# Do the trigonometric functions preserve units?

I saw an exercise where you had to calculate the units of $C_i, i=1,2$ from an equation like this:

• $v^2=2\cdot C_1x$ and
• $x=C_1\cdot \cos(C_2\cdot t)$

where

• $x$ means meters,
• $t$ means seconds and
• $v$ means velocity.

For $C_1$ I got $C_1=m/s^2$. But coming to $C_2$ the cosinus irritates me somehow:

$$x=C_1 \cdot \cos(C_2 t)\Rightarrow m=m/s^2 \cdot \cos(C_2 s)\Rightarrow s^2 = \cos(C_2 s)$$

Does this mean, that $C_2$ must have the unit $s$?

Thanks a lot!

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Not exactly a duplicate, but answered by Fundamental question about dimensional analysis. In short the argument to a trigonometric function must be dimensionless (or a pure angle if you're one of those strange people who like to work in degrees), so either a conversion has been elided or there is a mistake. – dmckee Mar 1 '12 at 23:06
@dmckee: I see know where I made my mistake. Since the result of cos must be dimensionsless, $C_2$ must be $1/s$ and therefore $C_1$ must be $m$. The two equations do not correlate, as I assumed by mistake. Thanks for the link, helped a lot! – Aufwind Mar 1 '12 at 23:22
An easy way to understand why functions such as $\sin$, $\cos$ or $\exp$ must have dimensionless arguments is to note that they have non-trivial power series, e.g. $\exp(x) = 1 + x + x^2/2 + \dots$. Since it doesn't make sense to add two quantities with different units, e.g. $1m + 3m^2$, we know that $x$ must be dimensionless. – Lagerbaer Mar 2 '12 at 4:12