most amazing problems

[ququqqu]

Points A and B are specified by the position vectors a and b. Then what's the
equation of the plane bisecting the segment AB perpendicularly ?



Use the relationship e^iB = cosB+isinB to express cos5B in terms of cosB
Hence show that x = COS(Pi/10)
is a root of the equation 16x^4 -20x^2 + 5 = O


Find y(x) if y(Pi)=2pi and the derivative of f(x)-y/x=xcosx

[krtxmrtz]

Points A and B are specified by the position vectors a and b. Then what's the
equation of the plane bisecting the segment AB perpendicularly ?

For now here's my answer to the first question:

R = (a + b) / 2 + mu₃ + nu₃ x (b - a)

where I have represented vectors bold-faced, "R" is the position vector of a point in the plane, "x" means the cross product, u₃ is a unit vector in the direction of the z axis and m and n are real parameters (i.e. by giving values to m and n all the points on the plane are obtained).

[krtxmrtz]

Use the relationship e^iB = cosB+isinB to express cos5B in terms of cosB

e^iB = cosB + isinB
e^i5B = cos5B + isin5B
e^i5B = (e^iB)⁵ = (cosB + isinB)⁵ =

(applying Newton's binomial formula)

= cos⁵B + 5icos⁴BsinB - 10cos³Bsin²B - 10icos²Bsin³B + 5cosBsin⁴B + isin⁵B

Grouping the terms without i:

cos5B = cos⁵B - 10cos³Bsin²B + 5cosBsin⁴B

[krtxmrtz]

Hence show that x = COS(Pi/10)
is a root of the equation 16x^4 -20x^2 + 5 = O

cos5B = cos⁵B - 10cos³Bsin²B + 5cosBsin⁴B =
cosB [cos⁴B - 10cos²Bsin²B + 5sin⁴B] =
cosB [cos⁴B - 10cos²B (1 - cos²B) + 5(1 - cos²B)²] =
cosB (16cos⁴B - 20cos²B +5)

For B=Pi/10:

cos(5*Pi/10) = cos(Pi/2) = 0

Hence, 16cos⁴B - 20cos²B +5 must be 0 for B = Pi/10
or, equivalently, 16x⁴ - 20x² +5 must be 0 for x = cos(Pi/10) =
{Sqrt[5 + Sqrt(5)] / 2]} / 2

[krtxmrtz]

Find y(x) if y(Pi)=2pi and the derivative of f(x)-y/x=xcosx

I'm not sure I understand this. Do you mean:
f'(x) - f(x)/x = xcosx ?
or
[f(x) - f(x)/x]' = xcosx
or...?