I love the devious laugh when he figured it out 😂

Since tan(3*theta) is given as a positive value, the angle (3*(theta)) lies in 1st quadrant or 3rd quadrant, therefore, 3*theta = (pi/3)+(2k+1)*(pi), where k=0,1,2,3,... etc., and therefore theta = (pi/9)+(2k+1)*(pi/3), where k=0,1,2,3,... etc.

I didn’t know about this tan 3θ formula, this problem would have defeated me. Genius, and then I thought it would only give one solution which was obviously wrong. Yes, I learned something today. This makes me wonder if there’s a collection of all these wonderful gems that are so useful. One such formula is ”The Fact” which says that the limit of this formula as n goes to infinity (1+a/n)^bn is e^ab

Its funny, how easily a question like this could slip you by. Thank you, Prime Newtons!

Fantastic approach !!! Thanks again for the interesting session

this was a beautiful solution. good job!

Brilliant and well presented.

This is powerful! Thank you, Prime Newton!

This problem is a bit of a giveaway for anyone well versed in trigonometric identities. If you want to participate succesfully in Math Olympiads you should know _inter alia_ the triple angle identities for the sine, cosine and tangent, and then you immediately recognize the pattern. I encountered this exact equation earlier in solving this Math Olympiad problem: _Prove that tan(20°) − tan(40°) + tan(80°) = 3√3_ The usual solutions rely on a lot of manipulation using various trigonometric identities. However, this problem can be done much more elegantly and easily using Vieta's formulas which relate the coefficients of a polynomial to its roots. First note that the expression to evaluate can be written as tan(−40°) + tan(20°) + tan(80°) since tan(−40°) = −tan(40°). Now note that the difference between −40° and 20° is 60° as is the difference between 20° and 80° and note that 60° is ⅓ of 180° which is the period of the tangent. This means that if we add another 60° to 80° to get 140° then tan(140°) is again equal to tan(−40°). It also means that the tangents of the triple of the angles −40°, 20°, 80° are all equal since these are each equal to tan(60°) = √3. Now recall the tangent triple-angle identity tan 3θ = (3·tan θ − tan³θ)/(1 − 3·tan²θ) Since the tangents of the triple of the angles −40°, 20°, 80° are each equal to √3 this identity implies that tan(−40°), tan(20°), tan(80°) are the roots of the equation (3x − x³)/(1 − 3x²) = √3 which is of course exactly the equation discussed in this video. Multiplying both sides by (1 − 3x²) and bringing all terms over to the left hand side this equation can be written as x³ − 3√3·x² − 3x + √3 = 0 The sum of the roots of this monic cubic equation is equal to minus the coefficient of the square term, that is 3√3, and since these roots are tan(−40°), tan(20°), tan(80°) it immediately follows that tan(−40°) + tan(20°) + tan(80°) = 3√3 or tan(20°) − tan(40°) + tan(80°) = 3√3 As an added bonus, we may note that the product of the roots of the cubic equation is equal to minus the constant term, that is −√3, so we also have tan(−40°)·tan(20°)·tan(80°) = −√3 or tan(20°)·tan(40°)·tan(80°) = √3

I greatly appreciate this amazing solution and the amazing way you used in solving this sir ❤️❤️❤️❤️

You are the master of simplifying those "trick" problems.

Great one!

Hey Prime Newtons! I think I have figured ot a way to differentiate a^x using first principles, using the exact way from your video on e^x (!) You'll get 1/log[basea](lim h-->0 (1+h)^1/h) and this becomes 1/log(basea)e = lna (!)

This is easy - x=tan(theta) then answer is x=tan(10 degrees)

sir please could you solve this jee advanced 2016 question? Paper 2: Maths Section 2 :- Question 44 ( multiple options correct ) its a limit question average time per question is 3-4 mins in some places its given as question 44 also

Thumb's up from the old geezer.

I watched some of your teachings 🎉🎉❤❤😊😊 Your style is cool and you are so handsome ❤🎉😊

really enjoyable

How come there are more than 3 solutions for different values of k?

