'tan (pi/4+x)/tan(pi/4-x) = [(1 + tanx)(1- tanx)] ^2' || tan(π/4+x)/tan(π/4-x)=(1+tanx/1-tanx)^2
cos(3pi/4+x)-cos(3pi/4-x)=-sqrt(2)sinx || Prove that `cos(3pi/4+x)-cos(3pi/4-x)=-sqrt(2)sinx`
曲線間の面積: y = cos(pi*x)、y = 4x^2 -1
なぜパイがここにいるのですか？そしてなぜ四角いのでしょうか？バーゼル問題に対する幾何学的な答え
cos(pi/4+x)+cos(pi/4-x)=root2.cosx || example15 chapter3 11th math
#Trigonometry all formulas
Prove that `cos(pi/4+x)+cos(pi/4-x)=sqrt(2)cosx` ....
lim〗┬(x→0)〖〖sin〗^2(π 〖cos〗^4x )/x^4 〗 is equal to:
Application of Derivatives | CET & JEE Competitive Exams | MATHEMATICS | Part 7 of 7
IIT Bombay CSE 😍 #shorts #iit #iitbombay
India vs japan || mathematics challenge || 😅🤣🤣🤭
Memorization Trick for Graphing Functions Part 1 | Algebra Math Hack #shorts #math #school
Graph curve with parametric equations x = 4 cos t, y = 2 sin t [0, 2 pi]. Orientation. Cartesian
Japanese Method #shorts
IIT Bombay Lecture Hall | IIT Bombay Motivation | #shorts #ytshorts #iit
Sketch region enclosed by curves y = 4x^2 -1 and y = cos pi x. Find area of region
That's Why Mohit Sir Called "God Of Mathematics"| Puzzle Brain teaser | #competishun #shorts #tricks
Jaldi Wahan Se Hato! IIT Delhi version! #iit #iitjee #iitdelhi
フーリエ級数: 1/n^4 の合計 = pi^4/90、1/(2n-1)^2 の合計 = pi^2/8
数学者が退屈するとき (ep1)
