How can I show the same HTML 5 Video twice on a website without loading it twice?

This can be done in some very easy steps via Javascript and the Canvas Element:

HTML:

<video autoplay id="previewVideo" data-videoid="JYpUXXD4xgc">
    <source src="video.php?videoid=JYpUXXD4xgc" type="video/mp4"/>
</video>    
<canvas id="bigVideo"></canvas>

JavaScript:

document.addEventListener('DOMContentLoaded', function() {
  var v = document.getElementById('previewVideo');
  var canvas = document.getElementById('bigVideo');
  var context = canvas.getContext('2d');
  var cw = Math.floor(canvas.clientWidth);
  var ch = Math.floor(canvas.clientHeight);
  canvas.width = cw;
  canvas.height = ch;
  v.addEventListener('play', function() {
    updateBigVideo(this, context, cw, ch);
  }, false);
}, false);


function updateBigVideo(v, c, w, h) {
  if (v.paused || v.ended) return false;
  c.drawImage(v, 0, 0, w, h);
  setTimeout(updateBigVideo, 20, v, c, w, h);
}

The canvas fetches the image of the video and displays it again on the BigVideo.
The updateBigVideo() function is called every 20ms, resulting in a framerate of about 50 FPS.

Read more

First, make the <video> element using JavaScript and then put it in the places you want.

var video1 = document.createElement("video");
video1["data-videoid"] = "JYpUXXD4xgc";
var sourceElem = document.createElement("source");
sourceElem.src = "video.php?videoid=JYpUXXD4xgc";
sourceElem.type = "video/mp4";
video1.appendChild(sourceElem);

var video2 = video1.cloneNode(true); //This makes a copy of the element, but makes sure it's not treated as the same element. This means you can add video1 AND this _different_ element to the document. However, unfortunately, everything still needs to get loaded again. I think this is the easiest way to copy an element over, though.
video2.id = "bigVideo";
video1.id = "previewVideo";

document.addEventListener("DOMContentLoaded", function() {
    //Now put video1 and video2 where you want.
});