Nice! But are there an expression for tan(pi/9) tan(4pi/9) and tan(7pi/9) ?

try this out: for which "a" the equation 6a*(9^x) + (9/8)*a = 81^x + 5a*(3^x) has only one solution. this one is from russian school graduation exam

Are the solutions for k=[4,5,6,...] somehow invalid? I know we only need three roots, but why not choose a different set of three? Im sure there is good reason. I just don't know it.

x = tan(theta) tan(3theta) = tan(pi/3) 3theta = pi/3 + kpi theta = pi/9 + k/3pi x = tan(pi/9 + k/3pi) x_{1} = tan(pi/9) x_{2} = tan(4pi/9) x_{2} = tan(7pi/9)

Why doesn’t k=3 give more solutions?

You say in your introduction that you thought about how to solve the equation (3x − x³)/(1 − 3x²) = √3 for almost a _year_ before you recognized that the tangent triple angle identity is the key to solving this equation easily. You also say that you considered and tried trig substitutions that didn't work. Let's consider this equation from the point of view of someone who _doesn't_ recognize that substituting x = tan θ reduces this equation to tan 3θ = √3, that is, tan 3θ = tan ¹⁄₃π, because solving this equation in a conventional way proves to be quite interesting. It is clear that what we have here is actually a cubic equation, because if we multiply both sides by (1 − 3x²) and then bring all terms over to the left hand side this equation can be written as x³ − 3√3·x² − 3x + √3 = 0 The first thing we notice here is that the coefficient of x² and the constant term (that is, the _even_ powers of x) are irrational because they contain a factor √3 but the coefficients of x³ and x (that is, the _odd_ powers of x) are rational. Now, since rational coeficients are easier to work with than irrational coefficients, a useful preliminary step is to rationalize the coefficients, which can be done here by substituting x = y/√3 which gives (y/√3)³ − 3√3·(y/√3)² − 3·(y/√3) + √3 = 0 y³/(3√3) − √3·y² − √3·y + √3 = 0 and multiplying both sides by 3√3 then gives y³ − 9y² − 9y + 9 = 0 The first step when solving a cubic equation conventionally is to reduce or _depress_ the cubic to a cubic that lacks a quadratic term by performing a suitable linear substitution. How can we do that? Well, in accordance with Vieta's formulas, the sum of the roots of our cubic in y is 9. So, if we reduce each of the three roots by 3 by means of a substitution z = y − 3, that is, y = z + 3, then the sum of the roots of the equation in the new variable z will be _zero_ meaning that the equation in the new variable will lack a quadratic term. And, indeed, if we substitute y = z + 3 then we obtain z³ − 36z − 72 = 0 Note that 72 = 2·36 and that 36 is a multiple of 4 and that 72 is therefore a multiple of 8 = 2³. This implies we can get a reduced cubic with smaller integer coefficients if we now substitute z = 2w which gives 8w³ − 72w − 72 = 0 and dividing both sides by 8 this gives w³ − 9w − 9 = 0 We now have a reduced cubic equation in w with small integer coefficients which looks simple, but appearances are deceptive, this equation is _not_ simple to solve. First, we can easily check using the rational root theorem that none of the potential candidates 1, −1, 3, −3, 9, −9 for a rational solution satisfies our equation in w so there are no rational roots. Secondly, if you examine the polynomial P(w) = w³ − 9w − 9 you easily find that P(−3) = −9, P(−2) = 1, P(−1) = −1, P(3) = −9, P(4) = 19, so, since P(w) is a continuous function of w, in acordance with the intermediate value theorem there is at least one real zero on each of the intervals (−3, −2), (−2, −1), (3, 