My Specialist Maths Mega-thread :D

[monkeywantsabanana]

ooh... i sort of understand... thanks TT

[monkeywantsabanana]

Given b = 3i+2j-k and c=2i-j+k, find the unit vector perpendicular to b and c

[nina_rox]

Can someone help me with these circular function problems?

solve these within -pi<x<equal to 2pi :
a) sin(2x)=sin(x)
b) sin^2(x)cos^3(x)=cosx
c) sin^2(x) - 1/2sin(x) -1/2=0

If anyone could help it would be very much appreciated!!

[Planck's constant]

Can someone help me with these circular function problems?

solve these within -pi<x<equal to 2pi :
a) sin(2x)=sin(x)
b) sin^2(x)cos^3(x)=cosx
c) sin^2(x) - 1/2sin(x) -1/2=0

If anyone could help it would be very much appreciated!!

Rule No 1: Never cancel any factors !

Therefore,

(a) sin2x = sinx
=> 2sinxcosx = sinx   ... (remember Rule No 1)
=> 2sinxcosx - sinx = 0
=> sinx(2cosx - 1) = 0
=> (i) sinx = 0 OR (ii) cosx = 1/2

(i) sinx = 0
=> x = -pi, 0, pi, 2pi

(ii) cosx = 1/2
=> x = -pi/3, pi/3, 5pi/3

(i), (ii) => x = -pi, -pi/3, 0, pi/3, pi, 5pi/3, 2pi
------------------------------------------------------------------------

b) sin^2(x)cos^3(x)=cosx
=> sin^2(x)cos^2(x)cos(x) = cos(x)
=> [1/4sin^2(2x)]cos(x) = cos(x)
=> sin^2(2x)cos(x) = 4cos(x)
=> cos(x)[sin^2(2x) - 4] = 0

=> (i) cos(x) = 0 etc
OR sin^2(2x) = 4  ... bzzzzzt! not possible as |sin(anything)| <= 1
--------------------------------------------------------------------

(c) sin^2(x) - 1/2sin(x) -1/2=0
=> 2sin^(x) - sin(x) - 1 = 0

solve this quadratic (u = sinx) and ..

(i) sinx = 1  etc   OR (ii) sinx = -1/2 etc

[nina_rox]

Thank you so much Argonaut!!! That all makes sense now!

[monkeywantsabanana]

Hello, complex numbers question. I've never really liked this...

How do I sketch

There are solutions, however it's not very clear.

I don't understand why one would complete the square then draw a line, on a complex plane,  from

Can someone please explain?

[paulsterio]

This isn't a question where you're solving, this is a graphing question.

Essentially it's asking you to plot the complex number Z, where Z = x^2 + x

[BubbleWrapMan]

The parabola x^2 + x has a minimum value of -1/4. So if you were to sketch all the values it can take on the complex plane, it would only be on the real axis since it only takes on only real numbers (because x is real), and it would be all real numbers greater than or equal to -1/4 since that is its minimum value.

[monkeywantsabanana]

The parabola x^2 + x has a minimum value of -1/4. So if you were to sketch all the values it can take on the complex plane, it would only be on the real axis since it only takes on only real numbers (because x is real), and it would be all real numbers greater than or equal to -1/4 since that is its minimum value.

Thank you.. this sort of makes sense but I'm still not really following, what happens if it was y instead of x?

What are the steps to draw one of these on a complex plane?

[Kaushik]

Z is an element of C, meaning that it has both real and imaginary parts right. But what the equation gave you is a function which is an element of R, meaning the range of (x^2 + x) gives you the x-values which the complex number takes on the complex plane. We know that the parabola can only take values greater than -1/4 (i.e. minimum) and so this gives you a straight line on the complex plane starting at -1/4 and going to infinite, since those are the values that the parabola can take. If it was for y then you would be given an equation then it would say where this is an element of i, meaning there is no real number part just a straight line on the y-axis.

 

[monkeywantsabanana]

Thanks guys!

Can someone help me out with this please?



Thanks in advance.

[Hancock]

Thanks guys!

Can someone help me out with this please?



Thanks in advance.

Let theta = x, just because writing theta each time is annoying.

