ADD. MATHS BLOG

Saturday, 2 November 2013

Solving Simultaneous Equations - Double Substitution Method

The following simultaneous equations may be quickly and easily solved by “double-substitution” method.

The conventional single substitution method is often times too lengthy and very prone to errors. Try your conventional method, before looking at my "double substitution' method and you will know what I mean.

So, always have a quick and closer look at the simultaneous equations before you - if they are amenable to “double-substitution” method, then use it. (Now no longer in Form 2 or Form 3, so use a more "skillful method" where possible. hahaha :))

1. Solve the simultaneous equations

3x + y = 1

x² + 3xy + y² = 5

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2. Solve the simultaneous equations

2x² + 9xy + 9y² = 0

3x + 3y = 2

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The “Double-Substitution” Method:

Q 1) 3x + y = 1 …Eqn (1)

x² + 3xy + y² = 5 …Eqn (2)

Fr. (1): y = 1 – 3x …Eqn (3)

Fr. (2): x² + y(3x + y) = 5 …Eqn (4)

Put (1) & (3) into (4):

x² + (1 – 3x)(1) = 5

x² + 1 – 3x = 5

x² – 3x – 4 = 0

(x + 1)(x – 4) = 0

x = -1 or, x = 4

Use (3): y = 1 – 3(-1) or, y = 1 - 3(4)

y = 4 or, y = -11

Answer: x = -1, y = 4

x = 4, y = -11

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Q2) 2x² + 9xy + 9y² = 0 …Eqn (1)

3x + 3y = 2 …Eqn (2)

Fr. (2): 3y = 2 – 3x …Eqn (3)

Fr. (1): 2x² + 3y(3x + 3y) = 0 …Eqn (4)

Put (2) & (3) into (4):

2x² + (2 -3x)(2) = 0

2x² + 4 – 6x = 0

2(x² + 2 – 3x) = 0

2(x - 1)(x – 2) = 0

x = 1 or, x = 2

Use (3): 3y = 2 – 3(1) or, 3y = 2 – 3(2)

y = -1/3 or, y = -4/3

Answer: x = 1, y = -1/3

x = 2, y = -4/3

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You can also use 'double substitution' to solve part (c) of the question below easily. Try your conventional method first before you click here for my "double-substitution" method.

1) 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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Wednesday, 30 October 2013

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)

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

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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)

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.)

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

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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.

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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^omust 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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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."

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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.)

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Saturday, 5 January 2013

TRIGONOMETRY (Edexcel Syllabus)

Edexcel Syllabus (Area 10):

10. Trigonometry	Notes
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.
The three basic trigonometric ratios of angles of any magnitude (in degrees or radians) and their graphs.	To include the exact values for sine, cosine and tangent of 30^o, 45^o, 60^o (and the radian equivalents), and the use of these to find the trigonometric ratios of related values such as 120^o, 300^o.
Applications to simple problem in two or three dimensions (including angles between a line and a plane and between two planes).
Use of the sine and cosine formulae.	General proofs of the sine and cosine formulae will not be required. The formulae are expected to be known. The area of a triangle in the form ½ ab sin C is expected to be known.
The identity cos² + sin² = 1.	cos² + sin² = 1 is expected to be known.
Use of the identity tan Ө = sin Ө / cos Ө	tan Ө = sin Ө / cos Ө will be provided when needed.
The use of the basic addition formulae of trigonometry.	Formal proofs of basic formulae will not be required. Questions using the formulae for sin (A+ B), cos (A + B), tan (A + B) may be set; the formulae will be provided, for example: sin (A + B) = sinAcosB + cosAsinB Long questions, explicitly involving excessive manipulation, will not be set.
Solution of simple trigonometric equations for a given interval.	Students should be able to solve equations such as: sin (x – π/2) = ¾ for 0 < x < 2π cos (x + 30^o) = ½ for -180^o< x < 180^o tan 2x = 1 for 90^o < x < 270^o 6 cos² x^o + sin x^o– 5 = 0 for 0^o< x < 360^o sin² (x + π/6) = ½ for –π ≤ x < π.