The trig solution is lovely. I wouldn't have thought of it myself, but worked it through after noticing it in Prakhar's answer. That said, the algebraic approach is nicer than it looks at first glance because of the way that (1 - sqrt(1-x^2))/x rationalises and inverts. And the y term simplifies nicely to Sqrt( (1+y)/(1-y)).
Sir, Thank you for the heart . Don't stop uploading difficult problems with simple solutions. I am eagerly waiting for the next weekly challenge number 71 .
Terrific job Prakhar Agarwal! I find it amazing that equation 1 can be rewritten as the equation of the unit circle using trigonometric identities such as the half angle of tangent function(tan(x/2).
i solved this without too many trig subtitution by turning the first equation into circle equation (turning sin^2 + cos^2 into y^2 + x^2 = 1), then solve for x/y in the 2nd eq and sub the value back to 1st equation
I'm going to use a for alpha and b for beta: tan a/2 = tan (45 - b/2) The following is a well known tangent identity: If tan x = tan y then x=y + k×360° where k is some whole number. Therefore: a/2 = 45 - b/2 + k×360° Rearranging gives you: a+b=90 + k×360° However, we set both a and b to be between 0° and 90° so we cannot add or subtract any whole number multiple of 360° except for 0. So: a+b=90° + 0×360° a+b=90° which is the definition of complementary angles.
I hope people can read this and respond. I just solved this problem without even using the first equation(the one with two radicals). If we take 25(1-y^2)=41-40sqrt(1-x^2) and let x=cos(θ) and y=sin(θ) them the equation turns into 25(1-y^2)=41-40y implying that y=4/5 and x=3/5. Done.
This is not something you can do immediately. Sinθ and cosθ are interrelated and substituting both of them in the equation for x and y is not right. In this question it just turned out to be correct. Thats not the case always as connection between x and y may not be the same as that of sinθ and cosθ.
Oh, wow. You are right. I got lucky. My solution assumes that (x,y) is a point on the unit circle which turned out to be correct in the end but was not necessarily true to begin with. Thank you for your feedback.
![I.N "HALLUCINATION" | [Stray Kids : SKZ-PLAYER]](http://i.ytimg.com/vi/n5B5q1Hwt_U/mqdefault.jpg)
Okay that’s epic.
The trig solution is lovely. I wouldn't have thought of it myself, but worked it through after noticing it in Prakhar's answer. That said, the algebraic approach is nicer than it looks at first glance because of the way that (1 - sqrt(1-x^2))/x rationalises and inverts. And the y term simplifies nicely to Sqrt( (1+y)/(1-y)).
Thank you very much for the solution. I am big fan of your channel.
Sir,
Thank you for the heart . Don't stop uploading difficult problems with simple solutions. I am eagerly waiting for the next weekly challenge number 71 .
I really like the use of tangent. It just makes you use all kinds of identities just to solve X and y.
I feel really dumb for not seeing the trig sub when I attempted this problem.
That was awesome! Love your channel
Very nice illustration.Thank you.
Terrific job Prakhar Agarwal! I find it amazing that equation 1 can be rewritten as the equation of the unit circle using trigonometric identities such as the half angle of tangent function(tan(x/2).
i solved this without too many trig subtitution by turning the first equation into circle equation (turning sin^2 + cos^2 into y^2 + x^2 = 1), then solve for x/y in the 2nd eq and sub the value back to 1st equation
very nice solution
Impresionante felicidades
Good mental exercise. Thank you sir
That was beautiful. .thanks!
Amazing
Hmm so basically polar co-ords but with r=1?
This was amazing!
Like next level clever...thanks mate!! :D
Seeing sqrt(1-x^2), I was thinking that the way to go would be to redo the whole thing in polar coordinates.
What is your reasoning that alpha and beta are complimentary?
I'm going to use a for alpha and b for beta:
tan a/2 = tan (45 - b/2)
The following is a well known tangent identity:
If tan x = tan y then x=y + k×360° where k is some whole number. Therefore:
a/2 = 45 - b/2 + k×360°
Rearranging gives you:
a+b=90 + k×360°
However, we set both a and b to be between 0° and 90° so we cannot add or subtract any whole number multiple of 360° except for 0. So:
a+b=90° + 0×360°
a+b=90° which is the definition of complementary angles.
Sea cucumber oh I understand now! Thanks :)
@@slightlokii3191 Happy to help :)
This was delightful
Nice Problem.
I can't help but notice That your "Hand"writing has become better over time; did you buy a drawing table?
What do you mean by exclusive?
Not inclusive.
Teacher please write up cause down in close by writing load
Ahhhh very nice. I didn't knew that tangent identity
Did you get my solution in email?
Only the solutions posted as RUclips comments will be considered. =)
I hope people can read this and respond. I just solved this problem without even using the first equation(the one with two radicals). If we take
25(1-y^2)=41-40sqrt(1-x^2) and let x=cos(θ) and y=sin(θ) them the equation turns into
25(1-y^2)=41-40y implying that y=4/5 and x=3/5. Done.
This is not something you can do immediately.
Sinθ and cosθ are interrelated and substituting both of them in the equation for x and y is not right. In this question it just turned out to be correct. Thats not the case always as connection between x and y may not be the same as that of sinθ and cosθ.
Oh, wow. You are right. I got lucky. My solution assumes that (x,y) is a point on the unit circle which turned out to be correct in the end but was not necessarily true to begin with. Thank you for your feedback.
Why has he used two different angles? Is it wrong to use only one? And if it is wrong, why is it wrong? Thanks for your time.
If you used the same angle then you would be assuming x and y are the same value because sine is bijective between 0 and 90.
@@superj9220 Thank you so much!
Face reveal please
why
I scream extremely loud in my mind when I see you sub
x = sin α
Love you 🥰
Better than watching stupid movies ...upload more
“gorgeous”
It’s just too much algebra just to get nice answer lol