Відео

Thank you 😊 for your help

like the music :)

Looks like a job for the shoelace formula.

What app do you use to write?

What program do you use to write?

Good teaching sir❤❤❤ From deevige classes bangalore

For grade school children who are not conversant with linear algebra, this problem can be solved graphically. A rectangle with sides parallel to the X and Y axes, bounding the given triangle, will have an area of 9x7=63. There are three triangles in the corners of this rectangle, exterior to the original triangle. These triangles will have areas of 1/2 x (9x2)=18/2, 1/2 x (5x2)=10/2, and 1/2 x (7x7)= 49/2. The original triangle then has an area of 63- (18/2 + 10/2 + 49/2) = 49/2.

Beautiful !

MathandBolt, Can we collab?

interesting

Keep it up

May we know what book are you following for this course (Classical Mechanics)?

this was very easy to understand.

Thanks man 👍

I'm glad we have finally worked out how to calculate the circumference of a bald head.

🙌🙌👍🏻

Is a reason for the trigonometry is confused. Thanks for demostrate the formula... 🤓

Well explained sir ,thanks ,

Well explained sir ,thanks ,

Promo sm

that was the best and must easiest proof for this formula great job

Pretty good, but it's not centrifugal, but centripetal. Also, this force points towards the center, not outwards

beautiful

Wow! This exercise was well explained and easy to understand. Please, keep it on.

Problems specking English !

What software do you use to write?

Very cool

I solved similarly to the first method The second method was genius 🔥

Pretty interesting identity Obviously the general form is tan(A+B)-tanA-tanB=tan(A+B)xtanAxtanB One thing I think you should do is do the problem with more of a general approach not just a single case of say 8x,6x,2x Still really cool video 🔥

I solved this using the trig identity and using the sinA=x,sinB=y CosA=p,cosB=q It just makes the writing less messy and at last I got sin^2 B as the final solution

... I enjoyed your clear and surprisingly simple/elegant process from L.H.S. to R.H.S. , by introducing just a sudden 1/SQRT(2) ... its simplicity is fascinating! I looked for another way and found ... ( COS8 - SIN8 )/( COS8 + SIN8 ) = TAN(37) ... also starting from L.H.S. and dividing top and bottom by COS8 ... ( 1 - TAN8 )/( 1 + TAN8 ) ... ( TAN45 - TAN8 )/( TAN45 + TAN8 ) .... applying TAN(A - B) = (TANA - TANB) / (1 + TANA*TANB) .... TAN(45 - 8) = ( TAN45 - TAN8 )/( TAN45 + TAN45*TAN8 ) ... TAN45 - TAN8 = ( TAN45 + TAN8 )*TAN(37) ... finally ( TAN45 + TAN8 )*TAN(37)/( TAN45 + TAN8 ) = TAN(37) ... the R.H.S. ... note: TAN45 = 1 , just for the record (lol) ... thank you sir for your continuing math efforts ... Jan-W

... Good day to you sir, With this trigonometric expression you have created a new representative for the number " 0 " ... quite impressive I must say (lol) ... we could also have written ... SIN(A - B) / [ COS(A - B) - SINA*SINB ] + SIN(B - C) / [ COS(B - C) - SINB*SINC ] + SIN(C - A) / [ COS(C - A) - SINC*SINA ] ... now we can observe that the quotients of the terms are not quite the tangents of their angle differences (lol) ... your presentation shows why trigonometry is always a lot of fun and interesting sir ... thank you ... best regards, Jan-W

tan56 can be converted to tan(45+11). From here it takes two steps to prove it.

Thanks for your trigonometric problem 🙏🙏🙏

Thanks for your proof problem of trigonometry.....🙏🙏🙏

Thanks for your quality problem of trigononetry. I always wait for your problem. I will give this problem to my students. Warm regards from Indonesia. 🙏🏻🙏🏻🙏🏻

X=(4/3)

Thanks for this❤

Nice

Can you solve this for me, please Q1/ Ψ(r, 0) = (1/√7)(2ψ100 + ψ210 + ψ211 + √3ψ21,-1) ,Calculate the value of the uncertainty (ΔLxΔLy) and explain in detail. Q2/For the particle of mass m in the one dimensional box with width a, the wave function of the particle at time (t = 0) inside the box is Ψ(x) = Asin(3πx/2a)cos(πx/2a) 1- Find Ψ(x, t > 0). 2- A measurement is made of the energy. What energies can be found? What is the probability of obtaining each value of the energy?

