I failed high school math used your videos to get my courses and now am headed off to become a electeician much love from waterloo canada

@vatsalmishra9240

4 ай бұрын

Wait, the same waterloo where napoleon lost the war??

@coocow9278

4 ай бұрын

congratulations

@anonymouscheesepie3768

4 ай бұрын

@@vatsalmishra9240no, different waterloo lol

The last one can be done by sec⁶x-tan⁶x sec⁶x-(tan²x)³ sec⁶x-(sec²x-1)³ sec⁶x-(sec⁶x-3sec⁴x+3sec²x-1) 3sec⁴x-3sec²x+1 1+3sec²x(sec²x-1) 1+3sec²xtan²x which is a more elegant solution

@cparks1000000

4 ай бұрын

I actually noticed the difference-of-cubes instantly then completed the square to handle the (sec^4 x + tan^4 x) portion.

@cparks1000000

4 ай бұрын

That said, I really like your solution.

@mohammedtassowuff1936

4 ай бұрын

thanks!

@valeriomorra5064

4 ай бұрын

Last One Is the easiest

@kinggamerz.official

4 ай бұрын

I did the same. (These questions are very easy for most JEE Aspirants)

Me watching even tho I couldn’t even do the first one

@turtlewolf34

4 ай бұрын

lol same

@pepperjohnns

4 ай бұрын

What how?

@The-wu6uo

4 ай бұрын

Thats just bad bruh😂

@xd0895

4 ай бұрын

and i thought not being able to do level 5 was bad

@wyffd

4 ай бұрын

there’s no way

This is very helpful and always fun to study along! Would it be possible to recreate this but in calculus version? Like “calculus problems but they increasingly get more difficult”?

@levels23

9 күн бұрын

i love this format too, i would love to see smth like "11 levels of integrals"

I enjoyed them all! Everyone on first try, the last one on my third. Thank you❤❤

This was quite a fun challenge

Thank you for the video! It was a much needed refresher on trig identities.

For the last one I think after the factorisation it’s simpler to add and subtract 1, leave the sec^2xtan^2x term alone and then you can quickly get the rest you need, the reason adding and subtracting one is obvious is because on the right side it means you can make the differing two terms the same with he extra one, this allowed me to solve the problem in my head very quickly without much ugly manipulation

I recently discovered your channel and it’s been very helpful. Thank you for your work!

The fact that the second one was with some SEC trigonometric function that I have never ever seen even though I am studying calculus 1 is insane😂😂

@extremelynoobgaming4742

4 ай бұрын

Tbh these questions were nowhere near high school level trig

@ishaankumar4587

4 ай бұрын

​@@extremelynoobgaming4742highschool was like till level 5

@ishaankumar4587

4 ай бұрын

Or level 7

All done! For the last identity we can use , a³-b³= (a-b)³ + 3ab(a+b).

Level 8 can be done easier by using cos^2(x)=1-sin^2(x) So we get (1-sin^2x)/sinx(1-sinx) -1 (1-sinx)(1+sinx)/sinx(1-sinx) -1 1+sinx/sinx -1 cscx+1-1 cscx

Thanks a lot professor I follow you from Algeria. Mes salutations. Bonne continuation professeur

For level 9, just divide the LS with cosxcosy separately

In the first one the LHS and RHS have different domains. It's only an identity for the values of x found in the domain of the RHS, not for all x.

franchement chanmé le compte ça fait bien bosser le cerveau là merci le boss

For level 4, I used the LHS. Factored out sin^2(x) and it became sin^2(x) ( sec^2(x) - 1) sec^2(x) - 1 is an identity, which is equal to tan^2(x). sin^2(x)tan^2(x)

Pls make p2, this is helpful revision for me🙏

Hope im cpaable of doing it! Great video!

Goth through all the levls 🔥🔥🔥

I think this one is more complex. Give this one a try: *Prove That* LHS =(tanA+cosecB)² - (cotB-secA)² RHS= 2tanAcotB(cosecA+secB)

The right side of #9, (1+tanx*tany)/(1-tanx*tany) is the tangent addition formula (a+b)/(1-a*b). Where one of them is 45 degrees and the other something else. It seems to come up when I'm playing with complex numbers in trig functions and change of basis. I wasn't aware of the cos(x-y)/cos(x+y) relationship.

