Trigonometry Help: Understanding 1/2^2 in Terms of Sine, Cosine, and Tangent

[seasnake]

Okay, I must admit, my trigonometry is rather awful...

anyway, I would like to write out what 1 / 2^2 is equal to in terms of sine, cosine, tangent, and the like

is the following correct, or how do I write it correctly (or what would be the correct figures for 0.5^0.5):

0.5^0.5 = a 45-degree angle = cos (45) = sin (45) = a tangent of 1

 

[Mentallic]

is the following correct, or how do I write it correctly (or what would be the correct figures for 0.5^0.5):

0.5^0.5 = a 45-degree angle = cos (45) = sin (45) = a tangent of 1

Yes this is correct. What you are looking for is a value \theta where
sin(\theta)=\frac{1}{\sqrt{2}}
And yes, you correctly noted that the isosceles right-angled triangle has adjacent and opposite sides (to the angle \theta) of value 1 and hypotenuse of value \sqrt{2}.

anyway, I would like to write out what 1 / 2^2 is equal to in terms of sine, cosine, tangent, and the like

Did you mean 1/2^(1/2)? If you actually meant 1/4 then you won't have a 'nice' simple value for \theta. Don't worry, this isn't uncommon.

The best answer you can give for \theta to
sin(\theta)=\frac{1}{4}

Is: \theta=arcsin(\frac{1}{4})\approx 14.48^o

this answer is just an acute angle, and I'm sure you're aware that there are more (actually, infinite) values of \theta that satisfy this result?

 

[seasnake]

thanks... but I did mean exactly what I wrote 0.5^0.5, which equates to a value around .71something or other

 

[Mentallic]

Yeah I thought so. It just put me off when you wrote:

I would like to write out what 1 / 2^2 is equal to

and the value .71 something IS \frac{1}{\sqrt{2}} and isn't 1/2^2