________ A 3. The COSINE RULE : In ABC: BC2 = __________________________________ B C 4. (xcosA-xsinA)2 = ____________________________________________ 5. The DISTANCE FORMULA: B(cosP; sinP) AB2 =_____________________________________ _____________________________________ _____________________________________ A(cosQ; sinQ) 7 6. P Q A B O X Consider the circle of radius 1 unit, with centre the point O at the origin. Let P and Q be the POˆ Q A B points on the circumference with POˆ X A and QOˆ X B . Complete: P is the point:(________; _________) Q is the point:(_________; ________) By the distance formula: (1) PQ2 = ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ By the Cosine rule: (2) PQ2 = ______________________________________________________________ ______________________________________________________________ From (1) and (2) above make cos(A-B) the subject of the formula. ___________________________________________________________________ ___________________________________________________________________ HENCE: cos(A-B) = _________________________________________ 8 DEDUCTIONS: Complete: 1. cos(A + B) = cos[A – (-B)] ___________________________________________________ ___________________________________________________ 2. sin(A – B) = cos[90° - (A –B)] = cos[(90° - A) + B] or cos[(90° + B) - A] ___________________________________________________ ___________________________________________________ ___________________________________________________ 3. sin(A + B) =___________________________________________________ ___________________________________________________ ___________________________________________________ ___________________________________________________ Using ordinary addition we have: Complete: sin2A = sin(A + A) = ___________________________________________________ = ___________________________________________________ cos2A = ___________________________________________________ = ___________________________________________________ = ___________________________________________________