Video wird geladen...

Video konnte nicht geladen werden

Beim Laden dieses Videos ist ein Problem aufgetreten. Dies könnte an einem vorübergehenden Netzwerkproblem liegen oder das Video ist möglicherweise nicht verfügbar.

sin²x + cos²x = 1’in görsel anlatımı.

Bilim Dünyası

1,036,556 subscribers

107,940 Aufrufe • vor 9 Tagen •via X (Twitter)

Bildung Wissenschaft & Technologie

Anya Rossi• Live Now

Private livecam show

3 Kommentare

Profilbild von Fully Charged Zone

Fully Charged Zonevor 9 Tagen

I asked you nicely to stop using my copyrighted material, and you chose not to. I have now received confirmation that a copyright violation occurred and action has been taken. At this point, I’m done asking. I’ll be turning this matter over to an attorney and pursuing the legal options available to me.

Profilbild von CareNotes

CareNotesvor 9 Tagen

rsel anlat.

Profilbild von حيدر أمير الداود

حيدر أمير الداودvor 9 Tagen

.

Ähnliche Videos

sin²x + cos²x =1

sin²x + cos²x =1

The Math Flow

44,420 Aufrufe • vor 5 Monaten

sin²x + cos²x =1

sin²x + cos²x =1

Math Hub

12,551 Aufrufe • vor 3 Monaten

(a+b)² formülünün görsel anlatımı.

(a+b)² formülünün görsel anlatımı.

Bilim Dünyası

39,354 Aufrufe • vor 1 Monat

cos²x+sin²x =1 (Pythagora in the unit circle) Visualization| Animation | Interactive applet GeoGebra #geogebra #math #science #iteachmath #mtbos #visualization #elearning Bernat ANCOCHEA

cos²x+sin²x =1 (Pythagora in the unit circle) Visualization| Animation | Interactive applet GeoGebra #geogebra #math #science #iteachmath #mtbos #visualization #elearning Bernat ANCOCHEA

Daniel Mentrard

244,933 Aufrufe • vor 3 Monaten

Hypnosis, 233 chars #つぶやきGLSL for (O *= m; m < 170.;O += .0007 / (abs(length( a + abs(sin(m* mix(.02,.07, sin(f)*.5 + .5) - f)) * vec2(cos(x = m*.05-f), sin(x)))-.5) + .02) * (1.+cos(m++*.1 + length(a)*6. - f + vec4(0,1,2,0))));

Hypnosis, 233 chars #つぶやきGLSL for (O = m; m < 170.;O += .0007 / (abs(length( a + abs(sin(m mix(.02,.07, sin(f).5 + .5) - f)) vec2(cos(x = m.05-f), sin(x)))-.5) + .02) (1.+cos(m++.1 + length(a)6. - f + vec4(0,1,2,0))));

kishimisu

38,594 Aufrufe • vor 2 Jahren

$t=0,d=5e-4,draw=_=>{t++||createCanvas(w=400,w);background(9).stroke(w,96);for(x=y=z=9,i=3e4;i--;point((q=x*(e=sin(t*PI/20-x*x/99+i%9)+1)+89)*cos(k=z/59-e/29+t*PI/480+i%9*8)+200,200-(q+60*cos(k/2))*sin(k)))[x,y,z]=[x+9*(y-x)*d,y+(x*(28-z)-y)*d,z+(x*y-z-z)*d]}//#つぶやきProcessing$

t=0,d=5e-4,draw=_=>{t++||createCanvas(w=400,w);background(9).stroke(w,96);for(x=y=z=9,i=3e4;i--;point((q=x(e=sin(tPI/20-xx/99+i%9)+1)+89)cos(k=z/59-e/29+tPI/480+i%98)+200,200-(q+60cos(k/2))sin(k)))[x,y,z]=[x+9(y-x)d,y+(x(28-z)-y)d,z+(xy-z-z)d]}//#つぶやきProcessing

