I have a real simple question about the Pythagoras theorem

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The discussion clarifies that the Pythagorean theorem specifically relates to the sides of a right triangle and its hypotenuse, rather than the sine and cosine functions. Participants note that the confusion arises from mixing these concepts, as the sine and cosine functions are defined based on the triangle's angles. One contributor highlights that the relationship between angles can be expressed through tangent functions. The conversation emphasizes the importance of distinguishing between the theorem and trigonometric functions. Understanding these distinctions is crucial for applying the Pythagorean theorem correctly.
rgtr
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Homework Statement
There are 2 angles c and d. If I pick angle d instead of angle c. Does opposite become adjacent and adjacent becomes opposite?
https://www.grc.nasa.gov/www/BGH/sincos.html
Relevant Equations
## tan(d) = \frac o a ##
Here is the link.
https://www.grc.nasa.gov/www/BGH/sincos.html

Sorry just a little rusty on Pythagoras theorem.

I mean the formula still holds but in order to find the opposite and the adjacent the opposite becomes the adjacent and vice versa .
 
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Yes.
 
Thanks.
 
@rgtr, your question has nothing to do the theorem of Pythagoras, which involves the two sides of a right triangle and the hypotenuse. It doesn't distinguish between the two sides.
What you're asking about is more relevant to how the sine and cosine functions are defined for a right triangle.
 
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Likes Delta2, SammyS and hutchphd
Sorry I was in school a few years ago. I just remember learning them all together.
 
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if the angles c and d are the two angles (besides the right angle) of a right triangle then yes it is as you say.

A generalization of your observation is that for two angles ##x## and ##y## such that ##x=\frac{\pi}{2}-y## it holds that $$\tan x=\frac{1}{\tan y}$$
 
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I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks

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