O-Level Additional Math

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Hi iFruit,


Many thanks for your help, will pass the solution to my DD.. :celebrate: :celebrate:

[YLH88]

Hi all,


Any good Maths text and assessment books to recommend for Sec 1 ?

Thank you!

[SKT]

Hi,


Find β in terms of α, where α < β, given that α and β are the roots, 0 ≤ x ≤ 360°, of the equation:
|6 sin x - 4| - 8 = 0

TIA.

[iFruit]

SKT:

Hi,


Find β in terms of α, where α < β, given that α and β are the roots, 0 ≤ x ≤ 360°, of the equation:
|6 sin x - 4| - 8 = 0

TIA.

|6 sin x - 4| - 8 = 0 ---> (6 sin x - 4) =8 or -(6 sin x - 4) =8

if (6 sin x - 4) =8,

sin(x) = 2 which is not possible.

if -(6 sin x - 4) = 8,

sin(x) = -2/3----> x = 180°+arcsin(2/3) or 360° - arcsin(2/3)

so if α = 180°+arcsin(2/3) , β = 360° - arcsin(2/3)

β = 360° + 180° -180° - arcsin(2/3) = 540° - α

Is this correct? α and β are absolute values. Not sure why it asks for β in terms of α

[SKT]

Hi iFruit,


Thanks. You’re truely gd!

[SKT]

Hi,


If 270° < x < 360°, simplify √[2 + √(2 + 2 cos x)].

TIA.

[iFruit]

SKT:

Hi,


If 270° < x < 360°, simplify √[2 + √(2 + 2 cos x)].

TIA.

cos 2a = cos a .cos a - sin a. sin a = cos² a - sin² a = cos² a - (1-cos² a) = 2cos² a -1

So,

√[2 + √(2 + 2 cos x)] = √[2 + √(2 + 2 (2cos² x/2 - 1)]

= √[2 + √(4cos² x/2)] = √[2 - 2cos x/2] (because cos x/2 is -ve)

= √[2 - 2 (2cos² x/4 -1)] = √(4 - 4cos² x/4) = 2√(1-cos² x/4) = 2√sin² x/4

= ±2sin x/4



If 270° < x < 360° is important because then cos x is +ve, cos x/2 is -ve and cos x/4 is +ve. So when taking root for (4cos² x/2), we must take value of - 2cos x/2


HTH

[SKT]

iFruit:

SKT:

Hi,


If 270° < x < 360°, simplify √[2 + √(2 + 2 cos x)].

TIA.

cos 2a = cos a .cos a - sin a. sin a = cos² a - sin² a = cos² a - (1-cos² a) = 2cos² a -1

So,

√[2 + √(2 + 2 cos x)] = √[2 + √(2 + 2 (2cos² x/2 - 1)]

= √[2 + √(4cos² x/2)] = √[2 - 2cos x/2] (because cos x/2 is -ve)

= √[2 - 2 (2cos² x/4 -1)] = √(4 - 4cos² x/4) = 2√(1-cos² x/4) = 2√sin² x/4

= 2sin x/4



If 270° < x < 360° is important because then cos x is +ve, cos x/2 is -ve and cos x/4 is +ve. So when taking root for (4cos² x/2), we must take value of - 2cos x/2


HTH

Hi iFruit,
Precise. Thank you. We have a question here, if the interval given was 0 < x < 360°, should there be two solutions 2sin x/4, 2cos x/4?

TIA.

[iFruit]

SKT:

iFruit:

[quote=\"SKT\"]Hi,


If 270° < x < 360°, simplify √[2 + √(2 + 2 cos x)].

TIA.

cos 2a = cos a .cos a - sin a. sin a = cos² a - sin² a = cos² a - (1-cos² a) = 2cos² a -1

So,

√[2 + √(2 + 2 cos x)] = √[2 + √(2 + 2 (2cos² x/2 - 1)]

= √[2 + √(4cos² x/2)] = √[2 - 2cos x/2] (because cos x/2 is -ve)

= √[2 - 2 (2cos² x/4 -1)] = √(4 - 4cos² x/4) = 2√(1-cos² x/4) = 2√sin² x/4

= 2sin x/4



If 270° < x < 360° is important because then cos x is +ve, cos x/2 is -ve and cos x/4 is +ve. So when taking root for (4cos² x/2), we must take value of - 2cos x/2


HTH

Hi iFruit,
Precise. Thank you. We have a question here, if the interval given was 0 < x < 360°, should there be two solutions 2sin x/4, 2cos x/4?

TIA.[/quote]Yes, then both 2sin x/4 and 2cos x/4 are valid solns. In fact, there should be four. ± 2sin x/4 and ±2cos x/4.

The original question should also have ±2sin x/4. I've corrected it.

[SKT]

Hi iFruit,


Refer to the original question, x/4 is at the first quadrant, why -2sin x/4 is valid?

TIA