Solutions of Trigonometric Equations

For a Given Interval, Solve the Trigonometric Equations

Solutions of Trigonometric Equations for a Given Interval

Edexcel Syllabus stipulates:

“Students should be able to solve equations such as:

1)    sin (x – π/2) = ¾,  for 0 < x < 2π

2)    cos (x + 30^o) = ½, for -180^o < x < 180^o

3)    tan 2x = 1,  for 90^o < x < 270^o

4)    6 cos² x^o  + sin x^o – 5 = 0, for 0  ≤ x < 360

5)    sin² (x + π/6) = ½,  for –π ≤ x < π.

6)   Past Year Question: Edexcel Pure Maths 7362/02, Q8 (15 May 2008)

Solve, to 3 significant figures, for 0 ≤ Ө ≤ π,

(a)  (4 sin Ө – 1)(2 sin Ө + 5) = 0                                                       (3)

(b)  tan(2Ө – π/3) = 2.4                                                                       (4)

(c)  9 sin² Ө – 9 cos Ө = 11                                                                (5)

7) Solve,   4 sin Ө cos Ө – 10 sin Ө - 2 cos Ө + 5 = 0, for - 90 ≤ Ө ≤ 90 (in degrees)

     

-----------------------------------------------------------------------------------------------------

1)      Solve sin (x – π/2) = ¾,  for 0 < x < 2π

To solve the trigo. equation is to find the value(s) of x for the given interval or domain: 0 < x < 2π

           

            To find x, you have to use the given relationship: sin (x – π/2) = ¾

Based on this relationship:

·           The Reference Angle, R = sin^-1 (3/4)

·           The Principle Angle P (x – π/2) is in

o    The Principle Domain: –π/2  < P < 2π – π/2

o    1^st quadrant (where, P = R): (x – π/2) = sin^-1 (3/4),

o    2^nd quadrant (where P = π – R): (x – π/2) = π - sin^-1 (3/4),

Hence,

1^st quadrant answer:

(x – π/2) = sin^-1 (3/4)

            x = π/2 + sin^-1 (3/4)

            x = 2.418 858 406

            x = 2.42 (to 3 s.f.)

                       

                        2^nd  quadrant answer:

                        (x – π/2) = π - sin^-1 (3/4)

                                    x = π + π/2 - sin^-1 (3/4)

                                    x = 3.864 326 901

                                    x = 3.86 (to 3 s.f.)

2)      Solve cos (x + 30^o) = ½, for -180^o < x < 180^o

Let:      R be the Reference Angle: R = cos^-1 (1/2)

P be (x + 30^o).

Therefore:

Domain for P: -150^o < P < 210^o

(since, -180^o + 30^o < (x + 30^o) or P < 180^o+ 30^o)

P (or, x + 30^o) must be in 1^st Qd or 4^th Qd

1^st Qd: P = R = cos^-1 (1/2) = 60^o

            x + 30^o = 60^o

                        x = 30^o

4^th Qd:             P = -60^o

              x + 30^o = -60^o

                        x = -90^o

6)    Past Year Question: Edexcel Pure Maths 7362/02, Q8 (15 May 2008)

Solve, to 3 significant figures, for 0 ≤ Ө ≤ π,

(a)  (4 sin Ө – 1)(2 sin Ө + 5) = 0                                                       (3)

(b)  tan(2Ө – π/3) = 2.4                                                                       (4)

(c)  9 sin² Ө – 9 cos Ө = 11                                                                (5)

Solutions:

(a)    (4 sin Ө – 1)(2 sin Ө + 5) = 0

(4 sin Ө – 1) = 0;                          or, (2 sin Ө + 5) = 0

sin Ө = ¼                                     or, sin Ө = -5/2

      Ө = sin^-1 (1/4)                        but, sin must be ≤ 1

      Ө = 0.252 680 255                 so, (2 sin Ө + 5) ≠ 0

     

      Ө = 0.253 (to 3 s.f.)

(b)    tan(2Ө – π/3) = 2.4

Reference Angle R = tan^-1 (2.4) 
                              ( = 1.176005207 rad. - but NO need to get. this value early)

Let P be (2Ө – π/3). Therefore,

Domain for P is: – π/3 ≤ P ≤ 2π – π/3

                      or, – π/3 ≤ P ≤ 5π/3

                    

For P’s domain, since tan = +ve value,

P must be in 1^st Quadrant;           or, 3^rd Quadrant:

P = R = tan^-1 (2.4);                       or, P = π + R = π + tan^-1 (2.4)

(2Ө – π/3) = tan^-1 (2.4)                or, (2Ө – π/3) = π + tan^-1 (2.4)

 2Ө = π/3 + tan^-1 (2.4)                 or, 2Ө = 4π/3 + tan^-1 (2.4)

   Ө = [π/3 + tan^-1 (2.4)] ÷ 2        or,  Ө = [4π/3 + tan^-1 (2.4)] ÷ 2                             

   Ө = 1.111 601 379                    or 2.682 397 706

 

  Ө = 1.11 or 2.68 (to 3 s.f.)