The 1st ten were very easy but for the last one it required some thinking until I arrived with both sides as sec^4(x)+sec²xtan²x+tan^4(x)

Solved the all

Can you do full video math for 7 and 8

Never thought i'd see them so fast after my math exam (they never showed up)

While cross multiplication is frowned upon in proving identities, the principle of addition in equations can still be used without any loss of validity or rigor. So in example 8, the problem would have been greatly simplified if 1 got added to both sides. Making a common denominator with that minus 1 does not add anything to the proof.

For the 11th one when u got sex⁴x + tan²xsec²x + tan⁴x [tan²x-sec²x=1] U can write tan²xsec²x as -2tan²xsec²x + 3tan²xsec²x And then it becomes sec⁴x - 2tan²xsec²x +tan⁴x +3tan²xsec²x We can see that sec⁴x - 2tan²xsec²x +tan⁴x is a perfect square [sec²x - tan²x]² And [tan²x-sec²x=1] so it becomes [ -1]² = 1 Therefore sex⁴x + tan²xsec²x + tan⁴x [tan²x-sec²x=1] this whole thing is just 1-3tan²xsec²x

well the last one was quite too easy 1=sec^2x - tan^2x cubing on both sides 1=sec^6x-tan6x-3tan^2xsec^2x(sec^2x-tan^2x) 1=sec^6x-tan^6x-3tan^2xsec^2x 1+3tan^2xsec^2x=sec^6x+tan^6x but overall a great video ;D

For 11 I solved it in a few steps, (1 + 3tan^2x * sec^2x) Simplify (put sec and tanx in terms of sin and cos (3sin^2x + cos^4x)/cos^4x Multiply top and bottom by cos^2x (3sin^2xcos^2x + cos^6x)/cos^6x Convert cos^6x on the top into (cos^2)^3 cos^2x = 1-sin^2x you then get (3sin^2x(1-sin^2x) + (1-sin^2x)^3)/cos^6x from here expand 3sin^2x(1-sin^2x) and expand (1-sin^2x)^3 into 1- 3sin^2x + 3sin^4x - sin^6x you get (3sin^2x - 3sin^4x + 1 -3sin^2x + 3sin^4x - sin^6x)/cos^6x 3sin^2x cancels, 3sin^4x cancels, leaving you with (1- sin^6x) / cos^6x also equal to 1/cos^6x - sin^6x/cos^6x or sec^6x - tan^6x

In level one, you can only cancel cosx only when: cosx is not equal to 0

try making 1+sinx+cosx to a product

You overdid Level 11: tan^2+1=sec^2; expanding left and right sides by replacing secants with tangents in a few steps grants equivalence

If you use other trig rules shouldn't they be proved also? There are values of x that clearly breaking the divide by zero rule. Should they be taken into account during the proof or just sta ted at the end of do we just assume they are part of the identity ?

Nice vid

for the last one LHS = s^6 - t^6 = (1 + t^2)^3 - t^6 = 1 + 3t^2 + 3t^4 = 1 + 3t^2( 1 + t^2) = 1 + 3t^2s^2 = RHS

@parie-_-12

3 ай бұрын

😮You did solve it in just few steps 😲. You are probably a mathematician.

Level 9: LHS: (cosxcosy+sinxsiny)/(cosxcosy-sinxsiny) Divide the numerator and denominator by cosxcosy you get, 1+tanxtany/1-tanxtany = RHS

Insane is geogery series of tantheta ,macularin exapansion of trigo, and de movires theorum these are the insane lvl of trigonometry

