4 unit assessment (1 Viewer)

[JulieClark]

Hey, I've been given an assignment that's weighted 25%, and there are a couple of interesting questions. Help would be most appreciated:

1. Find all x such that sinx=cos5x (I know how to do this, using demoivre's theorem to find an expression for cos5x in terms of cosx but I get stuck with a random sinx.)
2. AB is a fixed chord of a circle. P is a point on the major arc. In triangle APB, BD is the perpendicular bisector of side AP and AC is the perpendicular bisector of side BP (ie C and D are the midpoints of BP and AP respectively).
   • Prove that as P varies, the segment CD has constant length.
   • Find the locus of the midpoint of CD.
3. Let n be an integer greater that or equal to 2.
   • For i=1,2,3...n suppose xi is a real number 0<xi<pi. Use mathematical induction to show that there exists real numbers (a1, a2, a3...an) such that
     • sin(x1 + x2 +...+ xn) = a1sinx1 + a2sinx2 +...+ansinxn
   • Deduce that sinnx < nsinx whenever 0<x<pi

 

[shsshs]

for q1. its not asking to find cos5x in terms of sinx its asking for solutions of x
use the general formula

i.e sinx = cos5x
-------------------1
5x = 2nπ ± cos (sinx)
--------------------2
= 2nπ ± ( 1 - x )

and simplify

 

[shsshs]

i can barely understand what your asking in q.3

but if u let x1 = x2 =...=xn = x

then LHS = sin(nx) = RHS = a1sinx + a2sinx +...+ansinx

= (a1 + a2 +a3... + an)sinx

however that is only smaller than nsinx if a1,a2 etc is smaller than 1


-----

and for your q2

i again dont understand

how can a line perpendicular to a side also bisect it in a random triangle?

unless u are tlaking about the line bisecting the ANGLE not the LENGTH

 

[A l]

A rather sloppy method for the first question: lol
sin x = cos 5x
sin (π/2 - 5x) = sin x
.: π/2 - 5x = x
6x = π/2
.: x = π/12
x = aπ + (-1)ª(π/12) where a is an integer

 

[Raginsheep]

A rather sloppy method for the first question: lol
sin x = cos 5x
sin (π/2 - 5x) = sin x
.: π/2 - 5x = x
6x = π/2
.: x = π/12
x = aπ + (-1)ª(π/12) where a is an integer

Why do you need the last line?

Should we actually be providing answers though considering this thing is an actual assestment?

 

[Trev]

Why do you need the last line?

It states 'all x', it isn't restricted to a domain of x.

 

[Riviet]

Should we actually be providing answers though considering this thing is an actual assestment?

We shouldn't be giving someone the answers and doing their homework for them, I would suggest hints at the most.

 

[Sober]

We shouldn't be giving someone the answers and doing their homework for them, I would suggest hints at the most.

We shouldn't be giving someone the answers and doing their homework for them, I would suggest hints at the most.

To make an assessible homework task for maths is rediculous. What are the rules? Even if one does not plagiarise what is to stop them using more advanced calculators such as Mathematica, which would at least solve the first one correctly with no effort on the student's part.
Has anybody else ever had such a task? If I had to guess I would assume the original poster has made this up in the hope to get it answered faster.

 

[Raginsheep]

In uni you get similar tasks.

While I agree with you that a student may use a program such as Mathematica or Maple to find out the answers, should we be telling them ourselves? Also, Im pretty sure just using a program would not reveal any necessary working out that someone presumably needs for a maths test.

Plagrisim is discouraged on this site. Giving out answers to assesble items should be too.

 

[Raginsheep]

It states 'all x', it isn't restricted to a domain of x.

Ahh.......I see now, that was 'pi'. I thought that it was 'n'...

 

[shsshs]

how do u do the following

 

[Raginsheep]

LHS
= ∫(a, -a) f(x) dx
= ∫(a, 0) f(x) dx + ∫(0, -a) f(x) dx
= ∫(a, 0) f(x) dx + ∫(a, 0) f(-x) dx <---- for the second bit, sub in the values and you should see that they are equal
=∫(a, 0) (f(x) + f(-x)) dx
=RHS

 

[JulieClark]

No you don't understand! That was the point of the assessment.. well, not really but I am allowed to ask for help, as it is "all part of my study". The task itself doesn't really matter, just what I learn from it... btw it can't change my rank as I'm the only student so technically I could hand in anything... but I still want to know how to do these. I thought that only needing help for three wasn't asking for too much... thanks anyway!