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!

Solve 9 sin2 Ө – 9 sin Ө = 11

Last night at Mont Kiara, the above question was forwarded to me to solve.

Anyway, you will note from my proposed solutions below that different domains for Ө will mean different answers. 

For examples:

* For 0 ≤ Ө ≤ π radians, no value of Ө can satisfy the equation: 9 sin² Ө – 9 sin Ө = 11;

* For 0 ≤ Ө ≤ 2π radians, two values Ө satisfy the equation

My Proposed Solutions

                                  9 sin² Ө – 9 sin Ө = 11

-11:                   9 sin² Ө – 9 sin Ө – 11 = 0

                        (compare: ax² + bx + c = 0)

where:                   a = 9, b = -9 and c = -11

therefore, roots:     [α = (-b + √(b² – 4ac))/2a; or β = (-b - √(b² – 4ac))/2a]

imply:          sin Ө = -(-9) - √(-9)² – 4(9)(-11))/(2x9) = - 0.713351648; or,

                    sin Ө = -(-9) + √(-9)² – 4(9)(-11))/(2x9) =  1.713351648;

For 0 ≤ Ө ≤ 2π radians:

                    sin Ө = - 0.713351648 implies Ө in 3^rd or 4^th quadrants

                          Ө = π + sin^-1 (0.713351648) or 2π - sin^-1 (0.713351648)

                          Ө = 3.935861836 (3rd Qd) or 5.488916105 (4th Qd)

                          Ө = 3.94 or 5.49 rad (both to 3 sf)

But for 0 ≤ Ө ≤ π radians (1^st and 2^nd quadrants):

                    For all values of Ө, sin Ө must be +ve.

                    Therefore, sin Ө ≠ - 0.713351648.

Also for all values of Ө in any quadrant, sin Ө ≠ 1.713351648 because 1 ≥  sin Ө ≥ -1.

So remember: If you want to solve a trigonometric equation, you must know or be given the domain because the answers depend on the domain for Ө! In fact, Q 10(e) of Edexcel Further Pure Maths (4PMO) Yr 2013 May Paper 1 amply demonstrates this!!

Brain Teaser 6: Differentiate y = Ae^2x + Be^-x, where x ≥ 0,…

A curve has equation y = Ae^2x + Be^-x, where x ≥ 0. At the point where x = 0, y = 50 and dy/dx = -20.

i)             Show that A = 10 and find the value of B.                                             (5)

ii)           Using the values of A and B found in part (i), find the coordinates of the stationary point of the curve.                                                                 (4)

{My additional question: Express the exact values of the coordinates of this stationary point as     (C ln 2, D(2)^(2/3) ) and show that C = 1/3 and D = 30)

iii)          Determine the nature of the stationary point, giving a reason for your answer.     (2)

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In an earlier post, I have shown that e^{ln 2} = 2, 10^{lg 3} = 3 and generally b^log_b x = x. My additional question in part (ii) may require the use of this knowledge in order to get the exact value of the y-coordinate of the stationary point mentioned therein.

(Background to this Q: In the course of preparing a student to sit for her IGCSE Cambridge Add-Maths 0606 next Tuesday (10/06/2014), I came across the above Q as one of the past-year questions on "Calculus". After teaching her how to get the answers, I additionally asked her to get the exact values of the coordinates of the stationary point instead of just (0.231, 47.6). Thus, I have added an additional question to part (ii). :) )

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Answers:

i)                    A = 10; B = 40

ii)                   Stationary point = (0.231, 47.6) [or, ((1/3)ln 2, 30(2)^(2/3)); C = 1/3: D=30]

iii)                 d²y/dx² = 40e^2x + 40e^-x = +ve for all values of x – thus, minimum point.

Working leading to the answers will be posted in a week or so (please see below on "My Proposed Solutions"). You may post yours by way of comments if this question is of interest to you too. J

(Posted on 3/06/2014 from PJ SS2 Starbucks)

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My Proposed Solutions:

(i)             Given :    y = Ae^2x + Be^-x, where x ≥ 0

              At x = 0, y = 50 and dy/dx = -20

At x = 0, y = 50, therefore:          50 = Ae^o + Be^o

                                                   50 = A + B          ~~~ Eqn (1)

At x = 0, dy/dx = -20, therefore:  dy/dx = 2Ae^2x – Be^-x

                                                   -20  =  2Ae²⁽⁰⁾ – Be^-(0)

                                                   -20 = 2A – B      ~~~ Eqn (2)

Eqn (1) + Eqn (2):                       3A = 30,          hence, A = 10 (Shown)

Use Eqn (1):                               50 = 10 + B,    hence, B = 40 (Answer)

(ii)           Since,                 A = 10 and b = 40

therefore:            dy/dx = 2Ae^2x – Be^-x = 20e^2x – 40e^-x

At stationary pt, dy/dx = 0

therefore:            0 = 20e^2x – 40e^-x

x (e^x):                  0 = 20e^2x(e^x) – 40e^-x(e^x)

                           0 = 20e^3x – 40e^o

                           0 = 20e^3x – 40

                           e^3x = 40/20 = 2

ln both sides:       3x = ln 2.         Therefore,        x = (1/3)(ln 2) ~~ exact value

                                                                           x = 0.231 (3 sf) ~~ (Answer)

Put x = (1/3)(ln 2) into:   y = 10e^2x + 40e^-x

therefore,                        y = 10e^{2(1/3)(ln 2)} + 40e^{-(1/3)(ln 2)}

                                       y = 10e^{(ln 2^(2/3))} + 40e ^{(ln 2^(-1/3))}

[Use ln e^x = x:                 y = 10(2^(2/3)) + 40(2^(-1/3))]

All in terms of 2^(2/3):        y = 10(2^(2/3)) + 40(2^(-1/3))(2/2)

                                       y = 10(2^(2/3)) + [40(2^(-1/3))(2)]÷2

                                       y = 10(2^(2/3)) + [40(2^(1-(1/3)))] ÷2

                                       y = 10(2^(2/3)) + 20(2^(2/3)) = 30(2^(2/3))      ~~ exact value

                                       y = 47.6 (to 3 sf)               ~~ (Answer)                                       

(iii)          d²y/dx² = 40e^2x + 40e^-x = +ve value for all x ≥ 0

So, stationary pt is minimum.                                ~~ (Answer)

Please drop your comments if you have any view to express on the above. Have fun!

(Posted on 18/06/2014 from PJ SS2 Starbucks)