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C3 Chapter 7 FURTHER TRIGONOMETRIC IDENTITIES ADDITION FORMULAE Remember that: sin(90 – ) = cos cos(90 – ) = sin It is easily verified that sin (A + B) ≠ sin A + sin B One counter-example is all that is needed e.g. Let A = 30º and B = 45º Then sin(A + B) = sin 75º = 0.9659… and sin 30º + sin 45º = 1.2071… so sin(30º + 45º) ≠ sin30º + sin45º In fact: sin (A + B) = sin A cos B + cos A sin B sin (A – B) = sin A cos B – cos A sin B cos (A + B) = cos A cos B – sin A sin B cos (A – B) = cos A cos B + sin A sin B (You do not need to know how to derive these formulae) You can deduce the results from each other Example 1 In sin(A + B) = sin A cos B + cos A sin B, replace A with (90º - A) sin([90 – A] + B) = sin(90 – A)cosB + cos(90 – A)sinB sin(90 – [A – B]) = cosAcosB + sinAsinB cos(A – B) = cosAcosB + sinAsinB JMcC 1 C3 Chapter 7 tan(A + B) = sin(A + B) cos(A + B) = sinAcosB + cosAsinB cosAcosB – sinAsinB Divide top & bottom by cosAcosB tan(A + B) = tanA + tanB 1 – tanAtanB Replace B by –B tan(A – B) = tanA + tan( – B) 1 – tanAtan( – B) Hence we have: tan(A + B) = tan(A – B) = = tanA – tanB 1 + tanAtanB tanA + tanB 1 – tanAtanB tanA – tanB 1 + tanAtanB Example 2 2 Prove the identity 2 sin(A + B)sin(A – B) = sin A – sin B sin(A + B)sin(A – B) = (sinAcosB + cosAsinB)(sinAcosB – cosAsinB) 2 2 2 2 2 2 = sin A cos B + sinAcosAsinBcosB – sinAcosAsinBcosB – cos A sin B 2 2 = sin A cos B – cos A sin B 2 2 2 2 = sin A (1 – sin B) – (1 – sin A)sin B 2 2 2 2 2 2 2 2 = sin A – sin A sin B – sin B + sin A sin B = sin A – sin B JMcC QED 2 C3 Chapter 7 Example 3 Given that sin A = 3 5 and A is acute and cos B = – 5 13 and B is obtuse Find the exact value of sin(A + B) sin(A + B) = sinAcosB + cosAsinB A is acute 5 cosA = 3 4 5 A 4 B is obtuse 2nd Q sinB is positive sinB = + 13 12 12 13 B 5 3 5 4 12 sin(A + B) = + – 5 13 5 13 = – 15 65 + 48 65 = 33 65 Ex 7A p 99 JMcC 3 C3 Chapter 7 DOUBLE ANGLE FORMULAE In the formula sin (A + B) = sin A cos B + cos A sin B, let A = B Therefore, sin (A + A) = sin A cos A + cos A sin A sin 2A = 2 sin A cos A Hence the sine double angle formula: sin 2A = 2 sin A cos A In the formula cos (A + B) = cos A cos B – sin A sin B, let A = B Therefore, cos (A + A) = cos A cos A – sin A sin A cos 2A = cos2A – sin2A Since cos2A = 1 – sin2A, cos 2A = (1 – sin2A) - sin2A = 1 - 2 sin2A Or sin2A = 1 – cos2A, cos 2A = cos2A – (1 – cos2A) = 2 cos2A – 1 Hence the 3 versions of the cosine double angle formula: cos 2A = cos2A – sin2A or 1 - 2 sin2A or 2 cos2A – 1 In the formula tan(A + B) = tan(A + A) = tanA + tanB 1 – tanAtanB tanA + tanA 1 – tanAtanA = , let A = B 2tanA 2 1 – tan A Hence the tangent double angle formula: tan 2A = JMcC 2tanA 2 1 – tan A 4 C3 Chapter 7 Example 1 Write the following as a single trigonometric ratio: 1 2 2cos 37 – 1 2 2cos 37 – 1 = cos (2 37) = cos 74 1 1 = = sec 74 2 2cos 37 – 1 cos 74 Example 2 Without