ア

59,919 Aufrufe • vor 4 Monaten

x=u=v=f=0,n=200,draw=o=>{for(f||createCanvas(W=400,W),r=2*PI/n,background(0),i=0;i<n;i++)for(j=0;j<55;j+=.1)u=sin(i+v+f)-sin(R=r*i+x-99*f),v=cos(i+v+f)-cos(R),x=u+f,X=n+99*u,Y=n+99*v,stroke(W,2*j+66),line(X,Y,X+1,Y+1);f+=5e-5};//#つぶやきProcessing based on Basic code by ZXDunny

x=u=v=f=0,n=200,draw=o=>{for(f||createCanvas(W=400,W),r=2PI/n,background(0),i=0;i<n;i++)for(j=0;j<55;j+=.1)u=sin(i+v+f)-sin(R=ri+x-99f),v=cos(i+v+f)-cos(R),x=u+f,X=n+99u,Y=n+99v,stroke(W,2j+66),line(X,Y,X+1,Y+1);f+=5e-5};//#つぶやきProcessing based on Basic code by ZXDunny

Koma Tebe

26,068 Aufrufe • vor 1 Jahr

a=(x,y,d=5*cos(o=mag(k=x/8-25,e=y/8-25)/3))=>[(q=x/2+k/atan(9*cos(e))*sin(d*4-t))*sin(c=d/3-t/8)+200,(y/3+d+q)/2*cos(c)+200] t=0,draw=$=>{t||createCanvas(w=400,w);background(6,19);for(t+=PI/60,y=99;++y<300;)for(x=99;++x<300;stroke(x,y,y+x,36))point(...a(x,y))} #つぶやきProcessing

a=(x,y,d=5cos(o=mag(k=x/8-25,e=y/8-25)/3))=>[(q=x/2+k/atan(9cos(e))sin(d4-t))sin(c=d/3-t/8)+200,(y/3+d+q)/2cos(c)+200] t=0,draw=$=>{t||createCanvas(w=400,w);background(6,19);for(t+=PI/60,y=99;++y<300;)for(x=99;++x<300;stroke(x,y,y+x,36))point(...a(x,y))} #つぶやきProcessing

ア

41,075 Aufrufe • vor 1 Jahr

a=(x,y,d=abs(e=cos(k=x/8-12.5)+sin(y/24)+cos(k/2)))=>point((q=x/4+90+d*k*(1+cos(d*4-t*2+y/72)))*cos(c=y*e/594-t/8+d/6)+200,(q/2+99*cos(c/2))*sin(c)+e*6+200) t=0,draw=$=>{t||createCanvas(w=400,w);background(6).stroke(w,36);for(t+=PI/60,i=4e4;i--;)a(i%200,i/200)}#つぶやきProcessing

a=(x,y,d=abs(e=cos(k=x/8-12.5)+sin(y/24)+cos(k/2)))=>point((q=x/4+90+dk(1+cos(d4-t2+y/72)))cos(c=ye/594-t/8+d/6)+200,(q/2+99cos(c/2))sin(c)+e*6+200) t=0,draw=$=>{t||createCanvas(w=400,w);background(6).stroke(w,36);for(t+=PI/60,i=4e4;i--;)a(i%200,i/200)}#つぶやきProcessing

ア

98,863 Aufrufe • vor 1 Jahr

a=(x,y,o=mag(k=x/8-12.5,e=y/8-9)/4)=>stroke(255,36).point((q=x+y+e/k+o*k*cos(y/3)*sin(o*2-t)*o/2)/3*atan(3*sin(c=o*e/40-t/8))+200,o*sin(x/9)-y*cos(c)+q/4*cos(c)+200) t=0,draw=$=>{t||createCanvas(w=400,w);background(6);for(t+=PI/120,i=2e4;i--;)a(i%200,i/200<<1)}#つぶやきProcessing

a=(x,y,o=mag(k=x/8-12.5,e=y/8-9)/4)=>stroke(255,36).point((q=x+y+e/k+okcos(y/3)sin(o2-t)o/2)/3atan(3sin(c=oe/40-t/8))+200,osin(x/9)-ycos(c)+q/4*cos(c)+200) t=0,draw=$=>{t||createCanvas(w=400,w);background(6);for(t+=PI/120,i=2e4;i--;)a(i%200,i/200<<1)}#つぶやきProcessing

