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In session 8, the professor offers a geometric/visual proof of the fact that lim of sin(x)/x as x goes to 0 = 1. I think I'm missing some basic trig. At one point he mentions that the length of the right side of the triangle (the one next to the arc) is equal to sin(theta). I don't get how that's possible. If sin = opposite/hypotenuse, then how could the length of that opposite side be the same length as the ratio of o/h?
Is it simply because of the fact that the radius is 1, so the hypotenuse is 1 and since sin(theta) = o/h its just o/1 so therefore the opposite is = sin(theta)?
Which also explains why the arc length is equal to theta? Meaning, since we have radius of 1, and the angle theta is equal to arc length/r and r is 1, so then solving for arc length we have an equation of arc length = theta * r = theta?
 one year ago
 one year ago
In session 8, the professor offers a geometric/visual proof of the fact that lim of sin(x)/x as x goes to 0 = 1. I think I'm missing some basic trig. At one point he mentions that the length of the right side of the triangle (the one next to the arc) is equal to sin(theta). I don't get how that's possible. If sin = opposite/hypotenuse, then how could the length of that opposite side be the same length as the ratio of o/h? Is it simply because of the fact that the radius is 1, so the hypotenuse is 1 and since sin(theta) = o/h its just o/1 so therefore the opposite is = sin(theta)? Which also explains why the arc length is equal to theta? Meaning, since we have radius of 1, and the angle theta is equal to arc length/r and r is 1, so then solving for arc length we have an equation of arc length = theta * r = theta?
 one year ago
 one year ago

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