Let u = tan(x)
du = sec^2(x)dx or dx = du / sec^2(x) or dx = du / (1+u^2) because tan^2(x) + 1 = sec^2(x)

This makes the integral become:

int (1/(1+u)(1+u^2))du

Using partial fractions:

1/((1+u)(1+u^2)) = a/(1+u) + (bu+c)/(1+u^2)

1=a(1+u^2)+(bu+c)(1+u)
1 = a + au^2 + bu + bu^2 + c + cu
1 = u^2(a+b)+u(b+c)+(a+c)

a+b = 0
b+c = 0
a+c = 1

a = 0.5
b = -0.5
c = 0.5

So the integral is now:

int (0.5/(u+1) + 0.5(1-u)/(1+u^2))du

0.5 int 1/(u+1) du + 0.5 int (1-u)/(1+u^2))du

0.5ln(u+1) + 0.5 int 1/(1+u^2) du + 0.5 int -u/(u^2+1)du

0.5ln(u+1) + 0.5arctan(u) - 0.25ln(u^2+1) + C

0.5ln(tan(x)+1) + 0.5arctan(tan(x)) - 0.25ln(tan^2(x) + 1) + C

0.5ln(tan(x)+1) + 0.5x - 0.25ln(sec^2(x)) + C

0.5ln(tan(x)+1) + 0.5x - 0.5ln(1/cos(x)) + C

0.5ln([sin(x)+cos(x)]/cos(x)) + 0.5x + 0.5ln(cos(x)) + C

0.5ln(sin(x)+cos(x)) - 0.5ln(cos(x)) + 0.5x + 0.5ln(cos(x)) + C

0.5x + 0.5ln(sin(x)+ cos(x)) + C

Sorry for no latex, I don't know how to use that.

[brightsky]

1/(1+tanx)
= 1/(1+sinx/cosx)
= cosx/(sinx + cosx)
= 1/2 (2cosx/(sinx + cosx))
= 1/2 ((cosx + cosx + sinx - sinx)/(sinx + cosx))
= 1/2 ((sinx + cosx)/(sinx + cosx) + (cosx - sinx)/(sinx + cosx))
= 1/2(1+(cosx - sinx)/(sinx + cosx))
therefore, we require int [1/2(1+(cosx - sinx)/(sinx + cosx))] dx
= 1/2x + 1/2 int (cosx - sinx)/(sinx + cosx) dx
now sub u = sinx + cosx, du = cosx - sinx
= 1/2x + 1/2 int (1/u) du
= 1/2x + 1/2 ln|u| + c
= 1/2x + 1/2 ln|sinx + cosx| + c

hopefully no errors. there are other, more methodical techniques you can try as well, like weirstrauss substitution, although in this case, it will take forever and be quite messy.

EDIT: ...or you could do what hancock did.

[Hancock]

1/(1+tanx)
= 1/(1+sinx/cosx)
= cosx/(sinx + cosx)
= 1/2 (2cosx/(sinx + cosx))
= 1/2 ((cosx + cosx + sinx - sinx)/(sinx + cosx))
= 1/2 ((sinx + cosx)/(sinx + cosx) + (cosx - sinx)/(sinx + cosx))
= 1/2(1+(cosx - sinx)/(sinx + cosx))
therefore, we require int [1/2(1+(cosx - sinx)/(sinx + cosx))] dx
= 1/2x + 1/2 int (cosx - sinx)/(sinx + cosx) dx
now sub u = sinx + cosx, du = cosx - sinx
= 1/2x + 1/2 int (1/u) du
= 1/2x + 1/2 ln|u| + c
= 1/2x + 1/2 ln|sinx + cosx| + c

hopefully no errors. there are other, more methodical techniques you can try as well, like weirstrauss substitution, although in this case, it will take forever and be quite messy.

EDIT: ...or you could do what hancock did.

Didn't even think of doing that. Nice job 

[monkeywantsabanana]

When showing that 3 vectors are linearly INDEPENDENT, can you prove by contradiction? That is not being able to prove that they're linearly dependent?

Solutions state that If mu +nv+pw=0, only when m=n=p=0, then u, v, w are linearly independent.

I went the other way and said: If u, v, w are linearly dependent, u=mv+nw. After equating i, j, k co-efficients, finding the values and subbing back into the equations, they are not linearly dependent.

Is this a legit way to prove it? I've never seen the way in the solution before.

Thanks in advance.