O-Level Additional Math

[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

[iFruit]

SKT:

Hi iFruit,


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

TIA

Hi SKT,

In the original question, [2 + √(2 + 2 cos x)] is a +ve number because cos x is +ve.

we need to find the square root of √(a +ve number), which will have a +ve and a -ve root.

Sure, x/4 is in the first quadrant but that is not related to the value of √[2 + √(2 + 2 cos x)] at all. The sign of cos x/2 matters only when taking the square root of √(4cos² x/2), because we need to keep (2 + √(4cos² x/2)) > 2, so we must choose -ve root.

Just for argument's sake, let's say x=300, then [2 + √(2 + 2 cos 300 )] = 2 + √3 =3.732

so √3.732 = ±1.93

Hope this helps.

[Muffins]

woah..... these questions making my head spin already :faint: :faint:

[iFruit]

Muffins:

woah..... these questions making my head spin already :faint: :faint:

Welcome to the real world mate ! We have no Mohammads and Alis exchanging marbles or silly old mothers trying to pick up their daughters from the schools at constant speed every day here.

We just have beautiful x's and y's and before you blink sin As, cos Bs and Sec Cs.

[SKT]

Hi,


Find the equations of the tangents from (2, -3) to the curve y = x + x².

TIA

[iFruit]

SKT:

Hi,


Find the equations of the tangents from (2, -3) to the curve y = x + x².

TIA

let's say the point at which the line and curve meet is (x, y) = (x, x+x²).

Then, Slope of tangent = (x+x²+3)/(x-2)

but slope of tangent = dy/dx of curve = d( x + x² )/dx = 1+2x

so (x+x²+3)/(x-2) = 1+2x--> x+x²+3 = 2x² -3x -2-->x²-4x-5 = 0 -> (x+1)(x-5) = 0---> x = -1 or 5

when x =-1, y = x + x² = 0, m = 1+2x = -1
when x = 5, y = 30, m = 11

The lines of the equation

y = -x + C1, y = 11x +C2,

Solving for point ( 2,-3), we get the tangents of curves

y = -x-1,

y = 11x -25


HTH