using a calculator, find the value of 2 (cos 60 – sin 60) 2 2 2 (cos 60 – sin 60) = cos 60 – 2cos 60 sin 60 + sin 60 2 2 = (cos 60 + sin 60) – 2 cos 60 sin 60 = 1 – sin 120 = 1 – sin 60 = 1 – 3 2 = 2 – 3 2 Example 3 Given that sin θ = ¼ and that θ is obtuse, find the exact value of sin 2θ 4 1 Pythagoras x = 15 Obtuse cos is – ve x sin 2 = 2sin cos = 2 cos = – 15 4 1 15 15 = – – 4 4 8 Ex 7B p 103 JMcC 5 C3 Chapter 7 USING THE DOUBLE ANGLE FORMULAE TO PROVE IDENTITIES Example Prove the identity cot 2 = LHS = = 2 1 tan 2 1 sin 1 sin cosec θ – 2 cot 2θ cos θ = 2 sin θ 1 – tan = 2tan 2 – 2(1 – tan ) 2 tan 2 – 2(1 – tan )cos 2 sin 2 = cos 2 1 – (1 – tan ) cos sin cos 2 = 2 1 – cos + sin sin 2 = 2 sin sin = 2 sin QED Ex 7C Q 1, 2, 7-13 p 106 TRIPLE ANGLE FORMULAE sin 3A = sin(2A + A) = sin 2A cos A + cos 2A sin A 2 = 2 sin A cos A cos A + (1 – 2 sin A) sin A 2 2 = 2 sin A (1 – sin A) + (1 – 2 sin A) sin A 3 3 = 2 sin A – 2 sin A + sin A – 2 sin A 3 = 3 sin A – 4 sin A JMcC 6 C3 Chapter 7 cos 3A = cos(2A + A) = cos 2A cos A – sin 2A sin A 2 = (2 cos A – 1) cos A – 2 sin A cos A sin A 2 2 = (2 cos A – 1) cos A – 2 cos A (1 – cos A) 3 3 = 2 cos A – cos A – 2 cos A + 2 cos A 3 = 4 cos A – 3 cos A HALF-ANGLE FORMULAE 2 2 cos 2A = 2 cos A – 1 cos 2A + 1 = 2 cos A 2 cos A = 1 + cos 2A 2 2 2 cos 2A = 1 – 2 sin A 1 – cos 2A = 2 sin A 2 sin A = 1 – cos 2A 2 Replacing A with 2 cos x 2 = x 2 gives: 1 + cos x 2 and 2 sin x 2 = 1 – cos x 2 These are useful in integration. JMcC 7 C3 Chapter 7 USING THE DOUBLE ANGLE FORMULAE TO SOLVE EQUATIONS Example 1 Solve the equation 3 sin 2θ = 4 tan θ for 0º ≤ θ ≤ 360º 3 sin 2 = 4 tan 6sin cos = 2 4 sin cos 2 6 sin cos = 4 sin sin (3 cos – 2) = 0 sin = 0 or cos = 2 3 sin = 0 = 0, 180 , 360 cos = 2 has a solution in all 4 quadrants 3 = 35·3, 144·7, 215·3, 324·7 Hence = 0, 35·3, 144·7, 180, 215·3, 324·7, 360 Ex 7C Q 3, p 107 Example 2 Eliminate θ from the equations: 2 x = sin , y = 1 + cos 2 2 2 y = 1 + cos 2 = 1 + (1 – 2sin ) = 2 – 2 sin y = 2 – 2x Note: θ is known as a parameter Ex 7C Q 4-6, p 107 JMcC 8 C3 Chapter 7 THE FORM “a cos θ ± b sin θ” If you draw the graph of a cos θ + b sin θ it will have the same form as a sine or cosine graph. It can be expressed in the form R sin (θ ± α) or R cos (θ ± α) Example 1 Express 3 sin x + 2 cos x in the form R sin (x + ) where R > 0 and 0 < < 2 Let 3 sin x + 2 cos x = R sin (x + ) = R(sin x cos + cos x sin ) Equate coefficients of sin x R cos = 3 (1) Equate coefficients of cos x R sin = 2 (2) (2) (1) (1) 2 R sin R cos + (2) 2 = 2 3 2 tan = 2 2 2 3 = 33·7 2 R cos + R sin = 3 2 R = 9 + 4 = 13 R = Hence 3 sin x + 2 cos x = Remember: 2 + 2 2 13 13 sin (x + 33·7) Divide the equations to find tan α Square and add the equations to find R2 Ex 7D Q 1-7 p 111 JMcC 9 C3 Chapter 7 This method can be used to find maximum or minimum values of expressions and to solve certain types of trig equations. Example 2 Find the maximum