A creative trig problem

[@RAG981]

As a teacher it would have been nice to be able to edit out my errors and pretend they had not happened, but I always considered that if the class did not spot them, I had failed anyway.

[@andywhelan8608]

This is what I always tell my students, 'if you spot my errors, it means you're learning!"

[@agrajyadav2951]

@@andywhelan8608 my "teachers" say that if you spot their errors, you are insulting them

[@DaveJ6515]

@@agrajyadav2951 "Idiots" is the right definition in that case. Not "teachers".

[@MyOneFiftiethOfADollar]

As a teacher, you make around the same number of mistakes as your students. Teachers are rarely the smartest person in the room, but some carry on like they are. No need to look perfect in front of them. In fact, it's more honest and models the difficulties we all face in teaching and learning more about our craft.

[@shmuelzehavi4940]

@@agrajyadav2951 They are bad teachers.

[@carl13579]

Note that sin(x)+cos(x)=sqrt(2)(sin(x)cos(pi/4)+cos(x)sin(pi/4))=sqrt(2)sin(x+pi/4) which is why it lies between -sqrt(2) and sqrt(2).

[@samosamo4019]

Nice

[@oasisfan3149]

Neat

[@matheusjahnke8643]

yes, it even generalizes: A sin(x) + Bcos(x) =hypot(A,B) (~A sin(x) + ~B cos(x)) = hypot(A,B) [sin (x)cos(theta) + cos(x)sin(theta)] = hypot(A,B) sin(x + theta) Where hypot(A,B)=sqrt(A²+B²) ~A = A / hypot(A,B) ~B = B / hypot(A,B) theta = arcsin(~A) = arccos(~B)

[@michaelempeigne3519]

i like to turn 8x^3 - 24x + 11= 0 into a monic polynomial. Let u = 2x then you get u^3 - 12u + 11 = 0 now you can see that u = 1 by inspection. so you get ( u - 1 ) ( u^2 + u - 11 ) = 0 u = 1 2x = 1 x = 1 / 2 u^2 + u - 11 = 0 u = [ - 1 + or - sqrt ( 1 + 44 ) ] / 2 u = [ - 1 + or - 3 sqrt ( 5 ) ] / 2 so x = [ - 1 + or - 3 sqrt 5 ] / 4

[@prathikkannan3324]

Yes, and then simply note sinx + cosx = sqrt(2)(sinx * sqrt(2)/2 + cosx * sqrt(2)/2) = sqrt(2) * sin(x + 45) which is between -sqrt(2) and sqrt(2) inclusive, and reject all answers except 1/2

[@karolakkolo123]

@@prathikkannan3324 yes, that's a really clever trick I remembered from some stackexchange post I saw years ago, the sqrt(2)*sin(pi/4 + x) one

[@DavidSavinainen]

@@dondisco6399 No, (-1-3√5)/4 is about -1.927 and (-1+3√5)/4 is about 1.427, so neither is above 2 in absolute value.

[@mcwulf25]

@@dondisco6399 Not true because sin x and cos x are not independent. If sin x = 1 then cos x = 0.

[@Ron_Shvartsman]

Very clever trick with u=2x!!

[@stardust-r8z]

The rational root theorem is very useful.

[@leif1075]

Right and Inthink a lot of ppl myself included don't remember it so why not solve it without it? It's not that intuitive.

[@stardust-r8z]

@@leif1075 I think you'll realize it's intuitive if you read a proof for it. It's more systematic (at least to check if the cubic has any rational roots) than hoping to guess the solutions, or using the longer general ways of solving cubics.

[@tomhase7007]

In reality it is much less useful than one might think. It is extremely unlikely that a rational polynomial has rational roots - this happens only in the presence of very strong symmetries or in exercises in which the numbers have been doctored by the person designing the exercise. To guess a rational root and then to factor a cubic polynomial to be able to apply the formula for the roots of a quadratic polynomial is not a widely applicable method to solve cubic equations (unless one is explicitly only interested in rational roots). It is also completely unnecessary. There is a perfectly nice formula for the (real or even complex) roots of a cubic polynomial that should be taught and used more widely.

[@zeravam]

Great combination of algebra and trigonometry! What I liked the most was the point where algebra and trignometry sepparate each other, knowing that maximum value of sinx+cosx is +sqrt2 (1st Quadrant) and minimum value is -sqrt2 (3rd Quadrant). Great exercise Michael

[@carlosguerrayanez]

"That's a nice place to stop" is becoming my favorite catchphrase. As always, nice video Michael!

