SPM 2013 Paper 2 Q 4 on Trigonometry:

Q 4(a) involves using identities (including addition or double-angle formulae) to prove a given equation; while

Q 4(b) involves solving the proven trigonometric equation based on sound understanding of Reference Angle, Principle Angles and their Co-terminal Angles (for Principle Angles) up to 720 degrees. 

Remarks: Even if a candidate can't prove (the equation given in) part (a), he or she should proceed to solve part (b) as if the given equation is proven: Part (a) carries 2% marks while Part (b) carries 4% marks.

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SPM 2013 Paper2 (3472/2) Q4:

4.         (a) Prove that tan x sin 2x = 1 – cos 2x             (2)

            (b) Hence, solve the equation

tan x sin 2x = ¼, for 0^o ≤ x ≤ 360^o                    (4)

My Proposed Solutions

4          (a) To prove: tan x sin 2x = 1 – cos 2x

                        LHS = tan x sin 2x

                        Use identities (for 'double substitutions'):

                        tan x = sin x / cos x; and

                        sin 2x = 2 sin x cos x

                        LHS = (sin x / cos x)( 2 sin x cos x) = 2 sin² x

                        Use identities:

cos 2x = 1 - 2 sin² x; or,

2 sin²x = 1 – cos 2x

Therefore,

LHS = 2 sin²x = 1 – cos 2x = RHS (Proven)

            (b) Hence, tan x sin 2x = ¼, for 0^o ≤ x ≤ 360^o

                        Imply: 1 – cos 2x = ¼, for 0^o ≤ 2x ≤ 720^o

                                    cos 2x = 1 – ¼ = ¾ (+ve)

                       

Thus, for domain of 2x: 0^o ≤ 2x ≤ 720^o, 2x^o must be in

·        1^st Quadrant:

P₁ = 2x^o = Reference Angle (R) = cos^-1 (3/4)

x^o = (1/2)[cos^-1 (3/4)] = 20.70481106…^o

x = 20.7 (to 1 dp); or

(1^st Qd 1st co-terminal angle)

2x = 360(1) + P₁ = 360 + cos^-1 (3/4)

x = (1/2)[360 + cos^-1 (3/4)] =  200.7048111…

   = 200.7 (to 1 dp)

·        4^th Quadrant:

P₂ = 2x^o = 360^o – R = 360^o – cos^-1 (3/4)

x^o = (1/2)[ 360^o  - cos^-1 (3/4)] = 159.2951889…^o

x = 159.3 (to 1 dp); or

(4^th Qd 1st co-terminal angle)

2x^o = 360^o + P₂ = 360^o + [360^o – cos^-1 (3/4)]

x^o = (1/2)[720^o – cos^-1 (3/4)]

x = 339.2951889

x = 339.3 (to 1 dp)

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Questions on Logarithmic Functions and Indices

1.            Solve the following equations:

a)      i) 3^2x = 1000             (Cambridge IGCSE A-Maths 0606/23, May/June 2012 P2 Q5)

      ii) 36^2y-5/6^3y = 6^2y-1/216^y+6 

b)      4^x / 2^5-x = 2^4x / 8^x-3   (IGCSE Q)

c)      3^x+1 + 3^2-x = 28    (IGCSE Q)

d)      2^3x = 8 + 2^{3x – 1}   (SPM 2011 P1 Q7 @ pg 162)

e)      3^{x + 2} – 3^x = 8/9    (SPM 2010 P1 Q7 @ pg 136)

f)        16^2x-3 = 8^4x            (SPM 2008 P1 Q7 @ pg 83)

g)      9(3^{n – 1}) = 27ⁿ      (SPM 2007 P1 Q8 @ pg 58)

h)      3^{n – 3} x 27ⁿ =243   (SPM 2009 P1 Q7 @ pg 110)

i)        2^{x + 4} – 2^{x + 3} = 1   (SPM 2005 P1 Q7 @ pg 5)

j)        8^2x-3 = 1/√(4^{x + 2})  (SPM 2006 P1 Q6 @ pg 31)

k)      27(3^{2x + 4}) = 1     (SPM 2012 P1 Q7 @ pg 187)

2.      (i)  Write 125 = 5³ as a logarithmic equation

(ii) Express log_b32 = 5 in index form

 (iii) Evaluate:

(a)      b^log_b^x 

(b)     10^lg10

(c)      e^{ln x} for x = 3                                   

3.      For each of the following equations, sketch on separate axes its graph, showing clearly where the graph crosses the axis:

a)      y = e^x                                                                           (2)

b)      y = log₃ x                                                                      (2)

c)      y = 2^-x                                                                          (2)

d)      y = log₃ (-x)                                                                  (2)

                                      

4.      Solve the equations:

a)      log_x 128 = 7                                                                 (2)

b)      log₅ (7x -1) = 3                                                            (3)

c)      log₄ t = 6 log_t 16 -1                                                      (5)

5.      Solve

(a)   log_q 343 = 3                                                                             (2)

(b)   log₄ (5n + 9) = 3                                                                       (3)

(c)   log_m 4 + 8 log₄ m = 6                                                                (6)

(d)   2 log₃ x – 3x log₃ x + 6x = 4                                                     (5)

6.      Solve

(a) log_x 125 = 3                                                                             (2)

(b) log₄ (9y + 4) = 4                                                                       (3)

(c) 3 – log_p p = log_p 9                                                                     (6)

7.      Given that f(x) = log₅ 3 + log₅ 6 + log₅ 9 + log₅ 12 + log₅ 15

a)      Show that f(x) = 6 log₅ 3 + 3 log₅ 2 + 1                        (3)

b)      Solve f(x) = 1 + log₅ x + log₅ x²                                    (3)

8.      a) Given that log₈ 7 = m log₂ 7, find the value of m.                          (2)

      b) f(x) = 16x log₉ 10 – 12 log₉ 10 + 4x log₃ x – 3 log₃ x

i)        Factorise f(x) completely                                                     (3)

ii)   Hence, solve f(x) = 0                                                          (3)

9.       Solve the simultaneous equations:                                                  

(a)    2 log₃ x + 3 log₅ y = 7

                                 log₃ x – log₅ y = 1                                                (4)

(b)   (Given that p ≠ q):  log_p q + 3 log_q p = 4

                                          pq = 81                        (5)

(c)    3 log₂ x + 4 log₃ y = 10

   log₂ x – log₃ y = 1                                                (6)

10.  Solve:         (a) log_q 5 + 6 log₅ q = 5                                                (4)

                  (b) log₃ (5x + 12) + log₃ x = 2                                      (5)

11.  Given that log₉ 10 = k log₃ 10,

(a)   find the value of k.                                                                     (2)

      (b) Factorise completely

                                    4x log₃ x – 3 log₃ x + 16x log₉ 10 – 12 log₉ 10  (2)

(c)   Hence, solve the equation                                                          (5)

                        4x log₃ x – 3 log₃ x + 16x log₉ 10 – 12 log₉ 10 = 0       

                       

12.    Solve

(a)    log_p 343 = 3                                                                (2)

(b)   log₆(11q – 4) = 3                                                                     (2)

(c)    9 log_r 3 = log₃ r                                                                        (3)

(d)   Show that                                                                                (3)

           log₅ 3 + log₅ 6 + log₅ 9 + log₅ 12 + log₅ +15 = 6 log₅ 3 + 3 log₅ 2 + 1

(e)    Solve the equation                                                                   (3)

           log₅ 3 + log₅ 6 + log₅ 9 + log₅ 12 + log₅ +15 = 1 + log₅ x + log₅ x²

13.    (a) Solve the equations

(i)                  log_x 343 = 3                                                              (2)

(ii)                log₉(4y – 3) = 2                                                         (2)

(b)   Solve, to 3 significant figures, log_q 5 + 6 log₅ q = 5        (5)

(c)    Show that x log₂ x⁵ – log₂ x² ≡ (5x – 2) log₂ x.               (2)

(d)   Hence solve the equation x log₂ x⁵ – log₂ x² = 20x – 8   (4)

                                                           

14.    (a) Solve the equations log₄ 2 = p                                       (1)

Given that log₂ 3 = k log₄ 3

(b)   find the value of k                                                                   (2)

(d)      Show that                                                                               (4)

     5x log₄ x – 2 log₄ x – 10x log₂ 3 + 4 log₂ 3 = log₄ (x^{5x – 2}/3^{20x – 8})

(e)      Hence solve the equation                                                        (4)

5x log₄ x – 2 log₄ x – 10x log₂ 3 + 4 log₂ 3 = 0

15.    Solve

(a)    log_q 343 = 3                                                    (2)

