Tricky questions

[funkyducky]

is defined by . Find . Sketch on an Argand diagram.

is defined by
Find and such that is a subset of

and:

and has only one solution.
Find the possible values of 

Have fun

Here's another:

If the position vectors of A, B and C are , and respectively, determine whether the lines and intersect each other.

[jane1234]

is defined by . Find . Sketch on an Argand diagram.

Is the answer (x+1)^2 + y^2 = 25 for y E [-5,0)??

EDIT: On second thoughts, I don't think that's it...

EDIT2: Seems to be (x+1)^2 + y^2 = 25 for y E (0,5] instead... I'm probably completely wrong though

[dc302]

is defined by . Find . Sketch on an Argand diagram.

Is the answer (x+1)^2 + y^2 = 25 for y E [-5,0)??

EDIT: On second thoughts, I don't think that's it...

Could be wrong but it might be (x-1)^2 + y^2 = 25?

[syn14]

I also got (x+1)^2 + y^2 = 25. Where does the restriction on y come from?

Yeh nvm it's just the top half.

[dc302]

Oops you guys are right, it is x+1.

[jane1234]

I also got (x+1)^2 + y^2 = 25. Where does the restriction on y come from?

Doesn't work for the bottom values if you substitute. tan^-1(y/x-4) - tan^-1(y/x+6) will give you pi/2, but these are not the proper arguments. I think the proper arguments only work for the top half of the circle...

[Greatness]

  anyone care to explain how to do this? hahaha i hate complex numbers...

you guys are too good!

[dc302]

anyone care to explain how to do this? hahaha i hate complex numbers...

you guys are too good!

Arg of a complex number is simply the inverse tan (arctan) of the real part/imaginary part.

So the equation actually looks like:

arctan( x-4 / y ) - arctan( x+6 / y) = pi/2

Then take the cosine of both sides, the RHS will become 0 and you use the double angle formula for the LHS.

Also, would the hyperbola (x+1)^2 - y^2 = 25 not also work? Do you have the solutions?

edit: uhhhhh how did i get that wrong. it's y/x-4 and y/x+6 

[vea]

Can someone tell me how to simplify something of the form cos(arctanx) or sin(arctanx) ?  x__x

Much thanks!

[funkyducky]

Thanks
I've come across similar before where I've simply tan'd both sides, but obviously that doesn't work here, and I was stuck with a mess with this one.

[dc302]

Can someone tell me how to simplify something of the form cos(arctanx) or sin(arctanx) ?  x__x

Much thanks!

It might help to draw triangles.

If you let arctanx = P, where P is the angle of a triangle, and since tan = O/A, you know that tanP = x/1 so the opposite side has length x, and the adjacent side has length 1. So the hypotenuse has length, sqrt(1+x^2). Which means that sin(arctanx) = sinP = O/H = x/sqrt(x^2+1).

Hope this helps!

[vea]

Can someone tell me how to simplify something of the form cos(arctanx) or sin(arctanx) ?  x__x

Much thanks!

It might help to draw triangles.

If you let arctanx = P, where P is the angle of a triangle, and since tan = O/A, you know that tanP = x/1 so the opposite side has length x, and the adjacent side has length 1. So the hypotenuse has length, sqrt(1+x^2). Which means that sin(arctanx) = sinP = O/H = x/sqrt*x^2+1).

Hope this helps!

Thanks for that, I was wondering if there was an algebraic way though?

[dc302]

Can someone tell me how to simplify something of the form cos(arctanx) or sin(arctanx) ?  x__x

Much thanks!

It might help to draw triangles.

If you let arctanx = P, where P is the angle of a triangle, and since tan = O/A, you know that tanP = x/1 so the opposite side has length x, and the adjacent side has length 1. So the hypotenuse has length, sqrt(1+x^2). Which means that sin(arctanx) = sinP = O/H = x/sqrt*x^2+1).

Hope this helps!

Thanks for that, I was wondering if there was an algebraic way though?

Sure is.

artanx = P

so tanp = x, sinp/cosp = x

sinp = xcosp

sinp ^2 = x^2 cosp ^2

sinp ^2 = x^2 (1-sinp ^2)

rearrange to get sinp = x/ sqrt(1+x^2)


edit: keep in mind that when you take the sqrt, you have to be wary of the signs, which depend on which quadrant your triangle is in.

[vea]

Can someone tell me how to simplify something of the form cos(arctanx) or sin(arctanx) ?  x__x

Much thanks!

It might help to draw triangles.

If you let arctanx = P, where P is the angle of a triangle, and since tan = O/A, you know that tanP = x/1 so the opposite side has length x, and the adjacent side has length 1. So the hypotenuse has length, sqrt(1+x^2). Which means that sin(arctanx) = sinP = O/H = x/sqrt*x^2+1).

Hope this helps!

Thanks for that, I was wondering if there was an algebraic way though?

Sure is.

artanx = P

so tanp = x, sinp/cosp = x

sinp = xcosp

sinp ^2 = x^2 cosp ^2

sinp ^2 = x^2 (1-sinp ^2)

rearrange to get sinp = x/ sqrt(1+x^2)


edit: keep in mind that when you take the sqrt, you have to be wary of the signs, which depend on which quadrant your triangle is in.

nice!

thanks

[funkyducky]

There's more to this question, if you want:

is defined by
Find and such that is a subset of

and:

and has only one solution.
Find the possible values of 

Have fun

EDIT: I typed a mistake, was meant to be find (not )

EDIT 2: EEP another typo! Find b for the second part, not a lol.