Unsolvable tripology equation?

[Andrax]

We have (cos(x)) ^3 + (sin(x)), ^3 = (1/sqt 2)
I need to solve it in the R so basically been working on it 4 hours straight nothing seems to be, working tried simplify Inc to a lot of forms (cos x + sinx) (1-cosx sinx), and this is as far as I've gotten when I om working I try to get rid of xos and keep sin. Or vice versa but I always fail any help would be appreciated.

 

[Andrax]

apparently, pi/4 is the solution..tried randomly with calculator ,

 

[Ray Vickson]

apparently, pi/4 is the solution..tried randomly with calculator ,

Maple finds three real solutions (apart from their periodic extensions). Here is the input:
f:=sin(x)^3 + cos(x)^3;
sol:=solve(f=1/sqrt(2),x):

Here is some of the output:

sol[1];
Pi/4;

lprint(sol[2]);
arctan((-2*(-1/4*2^(1/2)+1/4*6^(1/2)+1/4*2^(1/2)*(2^(1/2)*6^(1/2))^(1/2))^3+3/4*2^(1/2)-2^(1/2)*(-1/4*2^(1/2)+1/4*6^(1/2)+1/4*2^(1/2)*(2^(1/2)*6^(1/2))^(1/2))^2+1/4*6^(1/2)+1/4*2^(1/2)*(2^(1/2)*6^(1/2))^(1/2))/(-1/4*2^(1/2)+1/4*6^(1/2)+1/4*2^(1/2)*(2^(1/2)*6^(1/2))^(1/2)))

which = -.4106637302

lprint(sol[3]);
arctan((-2*(-1/4*2^(1/2)+1/4*6^(1/2)-1/4*2^(1/2)*(2^(1/2)*6^(1/2))^(1/2))^3+3/4*2^(1/2)-2^(1/2)*(-1/4*2^(1/2)+1/4*6^(1/2)-1/4*2^(1/2)*(2^(1/2)*6^(1/2))^(1/2))^2+1/4*6^(1/2)-1/4*2^(1/2)*(2^(1/2)*6^(1/2))^(1/2))/(-1/4*2^(1/2)+1/4*6^(1/2)-1/4*2^(1/2)*(2^(1/2)*6^(1/2))^(1/2)))+Pi

which = 1.981460057.

Maple finds 5 solutions altogether: the three real ones given above and two other complex roots, which are complex conjugates of one another.

For real x you can let $\sin(x) = y$ and $\cos(x) = \pm \sqrt{1-y^2}$ to get the two functions $g_1(y) = y^3 + (1-y^2)^{3/2}$ and $g_2(y) = y^3 - (1-y^2)^{3/2}$ on $-1 \leq y \leq 1$. By plotting these we see that g1(y) = 1/sqrt(2) has the two roots, while g2(y)=1/sqrt(2) has a single root.

 

[epenguin]

A good example of Polya's last principle - when you have got the right answer the job isn't finished.

1/√2 is a suggestive number. However if on the RHS you didn't have that but another number, how could you do it?

Symmetry!

(Do you mind if I call your variable θ and then use x for something else?).
Draw a pic of a quarter unit circle in Cartesian axes and a line from origin to the circle, then the projections on those axes are cos θ and sin θ but are also x and y of the axes. So the equation is x³+ y³ = 1/√2 .

This equation is unaffected by interchanging x and y. That tells you if you think about it that the thing is symmetric about the x = y line, meaning if θ is a solution so is (π/2 - θ). You have hit on a pair of coincident solutions which fit exactly what I just said; now i see they tell us there are others hopefully the symmetry should enable you to slog them out.

 

[Andrax]

can anyone try expanding the sin^3x+cos^3x i don't think it needs that much of thinking since it's exercise 80 (not that hard) i did like 2 before it and they were correct butg igot stuck here..

 

[Ray Vickson]

can anyone try expanding the sin^3x+cos^3x i don't think it needs that much of thinking since it's exercise 80 (not that hard) i did like 2 before it and they were correct butg igot stuck here..