(b)   log₄(5n + 9) = 3                                               (3)

(c)    log_m 4 + 8 log₄ m = 6                                       (6)

(d)   2 log₃ x - 3x log₃ x + 6x = 4                             (5)

16.  Solve

(a)      log_p 243 = 5                                                               (2)

(b)      log₄ (3q + 4) = 3                                                        (2)

f(x) = 2x log_x 3 – 5 log_x 9 – x + 5

(c)      Find the value of a and the value of b so that

         

          f(x) = (x – 5)(a log_x 3 – b)                                (3)

(d)      Hence solve the equation f(x) = 0                                (3)

17.    (a) Solve the equations

(i)                  log₅ 625 = x                                                              (2)

(ii)                log₃(5y + 3) = 5                                                         (2)

(b)   (i)       Factorise          5x ln x + 3 ln x – 10x – 6                     

(ii)      Hence find the exact solution of the equation

5x ln x + 3 ln x – 10x – 6 = 0                            (5)

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Simultaneous Equations - Edexcel Past-Year Questions

Simultaneous Equations – Past-Year Questions:

1.        Solve the simultaneous equations

3x + y = 1

   x² + 3xy + y² = 5                                     (6)                    …24/03/2004, P1, Q4

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2.        Solve the simultaneous equations

                    2x² + 9xy + 9y² = 0

                    3x + 3y = 2                                          (6)                    …27/05/2004, P1, Q2

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3.        Solve the simultaneous equations

             y = x² – 7x + 10

             x – y + 3 = 0                                              (5)                    …26/01/2005, P1, Q3

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4.        Solve the equations

y – 2x = 5

     2x² + 2xy + y² = 5                                (6)                    …25/05/2005, P1, Q2

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5.        Solve the equations

                        x² + 4x – xy = 10

                 2x – y = 3                                            (6)            …24/05/2006, P1, Q1

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6.        Find the coordinates of the points of intersection of the curve with equation y = 3x² – 4x + 2 and the line with equation 7x + y = 8.        (5)                 …19/01/2007, P1, Q3

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7.        (a)  Find the coordinates of the points where the line with equation y = 3x + 3 intersects the

       curve with equation y = x² + x – 12.                                                                  (5)

(b)   Find the set of values of x for which x² + x – 12 ≥ 3x + 3.                                 (2)

                                                                                                  …17/5/2007, P2, Q4

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8.        Find the coordinates of the points of intersection of the curve with equation y = x² – 8x + 11 and the line with the equation x + y = 5.         (5)         …12/5/2008, P1, Q2

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9.         Solve the equations

x – 2y = 3

     2y² + 2xy + x² = 1                                (6)                       …13/5/2010, P1, Q3

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10.    Solve the equations

    xy = 6

     xy + x + y = 11                                    (6)                          …15/12010, P1, Q4

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11.    Find the coordinates of the points where the line with equation y = 2x - 5 meets the

curve with the equation xy = 12.                 (5)                                …14/5/2009, P2, Q11            

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12.    The grid opposite shows the graph of y = 3x – 4 + 6/x²

The line with equation y = 5x – 4 intersects the curve with equation y = 3x – 4 + 6/x²

(a)    Using algebra, show that the x-coordinate of P satisfies x³ = 3.          (3)

(b)   By drawing a suitable straight line on the grid, obtain an estimate, to 1 decimal place, for the value of 3^1/3                                                                                                                                       
                                                                                                           (13/05/2010, P1, Q1)
                                                                                (+ many more such questions)

                                                                                             

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Trigonometric Equations Involving Addition Formula

Solving Trigonometric Equations Involving Addition Formula

 Q1) Using cos(A + B) ≡ cos A cos B – sin A sin B,

(a)    show that

(i)    sin² Ө = (1/2)(1 – cos 2Ө),

(ii)   cos² Ө = (1/2)(cos 2Ө + 1)                                                            (4)

f(Ө) = 1 + 10 sin² Ө - 16 sin⁴ Ө .