I was able to solve all them in my head. Level 8 had me stumped a bit without paper. Level 11 was also super difficult because I couldn't initially remember the identity: a^3-b^3=(a-b)(a^2 + ab + b^2). I do have a PhD in math though. I also noticed some simplified solutions which I'll post below. For level 4, tan^2 x-sin^2 x = tan^2 x sin^2 x * cos^2 x/ cos^2 x = (1-cos^2 x)tan^2 x For level 9, use (1 + tan x tan y)/(1-tan x tan y) = [ (cos x cos y)(1 + tan x tan y)]/[(cos x cos y)(1 - tan x tan y)] then distribute and use sum/difference identies. For level 10, use the complete-the-square technique: tan^4 x + tan^2 x + 1 = [ sin^4 x + 2 sin^2 x cos^2 x + cos^4 x - sin^2 x cos^2 x] / cos^4 x = [ (sin^2 x + cos^2 x) ^2 - sin^2 x cos^2 x] / cos^4 x For level 11, use the formula above (with a=sec^2 x and b=tan^2 x) together with completing-the-square again: sec^6 x - tan^6 x = (sec^2 x - tan^2 x)(sec^4 x + sec^2 x tan^2 x + tan^4 x) = sec^4 x - 2sec^2 x tan^2 x + tan^4 x + 3 sec^2 x tan^2 x = (sec^2 x - tan^2 x)^2 + 3 sec^2 x tan^2 x

can you give the proof of double angle identities and triple ones please

@expl0s10n

4 ай бұрын

Put A=B into compound angle formulas, say cos(A+B)=cos(A)cos(B)-sin(A)sin(B)

Pretty easy just last one was a had bit of length to it.

The first 'identity' is not true. At Pi/2, the LHS is 1, while the right is undefined and is undefined. It is undefined for Pi/2 + n * Pi, where n is an integer; while the LHS is always defined.

@MrJensenMath10

4 ай бұрын

Good noticing that! When proving identities we are only proving they are equal for all values of the variable within their common domain. Which holds true for this example.

@ingiford175

4 ай бұрын

@@MrJensenMath10Now the analytic continuation of each (filling up /0 with the limits at that point)

It was easy till level 10 , took some time for level 11

i went straight for last one did it i in less than 1.5 min u just need to put 1^3=(sec^2-tan^2)^3 and bam its just done

limits but they keep getting harder

managed to do all first try but the last one. just going to pray I don't get something like that on my real A-levels exam because it will be a waste of time.

I was able to do level 11 but not able to do level 8 and 9

Last One is the simplest sec^6x-tan^6x (sec^2x)^3-(tan^2x)^3 a^3-b^3=(a-b)*(a^2+b^2+ab) (sec^2x-tan^2x)(sec^4x+tan^4x+sec^(2)x*tan^2x) Sec^2x-Tan^2x=1 (sec^2x)^2+(tan^2x)^2+sec^2x*tan^2x Now I am ignoring the angle. a^2+b^2=(a-b)^2+2ab (sec^2-tan^2)^2+2sec^2*tan^2+sec^2*tan^2 (1)^2+3sec^2*tan^2 1+3tan^(2)x*sec^(2)x LHS=RHS Hence proved. btw I am in 10th standard, love from India!

@parie-_-12

3 ай бұрын

How was the maths paper of board exams?

@parie-_-12

3 ай бұрын

You can put the value sec²x= (1+tan²x) in LHS in then apply (a+b)³ and your steps will be reduced. I too thought the same way as you did but here I discovered that you can easily reduce the number of steps by doing this as someone has solved like this here.

In EE major and couldn’t even do number 1

Prove that cos(x+y) = CosxCosy - SinxSiny

got all (even level 11) except level 10 😀😀

Level 7 is quite similar to level 2.

Till 10

2:40 can't we just simplify the eqn to cos^2x=(1+sinx)(1-sinx)

14:25 currently in calculus and still would not get that

The last one's level was comparable to my school level question asked in 10th grade cbse board 😂😂 Proud to be indian

i did em all in my head. next time use some radians and more identities like transformation formulae, sum difference formulas for tan and cot etc

That last one took me a while, but all of em were easy

@123ucr

4 ай бұрын

First 9 were very easy. The last two took some work (tedious) but the problems were still simple. These two problems would be Honors Pre-Calculus level problems. Of course, I took Calculus 3, Linear Algebra and Differential Equations so this is far below my math level.

Isn’t the first one not an identity because it doesn’t hold for sin of pi over 2?

@cparks1000000

4 ай бұрын

You're basically claiming that x^2/x = x is not valid because the left expression is undefined at 0 while the right one is; however, the equation dies hold at every nonzero value of x. Technically speaking, the right hand side of problem 1 has a removable singularity at all values given by pi/2 + k pi for any integer k. This function has a unique, continuous extension (i.e., expanding the domain). This extensions is equal to sin(x). More simply put, assume that the equal sign actually means "equal on the largest set of points for which both sides are defined".

