O-Level Additional Math

[iFruit]

small:

Hi all,


Please help to solve this Sec 1 question and many thanks in advance.

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.

Pi = π

The three vertices (lets say A, B, C) forming the shaded area are the centers of the three circles.

So if you join the three vertices by straight lines, it will be an equilateral triangle of radius 1cm

The area of the segment BC = Area of sector ABC - Area of triangle ABC
= 1/6 (area of circle) - 1/2 (base)(height)

=π/6 -1/2 x 1x √(1-1/4) = π/6 - (√3)/4

So area of the shaded part = 3x Segment BC + Area of triangle

= 3(π/6 - (√3)/4) + (√3)/4 = π/2 -(√3)/2 = 0.5(π-√3)


HTH

[small]

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.