Very difficult Double angle questions needed. (2 Viewers)

[shaon0]

Hello, I need very difficult trigonometric function questions including proving and solving.
Thanks
.....How can i simplify identities like; cos6x and sin8x etc. really quickly. Could someone teach me the method.

 

[vds700]

Hello, I need very difficult trigonometric function questions including proving and solving.
Thanks
.....How can i simplify identities like; cos6x and sin8x etc. really quickly. Could someone teach me the method.

arent u the guy that people showed how to do expand those using De Moivre's theorem? (4 unit technique).

Say u want to exapnd cos 6x

(cosx + isinx)^6 = cos6x + isin6x (using De Moivre's theorem)

now expand the LHS using the Binomial Theorem, and the real part willl be the expansion of cos6x.

And are these the sort of questions you're looking for?

http://www.counton.org/alevel/pure/purexatriide.htm

 

[shaon0]

arent u the guy that people showed how to do expand those using De Moivre's theorem? (4 unit technique).

Say u want to exapnd cos 6x

(cosx + isinx)^6 = cos6x + isin6x (using De Moivre's theorem)

now expand the LHS using the Binomial Theorem, and the real part willl be the expansion of cos6x.

And are these the sort of questions you're looking for?

http://www.counton.org/alevel/pure/purexatriide.htm

I don't think it was me. Crap, i haven't done binomial theorem.
I learn it...thanks for your help

 

[vds700]

I don't think it was me.

well, I dunno whether u get this method. If you're in year 11, u probably wont have learnt how to do binomial expansions??? (its a HSC topic)

 

[shaon0]

well, I dunno whether u get this method. If you're in year 11, u probably wont have learnt how to do binomial expansions??? (its a HSC topic)

Is there another way without using binomial theorem?

 

[vds700]

no not really, if binomial is in your textbook, have a look its not that difficult. Have a read of this:

http://en.wikipedia.org/wiki/Binomial_theorem

gives an overview

 

[shaon0]

no not really, if binomial is in your textbook, have a look its not that difficult. Have a read of this:

http://en.wikipedia.org/wiki/Binomial_theorem

gives an overview

So is this kind of like pascal's triangle?
Does that mean:
(cosx+i.sinx)^6
= cos^6(x)+i.sinxcos^5(x)-sin^2(x)cos^4(x)+i.sin^3(x)cos^3(x)-sin^4(x)cos^2(x)+i.sin^5(x)cos(x)-sin^6(x)

 

[shaon0]

What does cos(A-B-2@) equal?

 

[lolokay]

cos(A-B-2@)
=cos(A-B)cos2@ + sin(A-B)sin2@
= (cos²@ - sin²@)(cosAcosB + sinAsinB) + 2sin@cos@(sinAcosB - sinBcosA)

I wouldn't imagine there'd be a simpler way to do that one than regular expanding

 

[tommykins]

Hello, I need very difficult trigonometric function questions including proving and solving.
Thanks
.....How can i simplify identities like; cos6x and sin8x etc. really quickly. Could someone teach me the method.

Pascals triangle (you should know this, if not don't even bother trying these q's) and expand (cosx+isinx)^n for cos/sin(nx)

Use pascals triangle and also the fact that i^2 = -1, then just collect real terms for cos(nx) and imaginary (those with i's at the front) for sin(nx).

Don't see why you need to expand such high values in preliminary though.

 

[lolokay]

So is this kind of like pascal's triangle?
Does that mean:
(cosx+i.sinx)^6
= cos^6(x)+i.sinxcos^5(x)-sin^2(x)cos^4(x)+i.sin^3(x)cos^3(x)-sin^4(x)cos^2(x)+i.sin^5(x)cos(x)-sin^6(x)

I think this should be -i.sin^3(x)cos^3(x) +sin^4(x)cos^2(x)
since i³ = -i, and i⁴ = 1

 

[tommykins]

I think this should be -i.sin^3(x)cos^3(x) +sin^4(x)cos^2(x)
since i³ = -i, and i⁴ = 1

Correct.

 

[shaon0]

Parametrics Q:
The points P(2ap,ap^2) and Q(2aq,aq^2) lie on the parabola x^2=4ay.
1) Find the co-ordinates of A, the point of intersection of the tangents to the parabola P and Q.
.......A is an external point where the tangents meet, right? So can't i just say A(x0,y0)?
2) Suppose further that A lies on the line containing the focal chord which is perpendicular to the axis of the parabola
a) show that pq=1
......Does A have to satisfy the equation of a focal chord, so do i just use m1.m2=-1?
b) show that the chord PQ meets the axis of the parabola on the directrix.

 

[shaon0]

I think this should be -i.sin^3(x)cos^3(x) +sin^4(x)cos^2(x)
since i³ = -i, and i⁴ = 1

sorry, i was just making sure it looked ok. (the indexes)

 

[tommykins]

Parametrics Q:
The points P(2ap,ap^2) and Q(2aq,aq^2) lie on the parabola x^2=4ay.
1) Find the co-ordinates of A, the point of intersection of the tangents to the parabola P and Q.

Find eqn of tangents at P and Q and solve simulatenously. You cannot simply say x0 and y0

2) Suppose further that A lies on the line containing the focal chord which is perpendicular to the axis of the parabola
a) show that pq=1
Find eqn of PQ and sub in (a,0) and you will get the relationship that pq = -1
b) show that the chord PQ meets the axis of the parabola on the directrix.

from 1), you have the points of intersection.

manipulate x and sub into y, taking note that pq = -1.

 

[shaon0]

Find eqn of tangents at P and Q and solve simulatenously. You cannot simply say x0 and y0

from 1), you have the points of intersection.

manipulate x and sub into y, taking note that pq = -1.

isn't pq=1?

 

[Aerath]

Why is De Moivre's in Prelim MX1? =\

 

[shaon0]

Why is De Moivre's in Prelim MX1? =\

It isn't.

 

[tommykins]

isn't pq=1?

I remember tangents at focal chords interesect at the directrix at right angles.

 

[shaon0]

I remember tangents at focal chords interesect at the directrix at right angles.

I did the question, and i got the answer pq=-1.
My it is a typo.
Thanks for your help