4). And since there can be no more that three zeros it follows that there is _exactly one_ real zero on each of these three intervals. So, we have a reduced cubic in w with integer coefficients with three real but irrational solutions. Even though all roots of this cubic are real, paradoxically these roots cannot be expressed algebraically without using cube roots of complex numbers. Any attempt to extract these cube roots of complex numbers algebraically will only lead to cubic equations which are equivalent with the cubic equation w³ − 9w − 9 = 0 we are trying to solve in the first place. This catch-22 is the famous _casus irreducibilis_ (irreducible case) of the cubic equation. A way out of this catch-22 was first shown by Vieta, using - you guessed it - a trigonometric substitution. The method is based on the triple angle identity for the cosine, which is cos 3θ = 4·cos³θ − 3·cos θ A modernized version of his method is as follows. We rewrite our cubic equation as w³ − 9w = 9 and substitute w = r·cos θ which gives r³·cos³θ − 9·r·cos θ = 9 What we really want to get at the left hand side is 4·cos³φ − 3·cos φ so we begin by multiplying both sides by 4/r³ which gives 4·cos³θ − (36/r²)·cos θ = 36/r³ Now we want to have 36/r² = 3 so we need to have r² = 12 and so we can choose r = 2√3 which gives us 4·cos³θ − 3·cos θ = 36/(24√3) and because 4·cos³θ − 3·cos θ = cos 3θ and 36/(24√3) = (36√3)/72 = ¹⁄₂√3 this gives cos 3θ = ¹⁄₂√3 Now, there is a single value of 3θ on the interval [0, π] which satisfies this, which is 3θ = ¹⁄₆π, but since the cosine is a periodic function with a period 2π we have 3θ = ¹⁄₆π + 2·k·π, k ∈ ℤ and therefore θ = ¹⁄₁₈·π + ²⁄₃·k·π, k ∈ ℤ Of course, precisely since the cosine is a periodic function with a period 2π, this gives only three different values for cos θ by selecting _any three consecutive integers_ for k. And since we substituted x = r·cos θ and since r = 2√3 we find using k = 0, k = −1, k = 1 that the equation w³ − 9w − 9 = 0 has the solutions w₁ = 2√3·cos(¹⁄₁₈·π) = 2√3·cos(10°) w₂ = 2√3·cos(¹⁄₁₈·π − ²⁄₃·π) = 2√3·cos(−¹¹⁄₁₈·π) = 2√3·cos(¹¹⁄₁₈·π) = 2√3·cos(110°) w₃ = 2√3·cos(¹⁄₁₈·π + ²⁄₃·π) = 2√3·cos(¹³⁄₁₈·π) = 2√3·cos(130°) Since we subsequently performed the substitutions x = y/√3, y = z + 3, z = 2w this implies x = (2w + 3)/√3 = √3 + (2/√3)·w so the solutions of the equation (3x − x³)/(1 − 3x²) = √3 can be expressed as x₁ = √3 + 4·cos(10°) x₂ = √3 + 4·cos(110°) x₃ = √3 + 4·cos(130°) which may seem surprising because using the substitution x = tan θ we end up with solutions which can be expressed as tan(¹⁄₉·π) = tan(20°), tan(⁴⁄₉·π) = tan(80°), tan(⁷⁄₉·π) = tan(140°). But it is possible to show that the expressions for the roots above are indeed equivalent with x₁ = tan(80°) x₂ = tan(20°) x₃ = tan(140°) It is a bit of a challenge to prove, for example, that √3 + 4·cos(10°) = tan(80°) but we can proceed as follows. √3 + 4·cos(10°) = 2·(¹⁄₂√3 + 2·cos(10°)) = 2·(¹⁄₂√3·sin(10°) + 2·sin(10°)·cos(10°))/sin(10°) = 2·(cos(30°)·sin(10°) + sin(20°))/sin(10°) = 2·(cos(30°)·sin(10°) + sin(30° − 10°))/sin(10°) = 2·(cos(30°)·sin(10°) + sin(30°)·cos(10°) − cos(30°)·sin(10°))/sin(10°) = 2·(sin(30°)·cos(10°))/sin(10°) = 2·¹⁄₂·cos(10°))/sin(10°) = cos(10°)/sin(10°) = cot(10°) = tan(80°) Similarly, we can prove that √3 + 4·cos(110°) = tan(20°) √3 + 4·cos(130°) = tan(140°)

I know abt this qn from a step 2 2023 paper xD

Pi/9 considering only the principle domain

Im first

But this is a substitution where you defined x as a function of theta so don't you have to calculate the values to get the actual answer