How can I teach maths online

... At about time 2:03 ... 2X - 3 = (X - 3) * A + (X + 6) * B we could also find the values of A & B by ... (1) Set X = 3 ... cancelling A , then 6 - 3 = 9 * B , so B = 1/3 ... (2) Set X = - 6 ... cancelling B , then - 12 - 3 = - 9 * A , so A = 5/3 ... etc ... but of course this is nothing new to you sir ! thank you ... Jan-W

... A good new year's day sir, I guess your presentation regarding partial fractions is completely clear to me, so I have nothing further to comment on and wish you all the best for 2024 with great instructive mathematics ahead, but above all a good HEALTH .... best regards and thank you sir, Jan-W

... Good day sir, [ SEC(T) - TAN(T) ] / [ SEC(T) - TAN(T) ] + [ TAN(T) - SEC(T) ] / [ TAN(T) - SEC(T) ] = 1 + 1 = 2 , that's my work ... and now on to watching your presentation sir .... thanking you for these very effective knowledge testing exercises ... Jan-W

... Good day to you, The 3rd integral can also be solved as follows ... INT[1/(e^X + 1)]dX .... rewrite the numerator as follows ... 1 = e^X + 1 - e^X ... INT[ (e^X + 1)/(e^X + 1) - e^X/(e^X + 1) ]dX .... INT[ 1 ]dX - INT[ e^X/(e^X + 1) ]dX ... applying INT[ F'(X)/F(X) ]dX = LN I F(X) I + C on the 2nd integral ... X - LN I e^X + 1 I + C ... fascinating to see more strategies to solve the same problem ... thanking you sir for showing me a new solution path for this integral ... Best regards, Jan-W

... My apologies sir, I made a few stupid typos when typing my trig. work from paper to my comment on your channel, so I corrected my mistakes; while I was evaluating my result ( must be - COS^2(T)/- COS^2(T) = 1 ) I found the typos ... again my apologies ... however my result on paper was indeed 1... anyhow again thank you for your instructive reply .... Best regards, Jan-W

... Good day sir, When observing your exercise I directly think of the " UNIT CIRCLE " ... NUMERATOR : (1) SIN(180 + T) = - SIN(T) (2) COS(90 + T) = - SIN(T) (3) TAN(270 - T) = COT(T) = COS(T)/SIN(T) (4) COT(360 - T) = - COT(T) = - COS(T)/SIN(T) ... DENOMINATOR : (1) SIN(360 - T) = - SIN(T) (2) COS(360 + T) = COS(T) (3) CSC(- T) = - 1/SIN(T) (4) SIN(270 + T) = - COS(T) ... finally multiplying all NUMERATOR factors (1) to (4) ... result - COS^2(T) & DENOMINATOR factors (1) to (4) .... result - COS^2(T) ... so, NUMERATOR result / DENOMINATOR result = - COS^2(T) / - COS^2(T) = 1 ... WOWW! ... all that work for just an outcome of 1 ?! (lol) .... to confirm this result, I will watch your presentation with some tension (lol) ... thank you sir for your instructive (labour intensive) videos ... Best regards, Jan-W p.s The UNIT CIRCLE is possibly the best tool for Trigonometry in my opinion ....

... Let T = Thêta ... In the given expression I recognize the identity ( A - B ) * ( A + B ) = A^2 - B^2 , where A = 1 + COT(T) & B = SEC(pi/2 + T) ... ( 1 + COT(T) )^2 - ( SEC(pi/2 + T) )^2 .... [ We know the world famous identity 1 + COT^2(T) = CSC^2(T) and COS(pi/2 + T) = - SIN(T) , so SEC(pi/2 + T) = - 1/SIN(T) ] ... then ( 1 + COT(T) )^2 = 1 + 2*COT(T) + COT^2(T) = CSC^2(T) + 2*COT(T) and ( SEC(pi/2 + T) )^2 = ( - 1/SIN(T) )^2 = 1/SIN^2(T) = CSC^2(T) ... finally CSC^2(T) + 2*COT(T) - CSC^2(T) = 2*COT(T) ... the answer I think ! And now I have to watch your presentation sir (lol) ... thank you for your efforts into instructive education ... best regards, Jan-W

... Good day sir, That's exactly how I got the derivative of X^X explained via LD, and later on I read in a math textbook ... Y = X^X = e^(LN(X^X)) = e^(X*LN(X)) ... and then of course Y'= e^(X*LN(X)) * (X*LN(X))' = e^(X*LN(X)) * (1 + LN(X)) = X^X*(1 + LN(X)) ... but of course the goal of your presentation is showing Logarithmic Differentiation ... math at its best .... thank you sir for reminding me LD ... best regards, Jan-W

... Great Trig. exercises sir ... COT^2(X) indeed (lol) ... I will definitely recommend your Channel " MathandBolt " to my tutoring students here in Holland ... thank you sir and happy December ... Jan-W