[@DrQuatsch]

You can then eventually solve for cos(x) and sin(x). They are 1/4[1 +/- sqrt(7)]. If you then solve for the angles theta for which this happens, then you get two angles between -180 degrees and 180 degrees, namely approximately equal to -24.295 degrees and +114.295 degrees. Which at first I found curious, because the two angles sum to 90 degrees. But then I remembered the whole solution was symmetrical. cos(theta1) = a and sin(theta1) = b, and the other solution is cos(theta2) = b and sin(theta2) = a. You can just simply replace cos(theta2) by sin(pi/2 - theta2) and similarly with the sin(theta2) = cos(pi/2 - theta2), and you get that theta1 = pi/2 - theta2, so the sum of the two solutions is equal to pi/2 which of course is 90 degrees.

[@ArthurvanH0udt]

At around 8m30 why not use synthetic division (another tool) to show 1/2 is a root and then immediately also have the parameters (double in this case wrt your example) of the quadratic!

[@goodplacetostop2973]

10:09 Homework? 11:12 Good Place To Stop

[@SuperYoonHo]

OK

[@karolakkolo123]

You're still going lol

[@txikitofandango]

Slightly different way: (sin + cos)^3 = sin^3 + cos^3 + 3sincos(sin + cos) (sin + cos)^2 = 1 + 2sincos (sin + cos)^3 = 11/16 + (3/2)((sin+cos)^2 - 1)(sin + cos) x^3 = 11/16 + (3/2)(x^2-1)x 8x^3 - 24x + 11 = 0 ...

[@zeravam]

I just did it the same way

[@NaN_000]

Great way:)

[@kiyagiles655]

I did it pretty much the same way: (Sinx + cosx)^2 = sin^2x + cos^2x + 2sinxcosx = 1 + 2sinxcosx (1 + 2sinxcosx)(sinx + cosx) = sinx + cosx + 2sin^2xcosx + 2sinxcos^2x = H + 2cosx(1 - cos^2x) + 2sinx(1 - sin^2x) = H + 2H - 2H3 = H^3, 3H - 2H3 = H^3 H^3 - 3H + 2H3 = 0 H^3 - 3H + 2(11/16) = 0, 16H^3 - 48H + 22 = 0, 8H^3 - 24H + 11 = 0

[@Robokapp]

i like how that 16 magically appears on the second board after you super missed it on first board... :)

[@manucitomx]

What a great problem. Thank you, professor.

[@prbprb2]

Let S= a+b; D=a-b. Then the factorization suggested by Michael gives: S(S^2 +3D^2)/4 = 11/16 But due to trig S^2 +D^2 = 2 Eliminate D^2 between these two equations. 0= 8S^3-24S +11= (2S-1)(4S^2 +2S-11) . And, since S

[@Roberto-REME]

Excellent job and very well explained. You're amazing Michael!

[@vishalmishra3046]

@Mike you can mention @ 10:00 in the video | cosT + sinT = √2 sin (T+ π/4) = √2 cos(T - π/4). Therefore, range of cos + sin becomes -√2

[@vishalmishra3046]

*More general rule* -D

[@mikesmithyes967]

Michael, love the videos. I would like to see some more Putnam problems and problem solving strategies. Thanks.

[@memeboy1057]

There is an easier way to solve the cubic - you just factor it as following - 8x^3 - 24x + 11 = 0 8x^3 - 2x - 22x + 11 = 0 2x( 4x^2 - 1 ) - 11( 2x - 1 ) = 0 2x( 2x + 1 )( 2x - 1 ) - 11 ( 2x -1 ) = 0 (2x - 1) ( 4x^2 + 2x - 11 ) = 0 This will save the trouble of going through the substitution process.

[@ojas3464]

In this example the leading coefficient 8 of the cubic happens to be perfect cube. Otherwise some substitution may be inescapable.

[@MyOneFiftiethOfADollar]

Nice find. Factoring by grouping cleaner, but quadratic still has to be done

[@mcwulf25]

I just thought: I wonder if x = 1/2 is a root, inspired by the 8x^3 term. Then the rest was easy.

[@MyOneFiftiethOfADollar]

Knowing the range -sqrt(2) < sinx + cosx < sqrt(2) a big time labor saver from having to check the two nasty surds! Would not have noticed that working it cold.

[@chaosredefined3834]

To speed up the RRT step, you can introduce a variable u = 2x. 8x^3 - 24x + 11 = 0 (2x)^3 - 12(2x) + 11 = 0 u^3 - 12u + 11 = 0 The possible rational roots of this are 1, -1, 11, and -11. You can quickly check that 1 fits. Then, since u=2x, then x = 1/2.

[@yuriandropov9462]

х(16-8(х^2-1))

[@rome8726]

That such a nice problem !!!

