Trigo Question: Edexcel 4PMO, 2012 May 17 Paper 1 Question 7 – Q and A

Question 7:                              cos(A +B) = cosA cosB – sinA sinB

a)      Express cos(2x + 45^o) in the form M cos2x + N sin2x, where M and N are constants, giving the exact value of M and the exact value of N.                       (2)

b)      Solve, for 0^o ≤ x ≤ 180^o, the equation cos2x –sin2x = 1                                    (5)

The maximum value of cos2x –sin2x is k.

c)      Find the exact value of k.                                                                                  (2)

d)      Find the smallest positive value of x for which a maximum occurs.                      (3)

My Proposed Solutions

a)      Use:                 cos(A +B) = cosA cosB – sinA sinB

Therefore:         cos(2x + 45^o) = cos2x cos45^o – sin2x sin45^o

[Use:                cos45^o = 1/√2 = sin45^o (trigo of convenient angle 45^o)]

Therefore:         cos(2x + 45^o) = (1/√2) cos2x – (1/√2) sin2x

                                                = (1/√2) cos2x + (-1/√2) sin2x

                                                ≡ M cos2x + N sin2x

Where, M = 1/√2 and N = -1/√2                                             (Answer)

b)      For 0^o ≤ x ≤ 180^o, solve cos2x –sin2x = 1.

cos2x –sin2x = 1

            x 1/√2:             (1/√2) cos2x – (1/√2) sin2x = 1/√2

Therefore:         cos(2x + 45^o) = 1/√2

Imply:               (2x + 45^o) = cos^-1 (1/√2) (= Reference Angle 45^o)

[Principle Angle Domain: 0^o ≤ x ≤ 180^o implies 45^o ≤ (2x + 45^o) ≤ 360^o + 45^o]

Therefore:         (2x + 45^o) is in 1^st qdrant (qd), 4^th qd and 1^st qd 1^st co-terminal

1^st qd:               2x + 45^o = 45^o;            2x = 0^o;            x = 0^o   (Answers)

4th qd:              2x + 45^o = 360^o -  45^o; 2x = 270^o;       x = 135^o

5^th qd:               2x + 45^o = 360^o +  45^o; 2x = 360^o;       x = 180^o

c)      Max. value of cos2x – sin2x = k

x 1/√2:             (1/√2) cos2x – (1/√2) sin2x = k/√2

Imply:               cos(2x + 45^o) = k/√2

Therefore:         k = (√2) cos(2x + 45^o)

k maximum, when cos(2x + 45^o) = 1; and, max. k = (√2)(1) = √2 (Answer)

d)      For principle angle domain: 45^o ≤ (2x + 45^o) ≤ 360^o + 45^o,

the maximum value occurs when cos(2x + 45^o) = 1 = cos 360^o.

(cos0^o = 1 too but 0^o  is outside the domain)

Therefore:         (2x + 45^o) = 360^o;        2x = 315^o;        x = 157.5^o        (Answer)

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Good luck to Ms Kim and J Min!