As I mentioned in my previous post, if we let $\sin(x) = y$ and take $\cos(x) = \sqrt{1-y^2}$ we get the equation $g_1(y) = y^3 + (1-y^2)^{3/2} = 1/\sqrt{2}$. The two roots on $-1 \leq y \leq 1$ are
y_1 = \frac{1}{\sqrt{2}} \text{ and }<br /> y_2 = - \frac{\sqrt{2} \sqrt{ \sqrt{12} } }{4} + \frac{\sqrt{6}}{4} - \frac{\sqrt{2}}{4} (thanks to Maple).

If, instead, we take $\cos(x) = -\sqrt{1-y^2}$ we get the equation $g_2(y) = y^3 - (1-y^2)^{3/2} = 1/\sqrt{2}$. The real root on $-1 \leq y \leq 1$ is
y_3 = \frac{\sqrt{2} \sqrt{ \sqrt{12} } }{4} + \frac{\sqrt{6}}{4} - \frac{\sqrt{2}}{4}<br />

Knowing the three values of y = sin(x) (as well as whether cos(x) > 0 or < 0) allows us to find the three values of x, using inverse trigonometric functions.

 

[Andrax]

As I mentioned in my previous post, if we let $\sin(x) = y$ and take $\cos(x) = \sqrt{1-y^2}$ we get the equation $g_1(y) = y^3 + (1-y^2)^{3/2} = 1/\sqrt{2}$. The two roots on $-1 \leq y \leq 1$ are
y_1 = \frac{1}{\sqrt{2}} \text{ and }<br /> y_2 = - \frac{\sqrt{2} \sqrt{ \sqrt{12} } }{4} + \frac{\sqrt{6}}{4} - \frac{\sqrt{2}}{4} (thanks to Maple).

If, instead, we take $\cos(x) = -\sqrt{1-y^2}$ we get the equation $g_2(y) = y^3 - (1-y^2)^{3/2} = 1/\sqrt{2}$. The real root on $-1 \leq y \leq 1$ is
y_3 = \frac{\sqrt{2} \sqrt{ \sqrt{12} } }{4} + \frac{\sqrt{6}}{4} - \frac{\sqrt{2}}{4}<br />

Knowing the three values of y = sin(x) (as well as whether cos(x) > 0 or < 0) allows us to find the three values of x, using inverse trigonometric functions.

looks legit but don't know if it's the right way of answering , we've never seen an x^3 roots before so i don't think I'm allowed to do that i'll think more about it tonight anyway , thank you :)

 

[haruspex]

(cos x + sinx) (1-cosx sinx) =1/√2

Cos+sin may remind one of the expansion of cos(A-B) etc. In fact, setting B=pi/4 we have cos(A-π/4) = (cos(A)+sin(A))/√2. So, change of variable, y = x-π/4, and for simplicity writing c, s for cos and sin of y:
c(√2)(1-(c-s)(c+s)/2) = 1/√2
c(2-c²+s²)=1
2c³-3c+1=0
By inspection, c=1 is a solution:
(c-1)(2c²+2c-1)=0

 

[Andrax]

Thanks you I read / thought that cos x-pi/4 eq cos x + sinx x I was forgetting the 1/ 2^1/2

 

[epenguin]

Tripology? Shurely that should be tripeology?

But it isn't - it is quite a useful problem I found. I spent more time on it than the OP originally but consider it not wasted.

The thing for students is to realize the naturalness of the process illustrated by harusepex and why you should expect
it to work as it does. (I had worked it out myself by this morning, changing the variable θ to [θ = (ψ + π/4) ] but when I switched PF on it was already there! )

The thing is as I said the diagonal is a line about which there is symmetry. I originally expressed this as the x = y line but it is better to do it all in polar co-ordinates, and then that is the θ = π /4 line.

[Somehow, between the OP's observation and wondering if there was a solution on the symmetry line I very easily got yesterday one of them. Along this line cos θ = sin θ = 1/√2. You very soon verify that is a solution. Because of the symmetry it has to be a double root.

I might have fallen for thinking that was the solution if it hadn't been that the OP and question indicated there were others, as did plotting the LHS expression, which in polar co-ordinates gives a nice heart-shape.

Then, not quite seeing right away how to exploit the symmetry I relapsed into algebra relative comfort zone. Putting cos θ = x , I got an equation in x like Ray's and squaring the square roots expanded it into a sextic.

