# Trigonometry help wanted

1. Jun 20, 2009

### 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

2. Jun 21, 2009

### Mentallic

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 $\sqrt{2}$.

Did you mean 1/2^(1/2)? If you actually meant 1/4 then you wont 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?

3. Jun 21, 2009

### seasnake

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

4. Jun 21, 2009

### Mentallic

Yeah I thought so. It just put me off when you wrote:
and the value .71 something IS $$\frac{1}{\sqrt{2}}$$ and isn't 1/2^2