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The Big 2009 AEA Maths Thread! :)

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rbnphlp
Show sin^4x+cos^4x = 1-1/2sin^2(2x)

It took me an absolute eternity to spot it


It is sometimes useful to start by rearanging first;

i.e. Sin4(x)+Sin2(2x)2=1−Cos4(x) \displaystyle Sin^4(x) + \frac{Sin^2(2x)}{2} = 1 - Cos^4(x) because whenever you see trig powers that are even and a lone 1 you need to think difference of 2 sqaures.

Sin4(x)+Sin2(2x)2=(1−Cos2(x))(1+Cos2(x))=Sin2(X)(1+Cos2(x)) \displaystyle Sin^4(x) + \frac{Sin^2(2x)}{2} = (1-Cos^2(x))(1+Cos^2(x)) = Sin^2(X)(1+Cos^2(x))

Expanding Sin(2x) as 2Sin(x)Cos(x) the identity is apparent [as we are left with 2Cos2(x)−1=Cos2(x)−Sin2(x) \displaystyle 2Cos^2(x) - 1 = Cos^2(x) - Sin^2(x)

You can now start with cos(2x)≡cos(2x) \displaystyle cos(2x) \equiv cos(2x)

<-> 2Cos2(x)−1≡Cos2(x)−Sin2(x) \displaystyle2Cos^2(x) - 1 \equiv Cos^2(x) - Sin^2(x) [note use of the <-> sign which means implied and implied by]

Multiplying by Sin^2(x)

<-> 2Cos2(x)Sin2(x)−Sin2(x)≡Cos2(x)Sin2(x)−Sin4(x) \displaystyle2Cos^2(x)Sin^2(x) - Sin^2(x) \equiv Cos^2(x)Sin^2(x) - Sin^4(x)

<->
Unparseable latex formula:

\displaystyleSin^2(x)[1+cos^2(x)] \equiv Sin^4(x) + 2Cos^2(x)Sin^2(x)



<-> Sin4(x)+Sin2(2x)2≡1−Cos4(x) \displaystyle Sin^4(x) + \frac{Sin^2(2x)}{2} \equiv 1 - Cos^4 (x)

As required
Reply 61
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rbnphlp
11
How about :
Sin^4x+cos^4x=(sin^2x+cos^2x)^2-2sin^2xcos^2x. (sin^2+cos^2x=1)
i.e 1-2sin^x cos^2x
= 1-1/2sin^2(2x)

It took me ages to see that (sin^4x+cos^x) can be written as (sin^2x+cos^2x)^2-2sin^2xcos^2x
Part (i)

So if the roots of ax3+bx2+cx+d=0ax^3 + bx^2 + cx + d = 0 are α,β\alpha, \beta and γ\gamma

then x3+bax2+cax+da≡(x−α)(x−β)(x−γ) x^3 + \frac{b}{a}x^2 + \frac{c}{a}x + \frac{d}{a} \equiv (x-\alpha)(x-\beta)(x-\gamma)

Expanding the LHS we get:

1⏟ax3−(α+β+γ)⏟bx2+(βγ+γα+βα)⏟cx+−(αβγ)⏟d\underbrace{1}_{a}x^3 \underbrace{-(\alpha + \beta + \gamma)}_{b}x^2 + \underbrace{( \beta \gamma + \gamma \alpha + \beta \alpha)}_{c}x + \underbrace{-(\alpha \beta \gamma)}_{d}

Therefore

−ba=−−(α+β+γ)1⇒α+β+γ\boxed{-\frac{b}{a}= - \frac{-(\alpha + \beta + \gamma)}{1}\Rightarrow \alpha + \beta + \gamma}

and

−da=−−(αβγ)1⇒αβγ \boxed{ -\frac{d}{a}= - \frac{-(\alpha \beta \gamma)}{1} \Rightarrow \alpha \beta \gamma}



32x3−14x+3≡(x−α)(x−2α)(x−β)32x^3 -14x + 3 \equiv (x-\alpha)(x-2\alpha)(x-\beta)

Using the proof from above:

3α+β=0 3\alpha + \beta =0

and 2α2β=−3322\alpha^2 \beta = -\frac{3}{32}

Solving this α=14\alpha=\frac{1}{4} and β=−34\beta = -\frac{3}{4}

Therefore the roots are 14,12,−34\boxed{\frac{1}{4}, \frac{1}{2}, -\frac{3}{4}}
rbnphlp
How about :
Sin^4x+cos^4x=(sin^2x+cos^2x)^2-2sin^2xcos^2x. (sin^2+cos^2x=1)
i.e 1-2sin^x cos^2x
= 1-1/2sin^2(2x)

It took me ages to see that (sin^4x+cos^x) can be written as (sin^2x+cos^2x)^2-2sin^2xcos^2x

this is my fav method. i love these threads :biggrin:
ohh... could you explain me why?
Part (ii)

I dnt know whether this is the best method...

