nhrock3
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-\frac{\pi}{4}=arctan(\frac{\frac{-2L}{3}}{R})\\
why it equals
\frac{\frac{-2L}{3}}{R}=-1
?
why it equals
\frac{\frac{-2L}{3}}{R}=-1
?
Last edited:
The equation $\frac{\pi}{4}=arctan(\frac{\frac{-2L}{3}}{R})$ holds true when the expression $\frac{\frac{-2L}{3}}{R}$ equals -1. This equivalence arises from the properties of the arctangent function, specifically that $\tan\left(-\frac{\pi}{4}\right) = -1$. By taking the tangent of both sides, it is confirmed that $\frac{-2L}{3} = -R$, leading to the conclusion that $R = \frac{2L}{3}$.
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Take the tangent of both sides.nhrock3 said:-\frac{\pi}{4}=arctan(\frac{\frac{-2L}{3}}{R})\\
why it equals
\frac{\frac{-2L}{3}}{R}=-1
?