ア

25,793 Aufrufe • vor 1 Jahr

a=(x,y,d=5*cos(o=mag(k=x/8-12,e=y/13-14)/2))=> point((q=x/2+10+1/k+k*cos(e)*sin(d*8-t))*sin(c=d/3+t/8)+sin(d*2+t)*k+200,(y/4+5*o*o+q*cos(c*3))/2*cos(c)+200) t=0,draw=$=>{t||createCanvas(w=400,w);background(6).stroke(w,66);for(t+=PI/60,i=2e4;i--;)a(i%200,i/200)}#つぶやきProcessing

a=(x,y,d=5cos(o=mag(k=x/8-12,e=y/13-14)/2))=> point((q=x/2+10+1/k+kcos(e)sin(d8-t))sin(c=d/3+t/8)+sin(d2+t)k+200,(y/4+5oo+qcos(c3))/2cos(c)+200) t=0,draw=$=>{t||createCanvas(w=400,w);background(6).stroke(w,66);for(t+=PI/60,i=2e4;i--;)a(i%200,i/200)}#つぶやきProcessing

ア

33,212 Aufrufe • vor 1 Jahr

Finding the roots of a polynomial is one of the oldest and most central problems in Algebra. Once the coefficients are allowed to move through the complex plane, the roots start revealing a much richer geometry. Here is the Mathematics, try it yourself and play around with the coefficients and polynomial structure: P(x) = x¹² + a x⁸ + b x⁶ + c x⁴ + d x + 1 a = (sin(t₁) + i cos(sin(t₂) + 0.5 sin(2t₂))) M b = 0.3 (t₁ + t₂) Mᶜ c = (exp(i t₁) - cos(t₂)) M d = (cos(t₁) - 0.5 i sin(sin(t₂) + 0.5 sin(2t₂))) Mᶜ M = 1 + ∑ₖ₌₁ᵐ Aₖ exp(i 2π sₖ n / N) Mᶜ = 1 + ∑ₖ₌₁ᵐ Aₖ exp(-i 2π sₖ n / N) Here t₁ and t₂ lie on the unit circle, n is the frame index, N is the total number of frames, and Aₖ, sₖ are the modulation amplitudes and speeds.

Finding the roots of a polynomial is one of the oldest and most central problems in Algebra. Once the coefficients are allowed to move through the complex plane, the roots start revealing a much richer geometry. Here is the Mathematics, try it yourself and play around with the coefficients and polynomial structure: P(x) = x¹² + a x⁸ + b x⁶ + c x⁴ + d x + 1 a = (sin(t₁) + i cos(sin(t₂) + 0.5 sin(2t₂))) M b = 0.3 (t₁ + t₂) Mᶜ c = (exp(i t₁) - cos(t₂)) M d = (cos(t₁) - 0.5 i sin(sin(t₂) + 0.5 sin(2t₂))) Mᶜ M = 1 + ∑ₖ₌₁ᵐ Aₖ exp(i 2π sₖ n / N) Mᶜ = 1 + ∑ₖ₌₁ᵐ Aₖ exp(-i 2π sₖ n / N) Here t₁ and t₂ lie on the unit circle, n is the frame index, N is the total number of frames, and Aₖ, sₖ are the modulation amplitudes and speeds.

Mathelirium

23,042 Aufrufe • vor 5 Monaten

a=(x,y,d=5*cos(o=mag(k=x/8-25,e=y/8-25)/3))=>[(q=x/2+k/atan(9*cos(e))*sin(d*4-t))*sin(c=d/3-t/8)+200,(y/4+5*o*o+q)/2*cos(c)+200] t=0,draw=$=>{t||createCanvas(w=400,w);background(6).stroke(255,96);for(t+=PI/60,y=99;++y<300;)for(x=99;++x<300;)point(...a(x,y))} //#つぶやきProcessing

a=(x,y,d=5cos(o=mag(k=x/8-25,e=y/8-25)/3))=>[(q=x/2+k/atan(9cos(e))sin(d4-t))sin(c=d/3-t/8)+200,(y/4+5oo+q)/2cos(c)+200] t=0,draw=$=>{t||createCanvas(w=400,w);background(6).stroke(255,96);for(t+=PI/60,y=99;++y<300;)for(x=99;++x<300;)point(...a(x,y))} //#つぶやきProcessing

