man0005's specialist question thread :)

[xZero]

oh okay, but how would you find the tangent? :/

and would you know how to do this one by any chance?
Let w = cis theta   and z = w + (1/w)
Show that z lies on ellipse with equation (x^2/25) + (y^2/9) = 1/4

you let the line passing through the origin and the graph be y=mx, sub that into S and use discriminant = 0 to solve for m (should get 2 solutions)

[m@tty]

change 1+i and z into polar form so and



well since the new line moved by 45 degrees, then the adjacent side would be which will get you 6, so we have 2 sides with magnitude of 6 and a 45 degree angle between the hypotenuse and the opposite, hence it is a isosceles right-angled triangle

Did you assume a right angle in your explanation?

Nope, the angle between the new line and z is 45 degrees (from the pi/4 + theta part) and work out the magnitude of the line connecting the new line and z, which is 6. Then use these facts to proof it

How did you deduce that the magnitude of the joining line must be 6? It appears to me that you have used trig (for right angled triangles) -- sorry if I'm wrong... just a little perplexed at this time...

[m@tty]

Firstly, proving that it has a right angle   [Yes, should have used polar, but started with this; so I stuck with it]







Switching to vector notation:





(see below)

Because ()



Now using the angle formula for two vectors, a, b.




Now, using cosine rule to determine the other side length (as we haven't shown a right angle yet):







Now, test for pythagorian realationship:





Therefore a^2+b^2=c^2 <=> right angle

And since two sides are equal, you have a right angled isosceles triangle.

[xZero]

Well I probably didn't explain it well enough, what I thought was I used z as the direction and if we extend or shrink that line it will eventually form a right-angled triangle with (1+i)z.

Then I use trig to work out the distance required from the point (1+i)z to the line z if the triangle were to be a right-angled. From this we can use Pythagoras theorem to work out the magnitude of the length in the direction of z, which is 6.

Hence z doesn't need to be shortened or extended to make a isosceles right-angled triangle

Edit: @OP use m@tty's way, it seems better than mine haha

Edit 2:to below, the length from (1+i)z to a line in the direction of z must be 6 to form a right-angled triangle with (1+i)z as hypotenuse. then use pythag, (6root2)^2 = 6^2 + c^2 and solve for c, which will give you 6. and since z has the magnitude of 6, then the line from (1+i)z will meet z and form a right-angled triangle

I know its a bit confusing :S

[m@tty]

Our ways are exactly the same lol, just you use polar which is smarter as it's quicker/easier to compute...

Same basic logic though.

EDIT: Wait, sorry, so how did you show that the length is six, as is required for a right angled triangle?

[VCEMan94]

any idea how to do this one?
Let w = cis theta   and z = w + (1/w)
Show that z lies on ellipse with equation (x^2/25) + (y^2/9) = 1/4

[man0005]

can someone please help me with these
16b) and 17)

[xZero]

16b)



(If a vector dot products with itself, the result will be the square of its magnitude)

now sub into the dot product





since



now sub in a,b and c



as required

17)
, since AB = OC,

, since AE = OD, ,









If half of DB + half of CE = OA, half of DB must share a common point with half of CE, and that point is in the middle of DB and CE, hence the diagonals bisects each other.

as required

as for acute angle, just find CE and DB and use dot product to find the angle

[enpassant]

Here is a simpler proof for Q17.

Let M and N be the midpoints of DB and CE respectively.

DB=OB-OD=OA+OC-OD
OM=OD+DM=OD+1/2 DB=1/2 (OA+OC+OD)

CE=OE-OC=OA+OD-OC
ON=OC+CN=OC+1/2 CE=1/2 (OA+OC+OD)

.: OM=ON

.: M and N are the same point and hence the two diagonals bisect each other.

[man0005]

what about this one ? :S
Prove that the diagonals of a parallelogram bisect each other

sorry im really bad at proofs ><

[enpassant]

Label parallelogram as OABC. Let OA=a, OC=c, and let M and N be the midpoints of OB and AC respectively.

OM=1/2 OB=1/2 (a+c)
ON=1/2 (a+c)

.: OM=ON

.: M and N are the same point and hence the two diagonals bisect each other.

[man0005]

How would i prove the cosine rule for any triangle? :S

[enpassant]

Label triangle as ABC. Let CA=a, CB=b, and AB=c.

c = b - a
c.c = (b - a).(b - a)
c.c = b.b + a.a - 2b.a
c2 = b2 +a2 - 2bacosC

[man0005]

why do you do c.c? what does it represent?

[vea]

c.c=|c|^2 which is the magnitude/length of the triangle SQUARED