Valid Units of Measure for Trig Argument

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SUMMARY

In trigonometric functions, the argument must be expressed in valid angular units, specifically radians or degrees. When passing a value like cos(3π * 15 seconds) into a trig function, seconds are not valid units for angles and should be canceled out. The expression 3π has units of radians per second, which allows for proper dimensional analysis. Radians are often treated as unitless in calculations, as they represent a ratio of distances.

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Ronnin
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Quick question. If i pass into a trig function something like cos(3pi*15 seconds), do I drop the seconds from the resultant answer since it's not a valid unit of measure for theta?
 
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that's right.
the inside of the bracket for any trig identity has units of rad (or degrees if you think that way)
it's common convention to just drop the units though, since rads aren't really a 'unit'.
 
The argument (the thing in brackets) of any trig function has to be an angle. Seconds are not valid units for angles. If in your example, the 3 \pi has units of radians/second, then the units would work out properly.
 
If your problem is well expressed you will find that the factor (in your case) of 3 \pi has units of 1/s . Thus you do not drop the seconds, they cancel out.
 
Integral said:
If your problem is well expressed you will find that the factor (in your case) of 3 \pi has units of 1/s . Thus you do not drop the seconds, they cancel out.

Are radians when used in this way normally expressed per second?
 
Radians per second are the units of angular speed (\omega), which is a very common quantity in problems dealing with rotational motion, simple harmonic motion, and waves.

Some people prefer not to name the radians explicitly when doing units-analysis because they're not "really" units in the same sense as meters, seconds, etc. Recall that the radian is defined as the (unitless) ratio of two distances: the length along an arc of a circle divided by the radius of the circle.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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