Help with Maths =] (1 Viewer)

[withyou]

Hi
I was wondering if anyone had solutions to 3U Fitzpatrick Yr11 Chapter 22E questions 31, 33 and 49?

much appreciated!

 

[AnandDNA]

It'll help if you can post the question as not all of us have Fitzpatrick

 

[lyounamu]

Hi
I was wondering if anyone had solutions to 3U Fitzpatrick Yr11 Chapter 22E questions 31, 33 and 49?

much appreciated!

31.

sin3@ + sin@ = 2sin((3@+@)/2)cos(([EMAIL="3@-@)/2"]3@-@)/2[/EMAIL]) = 0
2sin2@cos@ = 0
Therefore, sin2@ = 0 or cos@ = 0
sin2@ = 0 =sin 0
2@ = npi +- (plus and minus) (-1)^n . 0
2@ = npi
@ = npi/2

Or

cos@ = 0 cospi/2
@ = 2npi +- pi/2

33. cos4x+cos2x = 0
2cos((4x+2x)/2)cos((4x-2x)/2) = 0
2cos3xcosx=0
cos3x=0
cos3x = 0 = cospi/2
3x = 2npi +- (plus and minus) pi/2
x = 2npi/3 +- pi/6

or cosx = 0 = cospi/2
x = 2npi +- pi/2

49. sin2x + sin4x = sin3x
2sin3xcosx = sin3x
2sin3xcosx - sin3x
sin3x(2cosx-1) =0
sin3x = 0 or cosx =1/2
3x = npi
x=npi/3

or cosx = 1/2 = pi/3
x = 2npi +- pi/3

I did it nearly 6 months ago so I am not that good but I think I got them right.

By the way, you cannot technically do these questions using the formula from pg 33-34 from Fitzpatrick. These formulas aren't in the syllabus either. So you don't have to know and you won't get these questions. (hardly)

But I heard that it is useful to know them if you want to do 4 Unit maths.

 

[withyou]

thanks for the help

how would you graph 2sin[x-(pi/3)] ?

since the max value is 2 and min is -2, i want to try and find the smallest positive values of x for which they have these values.

this is basically Chpater22C Question 5 i]

 

[Continuum]

thanks for the help

how would you graph 2sin[x-(pi/3)] ?

since the max value is 2 and min is -2, i want to try and find the smallest positive values of x for which they have these values.

this is basically Chpater22C Question 5 i]

Just shift a 2sinx graph 60 degrees to the right.

 

[lyounamu]

thanks for the help

how would you graph 2sin[x-(pi/3)] ?

since the max value is 2 and min is -2, i want to try and find the smallest positive values of x for which they have these values.

this is basically Chpater22C Question 5 i]

Amplitute = 2 as you mentioned
The period = 2pi/n = 2pi/1 = 2pi (one cycle repeats every 2pi)
The maximum is 2 so maximum occurs when 2sin(x-pi/3) = 2
sin(x-pi/3) = 1
so sin(x-pi/3) = sin pi/2
x-pi/3 = pi/2
x = 5pi/6

Minimum is -2 so minimum occurs when 2sin(x-pi/3) = -2
sin(x-pi/3) = -1
sin(x-pi/3) = -3pi/2
x -pi/3 = -3pi/2
x = -7pi/2

A little bit of info here.

 

[withyou]

heys i have another question xD

Find the equation of the locus of a point that moves so that it's equidistant from 4x-3y+2=0 and 3x+4y-7=0

thanks!

 

[lyounamu]

heys i have another question xD

Find the equation of the locus of a point that moves so that it's equidistant from 4x-3y+2=0 and 3x+4y-7=0

thanks!

Using the perpendicular distance formula:

Let the point be (x,y)

So, PD between 4x-3y+2=0 and (x,y): absoltue value of (4x-3y+2)/(square root of (4^2 + (-3)^2)) = absolute value of (4x-3y+2)/5

PD between 3x+4y-7 = 0 and (x, y): absolute value of (3x+4y-7)/(square root of (3^2+4^2)) = absolute value of (3x+4y-7)/5

Since it is equidistant, absolute value of (4x-3y+2)/5 = absolute value of (3x+4y-7)/5

absolute value of (4x-3y+2) = absolute value of (3x+4y-7)
Square both sides:
16x^2 + 9y^2 + 4 - 12xy + 8x - 12xy - 6y + 8x - 6y = 9x^2 + 16y^2 + 49 + 12xy - 21x + 12xy - 28y - 21x - 28y
7x^2 - 7y^2 - 45 -48xy + 58x + 44y = 0

 

[lolokay]

Using the perpendicular distance formula:

Let the point be (x,y)

So, PD between 4x-3y+2=0 and (x,y): absoltue value of (4x-3y+2)/(square root of (4^2 + (-3)^2)) = absolute value of (4x-3y+2)/5

PD between 3x+4y-7 = 0 and (x, y): absolute value of (3x+4y-7)/(square root of (3^2+4^2)) = absolute value of (3x+4y-7)/5

Since it is equidistant, absolute value of (4x-3y+2)/5 = absolute value of (3x+4y-7)/5

absolute value of (4x-3y+2) = absolute value of (3x+4y-7)
Square both sides:
16x^2 + 9y^2 + 4 - 12xy + 8x - 12xy - 6y + 8x - 6y = 9x^2 + 16y^2 + 49 + 12xy - 21x + 12xy - 28y - 21x - 28y
7x^2 - 7y^2 - 45 -48xy + 58x + 44y = 0

I would suggest not squaring both sides. Just solve for (4x-3y+2) = (3x+4y-7)
and (4x-3y+2) = -(3x+4y-7), which gives you x-7y+9=0 and 7x+y-5=0.

 

[lyounamu]

I would suggest not squaring both sides. Just solve for (4x-3y+2) = (3x+4y-7)
and (4x-3y+2) = -(3x+4y-7), which gives you x-7y+9=0 and 7x+y-5=0.

Yeah, that's what I should have done.

I have a habit of squaring everything.