can you solve this “impossible” trig problem?
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- Опубликовано: 8 сен 2024
- can you solve this “impossible” trig problem? Here is an equation that might look unsolvable, but that is actually possible to solve! For this, you need to use your trig identity cos^2+sin^2=1 to change it to a quadratic equation, and then use the quadratic formula, exponents and logs, and then the unit circle. This is a must see for high school algebra 2 math students, as well as college algebra and precalculus students, and is perfect for math final and exam preparation.
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People like Dr. Peyam (also like Redpenblackpen, Michael Penn and Sybermath) have a gift for teaching mathematics.
I also watch all of them
Mind your decisions, 3b1b, numberphile,mathologer?
@@advaykumar9726 We are addicted to math channels😀
blackpenredpen not redpenblackpen
@@pneujai 😀😁🤣👍👍
1:59
When i was a student at school, all my math teachers showed themself as a person who knows everything in math and I always was worried and sadden by thinking that I’m not intelligent enough to learn math.. but after seeing people like Dr. Payem and Math Sorcerer I realized that even after having phd in math they couldn’t solve certain problems… (by the way most of my math teachers didn’t even have a bachelors degree in math) this encouraged me to learn math again.. now I realize math is way more interesting and anyone could learn it, if they really wanted to learn.
let a=81^sin²x and b=81^cos²x
a+b=30
ab=81
there you go a nice guessable linear equation
a=3,27 b=27,3 take one set of the solution and solve for sinx,cosx
Just for curiosity, if the equation was 81^(sin(x)^2)+82^(cos(x)^2) = 30, that is a fractional polynomial that can be solved by using the Lambert-Tsallis Wq function, one would get sin^2(x) = 0.7499 (close to 3/4) and sin^2(x) = 0.2524 (close to 1/4).
Wouldn't you end up saying that M mod 3 != 0 ? That phrasing makes more sense to me
In a computer science context, yea. "mod" aka % in most programming languages is an operator that returns the remainder, so you write it like that.
In maths however mod is more of an equivalence class concept. Say for mod 3, we talk about all the numbers that leave the same remainder when divided by 3. We write x ≡ 0 mod 3 or whatever, or put a line through it, x ≢ 0 mod 3. In this context ≡ is read as "is congruent to".
Hello! A slightly better way to write the solution would be:
x = (3n ± 1)π/6; where n is an integer
Hello, Dr. Peyam. I'd tried solving the problem with my own method
Here's how I solved it :
81^sin²x + 81^cos²x = 30
9^2sin²x + 9^2cos²x = 30
9^(sin²x + 1 - cos²x) + 9^(cos²x + 1 - sin²x) = 30
because cos²x - sin²x = cos2x
9^(- cos2x + 1) + 9^(cos2x + 1) = 30
9(9^(-cos2x) + 9^cos2x) = 30
Let p = 9^cos2x
Then p^-1 = 9^(-cos2x)
9(p^-1 + p) = 30
multiply all sides by p, then divide all sides by 9
1 + p² = 30p/9
p² - 30p/9 + 1 = 0
(p - 3)(p - 1/3) = 0
p = 3 V p = 1/3
9^cos2x = 3 V 9^cos2x = 1/3
cos2x = log_9(3) V cos2x = log_9(1/3)
cos2x = 1/2 V cos2x = -1/2
when cos2x = 1/2
2x = 60° + 360°k or 2x = 300° + 360°k
x = 30° + 180°k or x = 150° + 180°k
when cos2x = -1/2
2x = 120° + 360°k or 2x = 240° + 360°k
x = 60° + 180°k or x = 120° + 180°k
As always, your videos are amazing. Great job!
Neat use of the other, lesser used trig identities.
@@Chris.S Yea I know. That Naveen guy's solution is really simple and understandable.
i really appreciate you for what you're doing ♥️
Thank you!!! Always makes me refresh and strength some concepts.
At school in the Netherlands I very often had to use the abc-formula for solving quadratic equations in both math and physics, so I was surprised you didn't use it and turned to Wolfram. Maybe they don't teach that formula at school anymore? If that is the case, then that would be sad... Lovely piece of algebra though, as usual!
Mathematicians are sometimes totally lost with "easy" stuff. The woman who mentored our math exercises after the professor had lectured, sometimes completely blacked out when she had to calculate straightforward number in her head, for example, 9 times 6 = 54
...AFTER SHE HAD JUST SOLVED A BRUTAL PARTIAL DIFFERENTIAL EQUATION IN ELECTRODYNAMICS ...WHICH SHE ALSO MADE LOOK EASY AS FUCK WHILE OUR HEADS WERE SMOKING AND WE SPONTANEOUSLY DEVELOPED ACUTE MATH-FOR-NON-MAJORS-DEPRESSION! "Oh 9 times 6, ehm, is that 48, no, it's 56, right?"
Also very famous: David Hilbert, the greatest mathematician of the 20th century and the second greatest mathematician ever (after Carl Friedrich Gauß) was known for being really bad at doing simple arithmetics with numbers up to 100. His students and Ph.D. students notoriously mocked him for that ...until the exam...
Well done, Dr. Peyam
Nice and easy. Good job.
Elegantly solved & thoroughly
Trig as the trick to solve this problem
Amazing Dr!