(b)   Show that f(Ө) = 3 cos 2Ө – 2 cos 4Ө.                                               (4)

(c)    Solve the equation

1 + 10 sin² Ө^o - 16 sin⁴ Ө^o + 2 cos 4Ө^o = 0.25, for 0 ≤ Ө ≤ 180

             giving your solutions to 1 decimal place.                                       (4)  

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Possible Solutions:

Given: cos (A + B) ≡ cos A cos B – sin A sin B            …Eqn (1)

a)      To show:

(i)                  sin² Ө = (1/2)(1 – cos 2Ө)                    …Eqn (2)

RHS = (1/2)(1 – cos 2Ө) = (1/2)[1 – cos (Ө + Ө)]

        = (1/2)[1 – (cos² Ө - sin² Ө)] = (1/2)[1 – (cos² Ө) + sin² Ө]

Use identity: cos² Ө = 1 - sin² Ө

RHS = (1/2)[1 – (1 - sin² Ө) + sin² Ө]

        = (1/2)[1 – 1 + sin² Ө + sin² Ө]

        = (1/2)[2 sin² Ө]

        = sin² Ө = LHS

(ii)                cos² Ө = (1/2)(cos 2Ө + 1)                  …Eqn (3)

RHS = (1/2)(cos 2Ө + 1) = (1/2)[cos (Ө + Ө) + 1]

        = (1/2)[(cos² Ө - sin² Ө) + 1] = (1/2)[1 + cos² Ө – (sin² Ө)]

Use identity: sin² Ө = 1 - cos² Ө

RHS = (1/2)[1 + cos² Ө – (1 - cos² Ө)]

        = (1/2)[1 + cos² Ө – 1 + cos² Ө]

        = (1/2)[2 cos² Ө]

        = cos² Ө = LHS

Given: f(Ө) = 1 + 10 sin² Ө – 16 sin⁴ Ө                        …Eqn (4)

b)      To show: f(Ө) = 3 cos 2Ө – 2 cos 4Ө              …Eqn (5)

Fr. (4):             f(Ө) = 1 + 10 sin² Ө – 16 sin⁴ Ө

                        f(Ө) = 1 + 2 sin² Ө (5 – 8 sin² Ө)

Use (2):            f(Ө) = 1 + 2 [(1/2)(1 – cos 2Ө)][5 – 8 ((1/2)(1 – cos 2Ө))

                        f(Ө) = 1 + (1 – cos 2Ө)[5 – 4(1 – cos 2Ө)]

                        f(Ө) = 1 + (1 – cos 2Ө)[5 – 4 + 4 cos 2Ө)]

                        f(Ө) = 1 + (1 – cos 2Ө)(1 + 4 cos 2Ө)

                        f(Ө) = 1 + (1 + 3 cos 2Ө - 4 cos² 2Ө)

Use (3):            f(Ө) = 1 + [1 + 3 cos 2Ө - 4 ((1/2)(cos 4Ө + 1))]

                        f(Ө) = 1 + [1 + 3 cos 2Ө - 2(cos 4Ө + 1)]

                        f(Ө) = 1 + [1 + 3 cos 2Ө - 2cos 4Ө - 2]

                        f(Ө) = 1 + 1 + 3 cos 2Ө - 2cos 4Ө - 2

                        f(Ө) = 3 cos 2Ө – 2 cos 4Ө  (Shown Eqn 5 from Eqn 4)

c)      To solve, for 0 ≤ Ө ≤ 180:

1 + 10 sin² Ө^o - 16 sin⁴ Ө^o + 2 cos 4Ө^o = 0.25            …Eqn (6)

Fr. (5):              2 cos 4Ө = 3 cos 2Ө – f(Ө)                             …Eqn (7)

Put (7) and (4) into (6): (by “double substitution”)

                                   

                                    f(Ө) + (3 cos 2Ө – f(Ө)) = 0.25

                                                            3 cos 2Ө = 0.25

                                                   cos 2Ө = (0.25/3)

Domain for 2Ө: 2 x 0 ≤ 2Ө ≤ 2 x 180 or  0 ≤ 2Ө ≤ 360

Since cos 2Ө is +ve:     2Ө is in 1^st Qd or 4^th Qd

                                    2Ө = cos^-1 (0.25/3)      …1^st Qd

                                     Ө = (1/2) cos^-1 (0.25/3)

                                     Ө = 42.6099…

                                     Ө = 42.6 (1 dp)

                        Or,       2Ө = 360 - cos^-1 (0.25/3)         …4th Qd

                                     Ө = (1/2) [360 - cos^-1 (0.25/3)]

                                     Ө = (1/2) (274.78019..)

                                     Ө = 137.390…

                                     Ө = 137.4 (1 dp)

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