[@Ron_Shvartsman]

I was able to find the cubic polynomial, but had forgotten the Rational roots theorem and was unable to proceed further. Nice problem! Also, instead of finding x^2, which is a fast way of doing it, I came up with the clunkier version of finding x^3, since we already started with sin^3(theta) + cos^3(theta). This gives us 11/16 + 3x*sin(theta)cos(theta) = x^3, which we can then substitute in to get x^3 - 3x + 11/8 = 0, as in the video.

[@paologat]

You can get to the same cubic equation in a less convoluted way by expanding x^3 as the first step, instead of factoring the sum of cubes.

[@carlosmirandarocha8905]

I wished professor Penn gave me math classes

[@cernejr]

x = 2 (π n + tan^(-1)(1/3 (2 +- sqrt(7)))), n element Z -0.42 rad == -24.3 deg 1.99 rad == 114.3 deg Both solutions have either the sin or the cos a negative value.

[@cernejr]

abs(sin(x)) + abs(cos(x)) >= 1 for all real x, so to get a solution one of the summands must be negative.

[@Ny0s]

Very nice problem

[@TechyMage]

After solving till cubic equation i was putting random value in specified range [-√2,√2] thought about 1/2 due to odd and even integers Thanks for reminding about rational root theorem

[@DaveJ6515]

Neat and easy, this is a good exercise.

[@SQRTime]

Hi Andrea. If you are into problems from math competitions, our channel can be something you enjoy. Hope it helps. One sample video Solving a nice trigonometry problem kzbin.info/www/bejne/oIbFkHyPeKiJpqM

[@TedHopp]

Looking at 8x^3 - 24x +11 = 0, an obvious substitution is y = 2x. This gives us the monic polynomial y^3 - 12y + 11 = 0. And look at that! The sum of the coefficients is zero, so y = 1 is a root. The rest is then easy.

[@MyOneFiftiethOfADollar]

The necessary -sqrt(2)

[@niuhaihui]

sin(theta) + cos(theta) = sqrt(2) * sin(theta + pi/4), thus its range is in between -sqrt(2) and sqrt(2).

[@jursamaj]

Quick solution: fire up Geogebra. Put in sin³+cos³ and 11/16.Find the 2 intersections in the 0-2π range, and drop verticals from them. Put in sin+cos. Find the 2 intersections with the previous verticals. Both happen at y=.5.

[@ojas3464]

Some proctors during exams / tests may insist students logon to the wolframalpha site and figure improving an input like x^2 + y^2 = 1, and x^3 + y^3 = 11 /16, what is (x + y)?

[@jursamaj]

@@ojas3464 That system of equations is better if you replace 11/16 with √2. 😁

[@skit_inventor]

Checking the range of sin(θ) + cos(θ) can be done easily without any calculus if one uses the auxiliary angle (in this case π/4). At least that's what comes to my mind first

[@s4archie]

I have a wrong solution in which I can't identify the error. Help appreciated. Let s=sin(a), c=cos(a), r=11/16. We have s^3+c^3=r, so (s+c)^3=r+sc(s+c). Also (s+c)(s^2-sc+c^2)=r so (s+c)(1-sc)=r giving (s+c)-sc(s+c)=r and (s+c)=r+sc(s+c). This gives us (s+c)^3=(s+c). Since r0, we have (s+c)^2=1 Since this result is independent of the value of r, it's obviously wrong, but I can't see why.

[@phiefer3]

How exactly did you come to this: (s+c)^3=r+sc(s+c) expanding out the left side, and then substituting r in for the sum of the 2 cubed terms would leave the right side as r+3sc(s+c) I didn't bother working through the rest of your work, but anything based on that first step is going to be mistaken.

[@s4archie]

@@phiefer3 Ah, yes. I forgot the coefficient.

[@fariborzhessabi5273]

It's very good bravo.

[@hcgreier6037]

05:34 ....the 16 multiplies the 1 too, isn't it? Do I miss something here? Aah, fixed at 06:28!

[@tungstentoaster]

I thought I was going crazy until I read this comment.

[@Kumurajiva]

I had no idea before that I can increase my bicep measure by teaching math! Now I know

[@dominicellis1867]

But then, what is theta?

[@mishania6678]

5:33 you forgot to put 16 instead of 1 in brackets cuz you multiplied both sides on 16

[@Reza_Audio]

RIGHT

[@s4623]

5:21 you see it here first 16 * 1 = 1 ! [should have been x(16 - 8(x²-1)) instead] and corrected through TV Magic after ad.🤣

[@abrahammekonnen]

Really cool problem. I definitely want to go back over this. Oh btw I joined your Patreon earlier today at the $2 tier I plan on moving up to the $5 soon, but I just wanted to test things out and join the discord first. As usual thank you for the video.