ア

36,330 Aufrufe • vor 1 Jahr

a=(x,y,o=mag(k=x/4-12.5,e=y/9)/9)=>point((q=x/3+99+3/k*sin(y)+k*(1+cos(y)/3+sin(e+o*4-t*2)))*cos(c=o/5+e/4-t/8)+200,(q+49)*sin(c)*cos(c)-q/3+30*o+220) t=0,draw=$=>{t||createCanvas(w=400,w);background(6).stroke(w,46);for(t+=PI/90,i=2e4;i--;)a(i%100,i/350)}// #つぶやきProcessing

a=(x,y,o=mag(k=x/4-12.5,e=y/9)/9)=>point((q=x/3+99+3/ksin(y)+k(1+cos(y)/3+sin(e+o4-t2)))cos(c=o/5+e/4-t/8)+200,(q+49)sin(c)cos(c)-q/3+30o+220) t=0,draw=$=>{t||createCanvas(w=400,w);background(6).stroke(w,46);for(t+=PI/90,i=2e4;i--;)a(i%100,i/350)}// #つぶやきProcessing

ア

123,912 Aufrufe • vor 1 Jahr

a=(x,y,o=mag(k=x/4-12.5,e=y/9+5)/9)=>point((q=x+99+tan(1/k)+o*k*(cos(e*9)/4+cos(y/2))*sin(o*4-t))*.7*sin(c=o*e/30-t/8)+9*cos(y/19+t)+200,200+q/2*cos(c)) t=0,draw=$=>{t||createCanvas(w=400,w);background(6).stroke(w,46);for(t+=PI/90,i=2e4;i--;)a(i%100,i/100)} //#つぶやきProcessing

a=(x,y,o=mag(k=x/4-12.5,e=y/9+5)/9)=>point((q=x+99+tan(1/k)+ok(cos(e9)/4+cos(y/2))sin(o4-t)).7sin(c=oe/30-t/8)+9cos(y/19+t)+200,200+q/2cos(c)) t=0,draw=$=>{t||createCanvas(w=400,w);background(6).stroke(w,46);for(t+=PI/90,i=2e4;i--;)a(i%100,i/100)} //#つぶやきProcessing

ア

55,880 Aufrufe • vor 1 Jahr

a=(x,y,d=mag(k=x/8-25,e=y/8-25)**2/99)=>[(q=x/3+k*.5/cos(y*5)*sin(d*d-t))*sin(c=d/2-t/8)+e*sin(d+k-t)+200,(q+y/8+d*9)*cos(c)+200] t=0,draw=$=>{t||createCanvas(w=400,w);background(6).stroke(255,96);for(t+=PI/60,y=99;y<300;y+=5)for(x=99;++x<300;)point(...a(x,y))}#つぶやきProcessing

a=(x,y,d=mag(k=x/8-25,e=y/8-25)**2/99)=>[(q=x/3+k.5/cos(y5)sin(dd-t))sin(c=d/2-t/8)+esin(d+k-t)+200,(q+y/8+d9)cos(c)+200] t=0,draw=$=>{t||createCanvas(w=400,w);background(6).stroke(255,96);for(t+=PI/60,y=99;y<300;y+=5)for(x=99;++x<300;)point(...a(x,y))}#つぶやきProcessing

ア

366,607 Aufrufe • vor 1 Jahr

a=(x,y,k=x/8-12.5,d=abs(e=cos(k)+sin(y/3)+cos(k/2)))=>point((q=x/4+90+d*k*(.5+cos(d*4-t*2+y/12)))*.7*(cos(c=y*e/198-t/8+d/5))+200,(q+70)*.7*sin(c)+200) t=0,draw=$=>{t||createCanvas(w=400,w);background(0,96).stroke(w,26);for(t+=PI/90,i=4e4;i--;)a(i%200,i/500)}//#つぶやきProcessing

a=(x,y,k=x/8-12.5,d=abs(e=cos(k)+sin(y/3)+cos(k/2)))=>point((q=x/4+90+dk(.5+cos(d4-t2+y/12))).7(cos(c=ye/198-t/8+d/5))+200,(q+70).7*sin(c)+200) t=0,draw=$=>{t||createCanvas(w=400,w);background(0,96).stroke(w,26);for(t+=PI/90,i=4e4;i--;)a(i%200,i/500)}//#つぶやきProcessing

ア

22,186 Aufrufe • vor 1 Jahr