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This does not quite work if you define sin(x) in terms of the ratio of the opposite side and the hypotenuse of a right angled triangle because you cannot have a triangle in which one of the angle measures 0 degrees.

One approach is to consider the limiting value of sin(x) as x approaches 0 from below and from above. But you would have to first prove (or assume) that sine is a continuous function at 0.

Another way is to consider sin(x) as the sum of the infinite series

sin(x) = x/1! - x3/3! + x5/5! - ... (where x is measured in radians).

This is the Taylor series definition of the sine function.

Clearly, every term is 0 when x = 0 and so you have sin(0) = 0

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13y ago

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Q: How is the value of sin0 is 0?
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