But we know that x = 1/√2 is a double root, therefore (x - 1/√2)² is a factor. So I laboriously divided the sextic by
(x² - 2x/√2 + 1/2). In the spirit of show your working I had got as far as

2x⁶ - 3x⁴ -2x³/√2 + 3x² - 1/2

= (x - 1/√2)²(2x⁴ + 4x³/√2 - 4x/√2 - 1) .

You should know that whenever there is a relation between two roots it can be used to depress the degree of the equation by 1. In our case there is a relation between pairs of roots so I know I can depress the quartic to a quadratic. Actually I was able to remove a factor and obtain a quartic only because the equation was special and had a known double root - if the RHS had been anything else but 1/√ 2 I would just have had a sextic that I could reduce to a cubic. So, thinking more of the polynomial in terms of Cartesian axes I was planning to implement this foreknowlege by shifting the origin up by π/4, i.e. for x substitute (x' + π/4). If I do it right I know I won't have to calculate all the terms because the coefficients of odd powers have got to be 0. So I will get a quadratic in x'². Both x'² and cos ψ are even functions but the advantage of cos ψ is that relatively easily we have the equation in terms of this even variable and won't have to go round the houses squaring and shifting, which would have been even heavier with a different RHS.

I got (for the real solution) cos ψ = -1/2 + √3/2 giving θ = 1.98146 clearly according with what I can see graphically. The two terms in the cos ψ vaule each correspond to trig values of special angles, but I can't see that their sum has any special significance, perhaps it's just tripeology.]

 

[Andrax]

Tripology? Shurely that should be tripeology?

But it isn't - it is quite a useful problem I found. I spent more time on it than the OP originally but consider it not wasted.

The thing for students is to realize the naturalness of the process illustrated by harusepex and why you should expect
it to work as it does. (I had worked it out myself by this morning, changing the variable θ to [θ = (ψ + π/4) ] but when I switched PF on it was already there! )

The thing is as I said the diagonal is a line about which there is symmetry. I originally expressed this as the x = y line but it is better to do it all in polar co-ordinates, and then that is the θ = π /4 line.

[Somehow, between the OP's observation and wondering if there was a solution on the symmetry line I very easily got yesterday one of them. Along this line cos θ = sin θ = 1/√2. You very soon verify that is a solution. Because of the symmetry it has to be a double root.

I might have fallen for thinking that was the solution if it hadn't been that the OP and question indicated there were others, as did plotting the LHS expression, which in polar co-ordinates gives a nice heart-shape.

Then, not quite seeing right away how to exploit the symmetry I relapsed into algebra relative comfort zone. Putting cos θ = x , I got an equation in x like Ray's and squaring the square roots expanded it into a sextic.

But we know that x = 1/√2 is a double root, therefore (x - 1/√2)² is a factor. So I laboriously divided the sextic by
(x² - 2x/√2 + 1/2). In the spirit of show your working I had got as far as

2x⁶ - 3x⁴ -2x³/√2 + 3x² - 1/2

= (x - 1/√2)²(2x⁴ + 4x³/√2 - 4x/√2 - 1) .

You should know that whenever there is a relation between two roots it can be used to depress the degree of the equation by 1. In our case there is a relation between pairs of roots so I know I can depress the quartic to a quadratic. Actually I was able to remove a factor and obtain a quartic only because the equation was special and had a known double root - if the RHS had been anything else but 1/√ 2 I would just have had a sextic that I could reduce to a cubic. So, thinking more of the polynomial in terms of Cartesian axes I was planning to implement this foreknowlege by shifting the origin up by π/4, i.e. for x substitute (x' + π/4). If I do it right I know I won't have to calculate all the terms because the coefficients of odd powers have got to be 0. So I will get a quadratic in x'². Both x'² and cos ψ are even functions but the advantage of cos ψ is that relatively easily we have the equation in terms of this even variable and won't have to go round the houses squaring and shifting, which would have been even heavier with a different RHS.

I got (for the real solution) cos ψ = -1/2 + √3/2 giving θ = 1.98146 clearly according with what I can see graphically. The two terms in the cos ψ vaule each correspond to trig values of special angles, but I can't see that their sum has any special significance, perhaps it's just tripeology.]

Thank you really good way of thinking

 

[epenguin]

I return to this problem as there were several things about it that bugged me. I have no doubts that the solution shown by haruspex is as simple and efficient as it can get.

