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Silly question about radians and degrees

4 Answers

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sinedegrees(x) = sineradians(x\*π/180)

So integrating sinedegrees(x) over 0..90 if you change variables u = x\*π/180  you get integral sineradians(u) over 0..π/2, but you from du = dx\*π/180 you also get a factor of 180/π ≈ 57.3.

Or if you look at the graphs, the integral in degrees is stretched horizontally by a factor of 90/(π/2) = 180/π.
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Have you learnt about transformations?

Changing from radians to degrees can acts like a horizontal stretch with a factor of 180/pi (roughly 57.3).

This stretch also stretches the area under the curve.
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Let's check units

* 0° to 90° is in the unit **degrees**
* 0 to π/2 is ~~unitless~~ in the unit **radians**
* sin(x) is **unitless**
* dx is the same unit as **x**

When you calculate the area, you **multiply** sin(x) with dx

* Degrees: unitless·degrees = degrees
* Radians: unitless·radians = radians

So you have an **area of 57.3°** or an **area of 1 radians**

If you decide to use **cm** on both x-axis and y-axis, you technically need to adapt the function itself:

    1cm · sin(x · 1rad/1cm)
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Changing the mode in the calculator changes the scaling of the x-axis. If you are in radian mode and graph the curve, then you get one arch of the sine curve on \[0,π\]. If you do it in degrees, then that same arch is stretched over \[0,180\]. In both of these cases, the maximum value of the curve is 1.

Now forget about the sine curve for a minute and imagine doing this with a rectangle. Your height is always 1, but the width is changing from \~3.14 to 180. Naturally, the numeric value of the area is going to change.

This is similar to changing your width units from feet to inches. If you draw a rectangle that is 1 foot wide, then the numeric value of that area is going to be 1/12th of what you'd get if you think of that rectangle as 12 inches wide. But it's the same rectangle.

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