(c)       9 sin² Ө – 9 cos Ө = 11

9 (1 – cos² Ө) – 9 cos Ө = 11 ….(using: cos² Ө + sin² Ө = 1)

9 –9 cos² Ө – 9 cos Ө = 11

9 - 11 – 9 cos² Ө – 9 cos Ө = 0

X (-1): 9 cos² Ө + 9 cos Ө + 2 = 0

(3 cos Ө + 1)( 3 cos Ө +2) = 0

(3 cos Ө + 1) = 0                                   or, (3 cos Ө + 2) = 0

cos Ө = - (1/3)                                       or, cos Ө = - (2/3)

Given the domain for Ө:  0 ≤ Ө ≤ π, and cos Ө = -ve values

Ө values must be in the 2^nd Qd.

Hence, Ө = π – cos^-1 (1/3)        or, = π – cos^-1 (2/3)

          Ө = 1.910 633 236           or, = 2.300 523 983

Ө = 1.91            or, 2.30 (to 3 s.f.)

7) Solve,  4 sin Ө cos Ө – 10  sinӨ - 2 cos Ө + 5 = 0, for - 90 ≤ Ө ≤ 90 (in degrees)

            Factorise:         2 cos Ө (2 sin Ө – 1) – 5 (2 sinӨ – 1) = 0

                                    (2 sinӨ – 1)( 2 cosӨ – 5) = 0

            Imply:               (2 sinӨ – 1) = 0            or         (2 cosӨ – 5) = 0

                                    sin Ө = ½                      or         cos Ө = 5/2

                                    Ө = sin^-1(1/2)               but       cos Ө ≤ 1

                                    Ө = 30^o                        so,        (2 cos Ө – 5) ≠ 0

-------------------------------------------------------------------------------------------

Your Homework:

3)    Solve, tan 2x = 1,  for 90^o < x < 270^o

4)    Solve, 6 cos² x^o  + sin x^o – 5 = 0, for 0  ≤ x < 360

5)    Solve, sin² (x + π/6) = ½,  for –π ≤ x < π.   

---------------------------------------------------------------------------------------

SPM 2013 Paper 2 Q 4(b) on Solutions of Trigonometric Equation:

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)

Proposed Solutions

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

                        LHS = tan x sin 2x

                        Use identities:

                        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)

                          ---------------------------------------------------------------------

Trigonometry: Radian Measure

TRIGONOMETRY – EDEXCEL SYLLABUS SAYS:

"Radian measure, including use of arc length and area of sector. The formulae s = rӨ and A = ½ r²Ө for a circle are expected to be known."

---------------------------------------------------------

Radian Measure

1)      What is a radian? How to convert degrees to radian and vice versa?

a.       1 radian subtends an arc length of 1 radius, r, in a sector; or, conversely,
       an arc length of 1 radius in a sector subtends an angle of 1 radian at the centre of the circle.

b.      Since a circle has 2πr as its ‘arc length’ or, circumference,

                                                               i.      a circle of 360^o has (2πr ÷ r) radian or 2π radian;

                                                             ii.      which means 360^o = 2π radian

                                                            iii.      180^o = π radian (2π/2 radian)

                                                           iv.      90^o = π/2 radian (since 2 x 90^o = 180^o)

                                                             v.      60^o = π/3 radian (since 3 x 60^o = 180^o)

                                                           vi.      45^o = π/4 radian (since 4 x 45^o = 180^o)

                                                          vii.      30^o = π/6 radian (since 6 x 30^o = 180^o)

                                                        viii.      270^o = 3π/2 radian (since 3 x 90^o = 180^o)

c.       Generally, to convert x^o to radian, we use:

   

x^o = (x^o/180^o) π radian

d.       And, to convert Ө radian to degrees, we use:

   

Ө radian = (Ө/π) 180^o (since π radian = 180^o)

2)      Arc Length, s = Өr

a.       Arc Length: Since 1 radian subtends an arc length of 1 radius, r, Ө radian therefore subtends Өr arc length. Thus,

Arc Length, s = Өr

b.      Segment perimeter, p = Өr + 2r sin(Ө/2)

                                                              i.      Segment perimeter = arc length + chord length

                                                            ii.      Arc length = Өr

                                                          iii.      Chord length = (2)[r sin(Ө/2)] --- bisect Ө to find half-chord length, then multiply by 2 to get full-chord length

                                                          iv.      Thus, Segment perimeter, p = Өr + 2r sin(Ө/2)

3)      Area of sector, A_sec = (1/2)(Ө)r²

Sector Area

a.       Since Circle Area = πr²;

b.      A Circle has 360^o or 2π radian at its centre.

c.       So, its sector of Ө radian will have a sector area A = (Ө/2π) πr² proportionately. After cancelling down or simplifying, we get:

Sector area, A_sec = (1/2)(Ө)r²

 

                  Segment Area, A_seg = (1/2)r²(Ө - sinӨ)

d.      Segment area = Sector Area – Triangle Area (the triangle formed by the 2 radii and the chord of the sector)

e.       Sector area, A_sec = (1/2)(Ө)r² – please see above

f.        Triangle Area, A_Δ = (1/2)(r)(r)sinӨ – Sine Formula for Area of Δ

g.       Therefore,

Segment Area = Sector Area – Triangle Area

                        = (1/2)(Ө)r² - (1/2)(r)(r)sinӨ

                        = (1/2)r²(Ө – sinӨ)

(Practice Questions coming up soon...Next, the relationship between Reference Angle, Principal Angle and the angle that you want to find...Click and view this video as it will help you in this area.)

--------------------------------------------------------