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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