Area Under Sine Wave: Anti Diff of -cosx

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SUMMARY

The area under the sine wave from 0 to 90 degrees, when calculated using the anti-derivative of -cos(x), confirms that the area is indeed 1 unit, assuming the angle is measured in radians. Specifically, the calculation involves evaluating -cos(90) and subtracting -cos(0), resulting in an area of 0 - (-1) = 1. It is crucial to note that the derivatives d(sin x)/dx = cos x and d(cos x)/dx = -sin x hold true only when x is expressed in radians, highlighting the importance of unit consistency in calculus.

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  • Understanding of basic calculus concepts, including derivatives and anti-derivatives.
  • Familiarity with trigonometric functions, specifically sine and cosine.
  • Knowledge of radians as a unit of angular measurement.
  • Basic grasp of integral calculus and area under curves.
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philrainey
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A little question when working out the area under a sin wave from 0 degrees to 90 degrees or some lower angle if you take the anti diff of -cosx, does that give the area under the graph assumming the angle is in radians? so enter -cos90=0 subtract -cos0 or-1 the area under the sine wave is 0--1 or one unit (what is this unit) one radian squared?
 
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Yes, in a simplified sense, d(sin x)dx = cos x and d(cos x)/dx= -sin x are only true if x is in radians.

I say "in a simplified sense" because the ways in which sin(t) and cos(t) are defined in more advanced mathematics, t is not an angle at all.
 

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