However I think there is something to learn from other angles on it, and if not I offer some nice pictures.

1. Andrax asked about the factorisation of

cos^3\theta + sin^3\theta - \frac{1} {\sqrt{2}} (1)

the LHS of our equation in question

cos^3\theta + sin^3\theta - \frac{1} {\sqrt{2}} = 0(2)

I wanted to do it in the spirit of trigonometrical formulae as it seems Andrax tried. You could get lost in this, Andrax did and so did I for a time, so let me give ideas even if it is probably equivalent to Haruspex' algebra.

2.The first factor.
Andrax spotted a factor at the start. Or at least that cos θ = sin θ = π/4 was a solution. That suggests if it does not prove or enable to prove (comments?) that a factor of (1) is

cos\theta - \frac{1} {\sqrt{2}} + sin\theta - \frac{1} {\sqrt{2}} \\= cos\theta + sin\theta - 2\frac{1} {\sqrt{2}}(3)
= cos\theta + sin\theta - {\sqrt{2}}(4)

Without real mathematics and just dividing numerically we see that this is so Fig. 1 - the division takes out the point of double contact but leaves the single ones.
http://i1122.photobucket.com/albums/l536/epenguin/Tripeology Figs/image-3.jpg


2. In polar co-ordinates.

We can also do that in polar co-ordinates, which the problem rather suggests to consider using. I have been imprinted with Cartesian co-ordinates and never been keen on using polar ones - I must be a bit of a square - nor have I had much occasion for them. I do not even remember hearing, nor realized by myself till now that r = cos θ and r = sin θ are equations for circles of diameter 1/2 resting on the horizontal and vertical axes, see Fig.2 http://i1122.photobucket.com/albums/l536/epenguin/Tripeology Figs/image-2.jpg (there is instead something sounding like that that for parametric representation of a circle). Anyway using them is quite thought-provoking as well as pretty and surprising if you have not thought about it before, because it is close to diagrammatic representation of complex numbers.

However graphing f(θ) in polar co-ordinates is not good for showing roots of f. Rather you need to take the constant term of f(θ), in this case -1/√2, to the other side of the equation and graph the function deprived of this term and the circle representing the constant term, in this case r = 1/√2.
Thus in Fig. 3 http://i1122.photobucket.com/albums/l536/epenguin/Tripeology Figs/image-1.jpg the circle of radius 1/√2 is black and
cos^3\theta + sin^3\theta
(5)
is blue. and the green function is the original function divided by the factor, adjusted as just explained, giving a curve that goes through the same two intersection points as (5) gives. (It also gives a line that goes through the other root, not always; you will no doubt explain that. ) But

3. ...how to obtain the second factor mathematically? (Still within the self-imposed constraints of essentially trig. formulae) .

One idea was to express Eq. 1 as differences of two cubes:

cos^3\theta + sin^3\theta - \frac{1} {\sqrt{2}} <br /> <br /> = cos^3\theta - \frac{1} {2\sqrt{2}} + sin^3\theta - \frac{1} {2\sqrt{2}} \
\ = [(cos\theta)^3 - ( \frac{1} {\sqrt{2}}) ^3] + [(sin\theta)^3 - ( \frac{1} {\sqrt{2}}) ^3]
= (cos\theta - \frac{1} {\sqrt{2}})(cos^2\theta + \frac{1} {\sqrt{2}}cos\theta + \frac{1}{2}) + (sin\theta - \frac{1}{\sqrt{2}})(sin^2\theta + \frac{1}{\sqrt{2}}sin\theta + \frac{1}{2})

From this I obtained the first the factorisation much as below.

Instead the following will also do it:


cos^3\theta + sin^3\theta - \frac{1} {\sqrt{2}}
= cos\theta(1 - sin^2\theta) + sin\theta(1 - cos^2\theta) - \frac{1}{\sqrt{2}}
= cos\theta + sin\theta- cos\theta sin\theta(cos\theta + sin\theta) - \frac{1}{\sqrt{2}} (6)
we try to lock everything that looks like our already known factor:
= (cos\theta + sin\theta - \frac{2}{\sqrt{2}}) - cos\theta sin\theta(cos\theta + sin\theta) + \frac{1}{\sqrt{2}}
= (cos\theta + sin\theta - \frac{2}{\sqrt{2}}) - \frac{(cos\theta + sin\theta)^2 - (cos^2\theta + sin^2\theta)}{2}(cos\theta + sin\theta) + \frac{1} {\sqrt{2}}
= (cos\theta + sin\theta - \frac{2}{\sqrt{2}}) - \frac{(cos\theta + sin\theta)^2 - 1}{2}(cos\theta + sin\theta) + \frac{1} {\sqrt{2}}
= (cos\theta + sin\theta - \frac{2}{\sqrt{2}}) - \frac{(cos\theta + sin\theta)^2 - 2}{2}(cos\theta + sin\theta) -\frac{1}{2}(cos \theta + sin\theta) + \frac{1} {\sqrt{2}}
= (cos\theta + sin\theta - \frac{2}{\sqrt{2}}) - \frac{(cos\theta + sin\theta + \sqrt{2})(cos\theta + sin\theta -\sqrt{2})}{2}(cos\theta + sin\theta) -\frac{1}{2}[(cos \theta + sin\theta) - \frac{2} {\sqrt{2}}]
We now see our factor (3) in every term and can remove it and find that another factor is
1 - (cos\theta + sin\theta)(cos\theta + sin\theta + \sqrt{2} ) (7)
So we have an equation in terms of (cos\theta + sin\theta). Which is actually \sqrt{2}(cos\theta - \frac{\pi}{4}) so we soon get haruspex' equation.

This factor is shown in the green curves of Figs.4a,b http://i1122.photobucket.com/albums/l536/epenguin/Tripeology Figs/image.jpg
http://i1122.photobucket.com/albums...ogy Figs/3f41bb11df9b32da23dfe8dfca18af24.jpg looking essentially the same as Figs.1, 3 and we see it has the same zeros as Eq.(2)
We have obtained the factorisation as requested of (1) by a method not better but different from haruspex' and without any squarings, though the methods are no doubt equivalent.

And there is more...

 

[ehild]

It is well known that a³+b³=(a+b)(a²-ab+b²)
(In general, a+b is a factor of aⁿ+bⁿ)

cos³(x)+sin³(x)=(cos(x)+sin(x))(cos²(x)-cos(x)sin(x)+sin²(x))=(cos(x)+sin(x))(1-cos(x)sin(x)).

sin(x)+cos(x)=√2 sin(x+π/4)

So the original equation can be written as
sin(x+π/4)(1-sin(x)cos(x))=1/2

It can be rewritten in terms of sinθ=sin(x+π/4) and easy to solve.

ehild

 

[epenguin]

The variable 2θ

Ah yes, thank you, I thought there must be a quicker path. I now incorporate haruspex' equation in a Fig., see below, which I meant to do before.

My other idea was this: I think a key is symmetry, the function is symmetrical about a diagonal in the x, y Cartesian plane and we get a simple equation by rotating so that the line of symmetry becomes co-ordinate axis. Haruspex' transformation to variable cos(θ - π/4) makes it the horizontal axis. Making the variable instead cos(θ + π/4) would make it the vertical axis and should similarly simplify.

As well as this my other idea was we make the symmetry axis vertical by another transformation - that of doubling the angle. Making our variable 2θ.

Thus we can write eq.(7) above as
\sqrt{2}(cos\theta + sin\theta) = 1 - (cos\theta + sin \theta)^2 \\ = 2cos\theta sin\theta = sin2\theta
Squaring we get
2 (cos\theta + sin\theta)^2 = sin^2\theta and finally
sin^22\theta - 2sin2\theta - 2 = 0(8)

I have used our final equation (7) for the factor to derive this, but perhaps someone could get it from the original equation (2) (cosθsinθ occurs in both derivations inviting this). In Figs.5a http://i1122.photobucket.com/albums...ogy Figs/f2eaa897ab90a5d8342b84e3b95e2c4d.jpg 5b http://i1122.photobucket.com/albums...ogy Figs/914918f82ff1ffc49f75990abd784d48.jpg
the curves described by both haruspex' and eq. (8) are seen to have the zeroes that they should. Eq.(8) is very similar to haruspex’ equation, in fact they are the same if the cos( θ+ π/4) of the one is changed to 1/cos2θ of the other. Must this not mean that cos(θ + π/4) = 1/cos2θ is equivalent to (7) = 0 ?