For level 11, since you have known already that sec² x - tan² x = 1, sec⁶ x - tan⁶ x = sec⁴ x + tan² x sec² x + tan⁴ x = (sec² x - tan² x) + 3 tan² x sec² x = 1 + 3 tan² x sec² x

@cparks1000000

4 ай бұрын

The second equal sign is unclear and incorrect. Perhaps you missed a square.

I could do number two in probably two hours with a side of headache 😂

the insane one was actually the easiest. Only 3 statements were necessary

@JensenMath -- You are *not* to work on both sides! Pick *one* side only for a given problem and transform it until it becomes the other side.

@imconfused1712

3 ай бұрын

Does it really matter

Level 1-10 is too easy. Only level 11 is a medium difficulty problem

@123ucr

4 ай бұрын

Level 1-9 were easy as pi. Level 10 and Level 11 would be honors level Pre-Calculus exam problems or at medium difficulty, but I was still able to solve them with ease. Of course, I took Calculus 3, Linear Algebra and Differential Equations so this is far below my math level.

@extremelynoobgaming4742

4 ай бұрын

​@@123ucrin our country these were grade 10 level trig but still too easy

@thesoluter605

4 ай бұрын

​​@@123ucri am a 10th standard student from India(2nd year of high school according to US standards) and these are included in our school's course and yeah it's not that hard for us at this level It is quite shocking that u are regarding such a question to be at calculus or honours level

Where are the double angle identities?

@antonlee1890

4 ай бұрын

also usually I only solve from one side

bro these aren't difficult if the only two identities that you need is Pythagorean, sum and difference, and double angle(used ONE time) identities

where is eulers identity

One alternative sol for last question: write sec(x)⁶-tan(x)⁶ as (sec(x)²)³-(tan(x)²)³ then => (tan(x)²+1)³-(tan(x)²)³ Using a³-b³=(a-b)(a²+ab+b²) substitute a = tan(x)²+1 & b = tan(x)² (tan(x)²+1-tan(x)²)*((tan(x)²+1)^2+(tan(x)²+1)tan(x)²+(tan(x)²)^2) Simplifying we get 1*(tan(x)⁴+1+2tan(x)²+tan(x)⁴+tan(x)²+tan(x)⁴) Adding the terms we get 1+3tan(x)⁴+3tan(x)²-> taking 3tan(x)² common from both terms 1+3tan(x)²[1+tan(x)²] -> 1+3tan(x)²sec(x)² { Yea i know it looks kinda messy wish KZread supported mathjax or latex }

10 was super easy but the ones before were a bit hard for me. Especially 9

All completed very easily 😂under 15 sec

xd i failed at the first stage...

ez

this is too easy. My 10th grade midterm test are much harder

What is this

All are easy Level 11 took some time but done 👍🏻

As an indian , i confirm i solved them in my mind

@parie-_-12

3 ай бұрын

10th grade maths is enough except for few involving compound angles one in level 5 and 6 I directly skipped to the end and it was easy

level 8 easier than level 7 team ↓

Erm, I am a professor who has done much work in the field of hyperbolic geometry. These elementary equations simply are not up to my caliber. Why would you make a video teaching high schoolers how to do trigonometric identities when there are people who want to flaunt their knowledge?

@_PK777_

4 ай бұрын

Maybe because it is a challenge for the aforementioned high schoolers? The vid simply isn't for you, not everything revolves around you

@cparks1000000

4 ай бұрын

I'm calling bull.

@kevinmadden1645

4 ай бұрын

😊😢

@kevinmadden1645

4 ай бұрын

So flaunt already!

These are easiest identifies ever😂😂

When you are facing jee this is nothing

too easy

@communist_amoeba3923

4 ай бұрын

Yeah

@123ucr

4 ай бұрын

The last two problems were honors level pre-calculus problems and I still found them pretty easy. Of course, I took Linear Algebra, Differential Equations, and Calculus 3 so this is far below my math level.

Me in grade 8 , i can solve every problem in the vid lol . Me feeling like a god lol. Thx a lot teacher.

What happens to the first one when x = (2k+1)π/2

All 11 is easy. Sadly I am an Engineer

These are all basic.