Are These Radian Equations Correct?

[paulo84]

$$0 = 2π/3$$.

$$π/6 =3π/4$$

$$π/4 = 5π/6$$

$$π/3 = π$$

$$π/2 = 3π/2$$

$$2π = 2π$$

$$π/2 + 3π/2 = 4π/2$$

I find this a little confusing but quite wonderful. But, what is it?

 

[DrClaude]

But, what is it?

You tell us. Apart from the last two equations, what you wrote are false statements.

 

[BvU]

May be wonderful, but almost all are simply wrong ! (exceptions: the last two are correct)

 

[fresh_42]

But, what is it?

Under the assumption that $2/3 \pi$ is meant to be $\dfrac{2}{3}\pi$ ,
it is an unconventional way to write zero by the letter $\pi$.

 

[paulo84]

oh dear. still figuring out Latax...

 

[lekh2003]

I understand that you were experimenting with LaTeX, but what was this?

 

[paulo84]

I think it's fixed. Can someone explain?

 

[DrClaude]

I think it's fixed. Can someone explain?

Explain what?

For the last time, you need to explain what you mean or this thread will be closed.

 

[paulo84]

Explain what?

For the last time, you need to explain what you mean or this thread will be closed.

It's just it seems to indicate 1=1/3 and 1=3. That's all...

 

[DrClaude]

It's just it seems to indicate 1=1/3 and 1=3. That's all...

But take you first equality:

$$0 = 2π/3$$.

which is not true. Where does this come from?

 

[paulo84]

But take you first equality:

which is not true. Where does this come from?

https://cnx.org/contents/i4nRcikn@2.69:blZd_oDf@3/Trigonometric-Functions

 

[DrClaude]

I didn't see those "equalities" on that page. You might be confusing the fact that $f(x) = f(y)$ with $x=y$, but this of course not true. Consider
$$
f(x) = x^2
$$
or, even "worse,"
$$
f(x) = \text{const.}
$$

 

[lekh2003]

https://cnx.org/contents/i4nRcikn@2.69:blZd_oDf@3/Trigonometric-Functions

First of all, look at the table carefully. 120 degrees is $\frac{2π}{3}$.

Second, that is a conversion, there is no equal sign.

 

[paulo84]

I didn't see those "equalities" on that page. You might be confusing the fact that $f(x) = f(y)$ with $x=y$, but this of course not true. Consider
$$
f(x) = x^2
$$
or, even "worse,"
$$
f(x) = \text{const.}
$$

I misunderstood the table.

 

[lekh2003]

I misunderstood the table.

Yes, yes you did. Did you get that sleep which you said you would get? Please take some rest and return to the forum.

 

[paulo84]

Too much time on Google Analytics.

 

[Mark44]

As noted, all but the last two equations are untrue.
Thread closed.