value of 2 cos 2θ – 5 sin 2θ and the smallest positive value of θ for which it occurs. Let 2 cos 2 – 5sin 2 = Rcos(2 + ) = R(cos 2 cos – sin 2 sin ) Equate coefficients of cos 2 R cos = 2 (1) Equate coefficients of sin 2 R sin = 5 (2) (1) 2 r sin R cos (1) + (2) 2 = 2 5 2 tan = 5 2 = 68·2 2 2 R = 2 + 5 = 29 R = Hence 2 cos 2 – 5 sin 2 = (2) 29 29 cos (2 + 68·2) 29 Hence maximum value = This occurs when 2 + 68·2 = 0, 360, etc Smallest + ve 2 + 68·2 = 360 2 = 291·8 = 145·9 Note; if necessary use a sketch to determine the max/min values e.g. y x JMcC cos graph 10 C3 Chapter 7 Example 3 Solve the equation 4 cos x + 3 sin x = 1 for 0º ≤ x ≤ 360º Let 4 cos x + 3 sin x = R cos (x – ) = R (cos x cos + sin x sin ) Equate coefficients of cos x R cos = 4 (1) Equate coefficients of sin x R sin = 3 (2) (2) (1) (1) 2 tan = + (2) 2 3 4 2 = 36·87 R = 4 2 + 3 2 = 25 R = 5 5 cos(x – 36·87) = 1 cos (x – 36·87) = 0·2 + ve cosine 1st or 4th quadrant x – 36·87 = 78·46, 281·54 x = 115·3 or 318·4 Ex 7D Q 8-15 p 111 JMcC 11 C3 Chapter 7 FACTOR FORMULAE sin(A + B) = sin A cos B + cos A sin B sin(A – B) = sin A cos B – cos A sin B Adding gives: sin(A + B) + sin(A – B) = 2 sin A cos B Let P = A + B and Q = A – B A = P + Q 2 and B = P – Q 2 P + Q P – Q cos sin P + sin Q = 2 sin 2 2 Replacing Q with –Q gives: P + ( – Q) P – ( – Q) sin P + sin ( – Q) = 2 sin cos 2 2 P – Q P + Q cos sin P – sin Q = 2 sin 2 2 Hence: sin P + sin Q = 2 x sin (semi-sum) cos (semi-difference) sin P – sin Q = 2 x sin (semi-difference) cos (semi-sum) JMcC 12 C3 Chapter 7 cos(A + B) = cos A cos B – sin A sin B cos(A – B) = cos A cos B + sin A sin B Adding gives cos(A + B) + cos(A – B) = 2 cos A cos B Let P = A + B and Q = A – B A = P + Q 2 and B = P – Q 2 P + Q P – Q cos cos P + cos Q = 2 cos 2 2 Replacing Q with –Q gives exactly the same result, so this time we subtract the two starting equations: cos(A + B) – cos(A – B) = – 2 sin A sin B P + Q P – Q Hence, cos P – cos Q = – 2 sin sin 2 2 Hence: cos P + cos Q = 2 x cos (semi-sum) cos (semi-difference) cos P – cos Q = - 2 x sin (semi-sum) sin (semi-difference) JMcC 13 C3 Chapter 7 Hence the 4 factor formulae: P + Q P – Q sin P + sin Q = 2 sin cos 2 2 P + Q P – Q sin P – sin Q = 2 cos sin 2 2 P + Q P – Q cos P + cos Q = 2 cos cos 2 2 P + Q P – Q cos P – cos Q = – 2 sin sin 2 2 Note the minus sign in the last formula Example 1 Express as a sum or difference of sines: P + Q P – Q sin P + sin Q = 2 sin cos 2 2 Use Let sin 4x cos 6x P + Q 2 = 4x and P – Q 2 Adding gives P = 10x Subtracting gives Q = – 2x = 6x sin 10x + sin ( – 2x) = 2 sin 4x cos 6x sin 4x cos 6x = JMcC sin 10x – sin 2x 2 14 C3 Chapter 7 Example 2 Solve the equation cos 2θ + cos 3θ = 0 cos 2 + cos 3 = 2 cos cos 5 2 = 0 5 2 = and cos 2 = 0 2 = = 5 2 cos 2 for 0 ≤ θ ≤ 2π = 0 3 5 7 9 11 , etc. , , , , , 2 2 2 2 2 2 3 5 7 9 , , , , 5 5 5 5 5 3 , etc. , 2 2 = Hence = 7 9 3 , , , , 5 5 5 5 Ex 7E p 115 Mixed Ex 7F p 116 JMcC 15