🤭 Btw, Michael penn make a video on D!, what you think?
Had to use wolfram alpha to factor a basic polynomial 😂
You can use Poh Shen Loh’s method to factor…
Very very very nice!
I like it when he brings out the Pi*M.
Really amazing
fantastic thank you for video
2:45 „… same Spiel with …“. You like German. 🤗
Hi Dr. Peyam!
x and y *and* z? And we haven't even left one dimension!
Really fun!
Fun as usual . hey Doc can you give me a reference to good Twitter math sites ?
Hello ,good morning Dr. PEYAM,allways problems so interestings,,surelly You knows the sing of rolling stones musical group" ....and i try And i try...i can get no satisfaction" is the same,i'm trying to go up looking for the knowledge You allways give us,si thanks You,!!!!!
Lo et me tell You,Dr,may be i'm your oldest pupil,i'm 68,but My math heart still Beats hard,sometime i tell myself that i need two or three lives more to underestand the lawd of nature
Plz help sir
But I can't understand this circle method
It's a Unit Circle:
en.wikipedia.org/wiki/Unit_circle
The circle is just showing the unit circle where sin^2(x) can equal 1/4 or 3/4, take the square root of everything and you have sin can be plus/minus 1/2 or rt(3)/2 (ex: sin(pi/6) and sin(5pi/6) gives you 1/2, and sin(7pi/6) and sin(11pi/6) gives you -1/2)
That unit circle is just showing when sin is +/- 1/2, +/- rt(3)/2
This is the unit circle which helps with trig values of sine and cosine (as the sine of the angle is the height of the triangle made by the radius while the cosine of the angle is the width of the triangle made by the radius), here's an explanation for all the dots and crosses from cross on the x-axis anticlockwise in reference to θ, the anticlockwise angle made with the horizontal and I'll also do it by quadrant:
First Quadrant:
1 - This cross represents when θ = 0 or any multiple of 2π, and hence this is when sin(θ) = 0 and cos(θ) = 1 (As you can see from the length of the horizontal radius at that point being 1, while it does not go up at all)
2 - This dot represents when θ = π/6 + 2πk, and hence this is when sin(θ) = 1/2 and cos(θ) = √3/2
3 - This dot represents when θ = π/3 + 2πk, and hence this is when sin(θ) = √3/2 and cos(θ) = 1/2
(Note that both sin(θ) and cos(θ) are positive, as both the x and y-coordinates are positive in this region)
Second Quadrant:
4 - This cross represents when θ = π/2 + 2πk, and hence this is when sin(θ) = 1 and cos(θ) = 0
5 - This dot represents when θ = 2π/3 + 2πk, and hence this is when sin(θ) = √3/2 and cos(θ) = -1/2
6 - This dot represents when θ = 5π/6 + 2πk, and hence this is when sin(θ) = 1/2 and cos(θ) = -√3/2
(Note that cos(θ) is negative, as the x-coordinate is negative in this region)
Third Quadrant:
7 - This cross represents when θ = π + 2πk, and hence this is when sin(θ) = 0 and cos(θ) = -1
8 - This dot represents when θ = 7π/6 + 2πk, and hence this is when sin(θ) = -1/2 and cos(θ) = -√3/2
9 - This dot represents when θ = 2π/3 + 2πk, and hence this is when sin(θ) = -√3/2 and cos(θ) = -1/2
(Note that both sin(θ) and cos(θ) are negative, as both the x and y-coordinates are negative in this region)
Fourth Quadrant:
10 - This cross represents when θ = 3π/2 + 2πk, and hence this is when sin(θ) = -1 and cos(θ) = 0
11 - This dot represents when θ = 5π/3 + 2πk, and hence this is when sin(θ) = -√3/2 and cos(θ) = 1/2
12 - This dot represents when θ = 11π/6 + 2πk, and hence this is when sin(θ) = -1/2 and cos(θ) = √3/2
(Note that sin(θ) is negative, as the y-coordinate is negative in this region)
So there it is, all 12 angles and their respective sines and cosines explained, try and work through the logic of the cases when θ equals any multiple of π/4 yourself and if you do that you will have mastered fundamental trig!
Some extra notes:-
The respective angles in degrees (once again, by quadrant) are as follows if degrees is more familiar to you:
π/6 = 30°, π/4 = 45°, π/3 = 60°, π/2 = 90°
2π/3 = 120°, 3π/4 = 135°, 5π/6 = 150°, π = 180°
7π/6 = 210°, 5π/4 = 225°, 4π/3 = 240°, 3π/2 = 270°
5π/3 = 300°, 7π/4 = 315°, 11π/6 = 330°, 2π = 360°
Finally, the reason he put his final answer as x = πm/3, m ≠ 0 MOD 3 (This basically means m is not a multiple of 3) is because the solutions of sin(θ) and cos(θ) = ±1/2 and ±√3/2 correspond to every angle in my list earlier, apart from when the m can cancel the 3 (e.g. x = π is not a solution, which is 3π/3 in a sense).
I know that was a lot but I hope that helps, understanding where the trigonometry comes from was difficult for me as well so i hope this is a rigorous explanation, look for visuals online and stuff if you want - they're really useful.
4:18 he said it he said it he said it he said it