Velocity Reviews > Re: subexpressions

# Re: subexpressions

Steve Howell
Guest
Posts: n/a

 06-02-2007
>
> def f(x): y = x*x; return sin(y)+cos(y);
>

Although I know valid trigonometry is not the point of
this exercise, I'm still trying to figure out why
anybody would ever take the square of an angle.
What's the square root of pi/4 radians?

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Steven D'Aprano
Guest
Posts: n/a

 06-02-2007
On Sat, 02 Jun 2007 05:54:51 -0700, Steve Howell wrote:

>>
>> def f(x): y = x*x; return sin(y)+cos(y);
>>

>
> Although I know valid trigonometry is not the point of
> this exercise, I'm still trying to figure out why
> anybody would ever take the square of an angle.
> What's the square root of pi/4 radians?

Approximately 0.886 radians. It corresponds to the angle of a point on the
unit circle quite close to (sqrt(2/5), sqrt(3/5)), or if you prefer
decimal approximations, (0.632, 0.775).

Angles are real numbers (in the maths sense), so sqrt(pi/4) radians is
just as reasonable an angle as pi/4 radians. Both are irrational numbers
(that is, can't be written exactly as the ratio of two integers).

--
Steven

Stef Mientki
Guest
Posts: n/a

 06-02-2007
Steve Howell wrote:
>> def f(x): y = x*x; return sin(y)+cos(y);
>>

>
> Although I know valid trigonometry is not the point of
> this exercise, I'm still trying to figure out why
> anybody would ever take the square of an angle.
> What's the square root of pi/4 radians?

Maybe he meant
sin(x)^2 + cos(x)^2
which is well known demodulation technique if you create two signals 90 degrees out of phase.

Stef