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Mr.Macdonald
May 27th, 2008, 06:42 PM
This isn't a homework question but a PreCalculus Test Question that I couldn't answer for lack of understanding

Write sin^3(x) in first degree cosine ( cos(?) )

I tried to use this formula but failed

(1-cos(2x))/2=sin^2(x)

fractions are ok and leave anwser in calculator ready

Please no high level calculus answers i am in precalculus and only understand to about calc 1

By the way i want the steps and an explanation not just

cos(13x -2)/2

Can+~
May 27th, 2008, 08:48 PM
http://en.wikipedia.org/wiki/List_of_trigonometric_identities#Power-reduction_formulae

ifross
May 28th, 2008, 12:19 AM
Not sure that answered the question as he asked it in terms of cosine... Seems quite hard, are you sure that that is the question?

With my limited maths knowledge, I think that it would be tough getting it in terms of just first order cosines (short of saying cos(x) = sin(x+(pi/2))

CptPicard
May 28th, 2008, 12:31 AM
There is a formula for sin^3 in terms of sines as I think was on the Wikipedia page... but frankly, I have been messing with it for an hour now and can't coerce it into cosines... I would be prone to agree with the previous poster and suggest that you may have misunderstood the question?

ConMan318
May 28th, 2008, 01:06 AM
It seems like the answer would be

sin^3(x) = (1 - cos^2(x))^(3/2)

I don't know how it would end up being what you wrote it was, unless I read your original post wrong; is what you wrote there: "cos(13x -2)/2", the correct answer?

I've never heard of needing something to be "first-degree" cosine, does that mean that there can't be any squares/exponents besides one?

geirha
May 28th, 2008, 02:38 AM
It seems like the answer would be

sin^3(x) = (1 - cos^2(x))^(3/2)

That does not seem to be right. Try drawing the graphs with gnuplot (apt:gnuplot):

\$ gnuplot
gnuplot> plot sin(x)**3, (1-cos(x)**2)**(3/2)

You'll see they don't match.

I've never heard of needing something to be "first-degree" cosine, does that mean that there can't be any squares/exponents besides one?
That is what it means, yes.

Can+~ has allready posted the formula which will give you the answer in first degree sine:

sin^3(x)= (3sin(x)-sin(3x))/4

and converting between sine and cosine is simple:

cos(x) = sin(pi/2-x)
and
sin(x) = cos(pi/2-x)

So the answer is ...

OpenFish
May 28th, 2008, 02:33 PM
last time i checked

sin**3(x)
sin**2(x) * sin (x)
(1 - cos**2(x)) * sin(x)
(1 - cos**2(x)) * cos(x - 90)
cos(x - 90) - cos**2(x)cos(x - 90)

every line should be the same
and the last two lines both express sin**3(x) in turmes of cos(x)

is this what ur looking for ?? calc says they are the same any way

there will be a way of simplifing it down a lot but i cant remember

shifty2
May 28th, 2008, 04:13 PM
Just curious, what level is precalculus? ie age?

It seems like quite a lame question tbh, not a particularly nice answer. Did they want the answer in terms of cos(x) rather than having cos**y(x)? Don't know what level you are, but using de moivre would get it into single powers of cos and sine and from there you could just use a lot of cos(x-90)=sin(x) to get it into single powers of cosine.

OpenFish
May 28th, 2008, 05:15 PM
cos(x - 90) - cos**2(x)cos(x - 90)

well im sorry

cos(x - 90) - cos(x)*cos(x)*cos(x - 90)

or

cos(x - 90)*cos(x - 90)*cos(x - 90)

and

cos(A+-B) = cosAcosB +- sinAsinB

i love trig it just gose round and round in ever more complicated circles

by the way have u seen the matrix opengl uses for giving orientations it dosen't have any identity's in it but it keeps us on are toes

yes i do think there is a simple er way to show it but i havnt needed it so dont know it
but the curve that sin**3(x) isn't a linia transformation of sin(x) so it cant just be maded from cos(x*A)*B (i think!)

OpenFish
May 28th, 2008, 05:25 PM
sorry i hadn't read all of ur post

(3*cos(x-90)-cos(3*x - 90))/4

and yes yes this is better but the point of this post is to prove how u get hear and i cant explain how u get to this!!

although it is quite pretty

see what i mean about circles!!

Mr.Macdonald
May 28th, 2008, 06:25 PM
The age levels of this class is Junior in High School (15 - 17) but this teacher makes the class about college level.

The indented formula on my first post is psuedo and just an example.

Maybe I did misunderstand it because that test was brain rattling and I may have misread it. Wikipedia has power reduction formulae but none for that question. The Teacher also didn't teach us the formulae after 2nd power. Ill ask his and release the solution privately (by email at my discretion) as to avoid academic dishonesty. All i hope is that their was a typo and i will get credit.

galileon
May 28th, 2008, 06:32 PM
sine cubed x is also sine squared x times sine x

sine squared x is half ( 1 - cosine x )

sine x itself is cosine of ninety degrees minus theta

does that make sense?

This isn't a homework question but a PreCalculus Test Question that I couldn't answer for lack of understanding

Write sin^3(x) in first degree cosine ( cos(?) )

I tried to use this formula but failed

(1-cos(2x))/2=sin^2(x)

fractions are ok and leave anwser in calculator ready

Please no high level calculus answers i am in precalculus and only understand to about calc 1

By the way i want the steps and an explanation not just

cos(13x -2)/2

unutbu
May 28th, 2008, 08:25 PM
As has been previously noted, sin(x) = cos(x- pi/2). So if we can get sin^3(x) written in terms of first degree sines and cosines, then we are home free.

Important formulas for trig reduction are

sin(A+B) = sin(A)cos(B) + sin(B)cos(A).
cos(A+B) = cos(A)cos(B) - sin(A)sin(B).

Two particular cases of this are

sin(2x)=sin(x)cos(x)+sin(x)cos(x)=2sin(x)cos(x)
cos(2x)=cos(x+x)=cos^2(x)-sin^2(x)=(1-sin^2(x))-sin^2(x)=1-2sin^2(x).

Notice from the main trig reduction formulas there seems to be a way to "translate" from things like sin(3x) to products of sines and cosines ... things like sin^3(x). We're interested in sin^3(x), and we are interested in getting it down to something like sin(3x), so there is reason to pursue this line of thinking.

sin(3x)=sin(x+2x)=sin(x)cos(2x) + sin(2x)cos(x)
= sin(x)*(1-2sin^2x) + 2sin(x)cos(x)cos(x)
= sin(x)*(1-2sin^2x) + 2sin(x)cos^2(x)
= sin(x)*(1-2sin^2x) + 2sin(x)(1-sin^2(x))
= sin(x)*(1-2sin^2x) + -2sin^3(x)+2sin(x)
= sin(x)-2sin^3(x) + -2sin^3(x)+2sin(x)
= -4sin^3(x) + 3sin(x)

So

4 sin^3(x) = 3sin(x)-sin(3x)
sin^3(x) = (3sin(x)-sin(3x))/4 = (3cos(x-pi/2) - cos(3x-pi/2))/4

% gnuplot
gnuplot> plot sin(x)**3, (3*cos(x-pi/2)-cos(3*x-pi/2))/4