[@mathadventuress]

couldnt we just use complex exponentials?

[@johns.8246]

I did this for (cos z)^3 + (sin z)^3 = 1 where the real part of z is on the interval [0 , 2pi). I found 2 rather obvious real solutions and two pesky complex ones. Anyone up to the task?

[@roberttelarket4934]

Very nice!

[@jakolu]

It's funny as (-1+3sqrt(5))/4=1.427.... which is really close to sqrt(2)

[@carstenmeyer7786]

_Newton's Identities_ are another way to go! Define *x_1 := sin(𝞱), x_2 := cos(𝞱)* and the sums of powers *k ≥ 0: s_k := x_1 ^ k + x_2 ^ k with s_1 = ?* By definition we get *s_0* , _Pythagoras' Theorem_ yields *s_2* while the problem gives us *s_3* directly: *s_0 = 2, s_2 = 1, s_3 = 11 / 16* Now consider the polynomial *p(x) := ( x - x_1 ) * ( x - x_2 ) = x^2 - s_1 * x + x_1 * x_2* If we sum over the zeros, we get a recursive relation for *s_k* : *k ≥ 2: 0 = \sum_{m = 1}^2 x_m^{k - 2} * p(x_m)* *= s_k - s_1 * s_{k-1} + x_1 * x_2 * s_{k - 2}* We may evaluate the recursion for *k = 2* and *k = 3* to get two equations in *s_1* : *k = 2: 0 = 1 - s_1^2 + 2 * x_1 * x_2* *k = 3: 0 = 11/16 - s_1 + s_1 * x_1 * x_2* Use the first equation to eliminate *x_1 * x_2* from the second equation and obtain the cubic: *0 = x^3 - 12 * x + 11 | x := 2 * s_1* *= (x - 1) * (x^2 + x - 11)* Via _Rational Root Theorem_ we guess *x = 1* and get three real solutions: *x ∈ {1; (-1 ∓ 3√5) / 2 }* Use the addition theorem *a * sin(𝞱) + b * cos(𝞱) = √(a^2 + b^2) * cos(𝞱 - atan2(b; a))* to estimate *s_1* : *s_1 = √2 * cos(𝞱 - 𝝅/4) ∈ [-√2; √2]* Only the first solution *x = 1* leads to *s_1 = 1/2 ∈ [-√2; √2]*

[@SQRTime]

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[@Vidrinskas]

Probably could just solve for theta but I'd imagine it's messy.

[@abrahammekonnen]

5:38 Note: There should be a 16 where the 1 is.

[@abrahammekonnen]

6:26 As usual you're on top of it

[@leif1075]

Why write it as 2x -1 instead of x - 1/2..I don't think anyone would.ever think of that so why do it?

[@luciferbikkuangmin9528]

I think we can use Newton's Identity, the properties of symmetric polynomials.

[@Mystery_Biscuits]

nice problem

[@kevinmartin7760]

Did this actually prove 1/2 is a solution? It was not disproven by the range test, but I'm not sure that proves it isn't just another artifact of the calculations.

[@benjaminojeda8094]

5:22 You forgot to put 16 instead of 1

[@NeganLucilleForever]

this problem is a trig cove, so it is, but you my cullies can solve it

[@CM63_France]

Hi, Ok, great! Home work : given two numbers a and b, that we only know the 2 quantities a^3 - b^3 and a^2-b^2 , calculate a-b .

[@abdallahmohamedelhady6668]

Good job👏

[@mariomestre7490]

Genial, ets un crack

[@FedericoGuille]

Cool

[@SQRTime]

Hi Federico. If you are into problems from math competitions, our channel can be something you enjoy. Hope it helps. One sample video Solving a nice trigonometry problem kzbin.info/www/bejne/oIbFkHyPeKiJpqM

[@c0d3w4rri0r]

Question. If you alow theta to be complex do the 2 excluded solutions come back in.

[@christianimboden1058]

I was just going to suggest to solve for all complex solutions

[@nbourbaki6610]

You didn't show that 1/2 is a solution, but just that if a solution exists it has to be 1/2.

[@TJStellmach]

That's a trivial consequence sin θ + cos θ being continuous and ranging between a minimum of -√2 and a maximum of √2 (which were asked to accept as given when we rejected the other roots). Really, it only requires that sin θ + cos θ can clearly take on values both less than and greater than 1/2 (say, -1 and +1).

[@SuperYoonHo]

2nd