The kombination of trigonometry with an irrational equation is just cool.Excellent as always,you are in top.More such videos on the channel.Good Luck in everything Syber Math 👏👏👏👏👍👍👍👍👍👍👍👍
@SyberMath
2 жыл бұрын
Thank you!!! 💕🙂
@davidseed2939
3 жыл бұрын
An interesting disguised 3,4,5 triangle. The RHS can be rearranged to 1/c +1/s, where c,s are the cosine and sine of the 3,4,5 triangle ie 3/5 and 4/5. The LHS can be rearranged to 1/u+ 1/sqrt(1-u^). The video shows the interchangeable correspondence of these terms i.e. u = 3/5 or 4/5 Hence x=5/3 or 5/4
@SyberMath
3 жыл бұрын
That's very smart!
@scottleung9587
2 жыл бұрын
I had no idea how to solve this, but it was fun!
@SyberMath
2 жыл бұрын
Glad to hear that!
@scottleung9587
2 жыл бұрын
@@SyberMath I meant to say it was fun watching you solve it!
@fredericoapuleio9583
3 жыл бұрын
Eu resolvo os problemas primeiro, antes de assistir sua resolução. Resolvi esse problema da seguinte forma: fiz a = x, e b = x/sqrt(x² -1) daí notei que a² + b² = a²b² → (35/12)² = a²b² + 2ab, fazendo ab = t, temos t² + 2t - (35/12)² = 0 e resolvendo esta equação se tem os valores corretos.
@yennefer415
3 жыл бұрын
A few questions: 1) at 4:27 cant alpha be in 4th quadrant as well? 2) at 15:20 dont you have to check if for cosa = 3/5, sina is actually =4/5, sina and cosa arent independent of each other like lets say simple 'a' and 'b' Ad. 1) would it be that I can just choose 1st or 4th quadrant, because they both return values of x's range, namely (1,inf)? So for that matter could we- just choose any one of 1+4n or 3+4n --th quadrants where 'n' is an integer? Ad. 2) im mainly just curious if the 'potential solutions' will always be solutions? Dont we have to check if solution_1^2 + solution_2^2 = 1 ? Cuz that should be true since theyre sina and cosa
@angelmendez-rivera351
3 жыл бұрын
Actually, you totally can have x < 1, so long as you can use the absolute value operator to maintain equality. sqrt[sec(a)^2 - 1] = |tan(a)|, or -tan(a) if π/2 < a < π, tan(a) if 0 < a < π/2. This means you get 12·[sin(a) + cos(a)] = 35·sin(a)·cos(a) if 0 < a < π/2, 12·[sin(a) - cos(a)] = 35·sin(a)·cos(a), then proceed along for both cases individually.
@SyberMath
3 жыл бұрын
If -1
@angelmendez-rivera351
3 жыл бұрын
@@SyberMath Ah, you're correct. I must have been thinking of something different when I wrote this.
@SyberMath
3 жыл бұрын
@@angelmendez-rivera351 Not a problem at all!
@anggalol
3 жыл бұрын
instead substitute x = secan alpha, why not substitute x for wolfram alpha
@SyberMath
3 жыл бұрын
What did I substitute?
@aashsyed1277
3 жыл бұрын
@@SyberMath she\he is talking about wolframalpha!
@ashishpradhan9606
3 жыл бұрын
Fantastic as always . BTW, When you just say that word substitution I get certain that you are going to substitute by sec x.
@SyberMath
3 жыл бұрын
Great! Thank you, Ashish!
@050138
3 жыл бұрын
The moment I saw the question, sec and tan were kinda obvious! ☺️
@johnnath4137
3 жыл бұрын
I like this solution very much!
@SyberMath
3 жыл бұрын
Glad you like it!
@septembrinol1
Жыл бұрын
You could also do 12^2 (sin a+ cos a)^2 = 35^2 (sin a cos a)^2 => 144 (1+2 sin a cos a) = 1225 (sin a cos a)^2 and solve using that sin (2a) / 2 = sin a cos a
@Jared7873
3 жыл бұрын
What would happen if your started by multipling both sides by square root of (x^2+1)?
@karmkaria9601
2 жыл бұрын
2 nd approch sinacosa=12/25 than multiply with 2 than sin2a=24/25 so 2a in 0 to π so cos 2a is 7/25 or - 7/25 and use cos2a = 2cosa^2 -1 and we get ans.5/3 5/4
@tarunmnair
3 жыл бұрын
amazing approach and solution, what if we took the negative value for y=-5/7 ?
@andreyberezhnoy2560
3 жыл бұрын
it cant be negative. if 0 < alpha < pi/2 -> cos(alpha) > 0, sin(alpha) > 0 -> cos(alpha) + sin(alpha) = y > 0
@aliasgharheidaritabar9128
3 жыл бұрын
Marvellous pal.
@SyberMath
3 жыл бұрын
Thanks pal!
@holyshit922
3 жыл бұрын
This quartic is not so difficult to solve (I prefer solving by completing the square, introducing parameter, and reducing quartic to the difference of squares Method of undetermined coefficients in an alternative way of solving quartics for me)
@SyberMath
3 жыл бұрын
Absolutely!
@aashsyed1277
3 жыл бұрын
what is undetermined coefficients?
@holyshit922
3 жыл бұрын
@@aashsyed1277 You assume that quartic is a product of two quadratics multiply them , compare coeficients and this undetermined coefficients are coefficients of these quadtatics In this approach you will need substitution No matter when (before or after getting resolving sextic from system of equation)
@aashsyed1277
3 жыл бұрын
@@holyshit922 So does this work for every quartic with non zero a b c d? where a b c d are the coeficents.
@holyshit922
3 жыл бұрын
@@aashsyed1277 Yes but be careful because of possible division by zero (You have to choose solution of resolvent sextic wisely) On some Indians or Hindu videos this method is called Descartes method I wrote that in this method you need substitution but it doesn't matter when Substitution looks like x=y+h but when you use ths substitution in quartic equation h is different than you use this substitution for resolvent sextic (try & see)
@ricardovalentin5056
2 жыл бұрын
Fantastic ! Avoiding solving a quartic led us to deal with trig and quadratic twice instead, which is awesome !
@vbinsider
3 жыл бұрын
My favourite part is to use Vieta's formulas to solve the non-linear system of equations in 14:15 . That's hilarious. :)
@khayelihle1
3 жыл бұрын
How can I become smart like you?
@leecherlarry
3 жыл бұрын
straightforward for the computer hehe: *𝚂𝚘𝚕𝚟𝚎[𝚡 + 𝚡/𝚂𝚚𝚛𝚝[𝚡^𝟸 - 𝟷] == 𝟹𝟻/𝟷𝟸]*
@SyberMath
3 жыл бұрын
No cheating!!! 😂
@leecherlarry
3 жыл бұрын
@@SyberMath Ah, no Maple for me. Am giving up. I stick with Wolfram L, loving it. So consistent beautiful language. Yeah, no cheating!! haha :D
@-basicmaths862
3 жыл бұрын
35/12=5(7/12) 35/12=5(1/3+1/4)
@ilkinhsnov4463
3 жыл бұрын
Great...
@SyberMath
3 жыл бұрын
Thanks!
@ongvalcot6873
2 жыл бұрын
x is sec or x is minus sec
@Qermaq
3 жыл бұрын
Fun, but aww it's really just 12x^2 - 35x + 25 = 0,
@SyberMath
3 жыл бұрын
What do you mean?
@Qermaq
3 жыл бұрын
@@SyberMath Heh. It's just a binomial in cool clothes. ;D Dry humor.
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