I used the long devision for each of the ones to find the remainder ...

For (x−1)(x-1) i got :a+b+b=1: a + b + b= 1

(x+1):a−b+c=25(x+1): a - b + c= 25

(x−2):4a+2b+c=1(x-2): 4a + 2b + c= 1

I just solved these ones simultaneously and calculated

a=4,b=−12,c=9\boxed{a=4, b=-12, c=9}

so f(x) becomes

f(x)=4x2−12x+9f(x)= 4x^2 -12x +9

completing the square we get
⇒f(x)=(x−32)2\Rightarrow \boxed{f(x)=\left (x-\frac{3}{2}\right )^2}
thanks ... ! :smile:
You don't need to use long division. Just evaluate when x=1,-1 and 2.
DFranklin
You don't need to use long division. Just evaluate when x=1,-1 and 2.

:o: ohh of course...
Reply 69
Avatar for NesQuiK.
NesQuiK.
13
Woah you guys are already doing AEA papers!
I have way too many exams coming up to even consider starting on the papers yet!
Reply 70
Avatar for rbnphlp
rbnphlp
11
yh im in the situation 13 exms to be precise,but try to do aea qs wenevr u want a break:p:
Reply 71
Avatar for NesQuiK.
NesQuiK.
13
rbnphlp
yh im in the situation 13 exms to be precise,but try to do aea qs wenevr u want a break:p:

18 exams here, no time for breaks lol!
Reply 72
Avatar for NesQuiK.
NesQuiK.
13
Woah, i've 'just' got 8 maths exams. Atleast your doing them alongside your AS's and not alongside 5 other A2s like me! :work:
Reply 73
Avatar for NesQuiK.
NesQuiK.
13
You too, thats quite a lot of maths to be doing all at once!
Which modules are you sitting?
Reply 74
Avatar for rbnphlp
rbnphlp
11
The function f is given by:
f(x) =1/ t(x^2 - 4)(x^2 – 25),
where x is real and t is a positive integer.

(a) Sketch the graph of y = f(x) showing clearly where the graph crosses the coordinate axes.

(b) Find, in terms of t, the range of f.

(c) Find the sets of positive integers k and t such that the equation
k = |f(x)|
has exactly k distinct real roots.

I culd do the first two bits , but got stuck on the last bit ,any ideas???
shuld i consider the graph or try to solve it by by simltaneous eqs??
rbnphlp
The function f is given by:
f(x) =1/ t(x^2 - 4)(x^2 &#8211; 25),
where x is real and t is a positive integer.

(a) Sketch the graph of y = f(x) showing clearly where the graph crosses the coordinate axes.

(b) Find, in terms of t, the range of f.

(c) Find the sets of positive integers k and t such that the equation
k = |f(x)|
has exactly k distinct real roots.

I culd do the first two bits , but got stuck on the last bit ,any ideas???
shuld i consider the graph or try to solve it by by simltaneous eqs??


Unparseable latex formula:

\frac{1}{t}\times (x^2 - 4)(x^2 &#8211;25)

or

Unparseable latex formula:

\frac{1}{t\times(x^2 - 4)(x^2 &#8211; 25)}

?
ohhkay



i dnt know ... sometimes i just dnt get how this thing works -.-
Reply 77
Avatar for rbnphlp
rbnphlp
11
Yes it is a 2003 ppaer, yes I think 5 intersections with the graph
question 7 on that paper was a good one ... i started to struggle with part (e) ... but eventually i got to the answer ^^
rbnphlp
Yes it is a 2003 ppaer, yes I think 5 intersections with the graph

if you sketch mod(f(x)) you will notice that you will find various numbers of intersection ... think about the link y=k below the turning point and above it

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