# How does $\frac{\sqrt{3}}{2} = \sqrt{\frac{3}{4}}$?

I'm studying trigonometry and one of the key triangles (60 30 90) has a side of $\frac{\sqrt{3}}{2}$. I understand that it is derived from an equilateral triangle of sides 1, however I can't see how $\frac{\sqrt{3}}{2} = \sqrt{\frac{3}{4}}$ and why $\frac{\sqrt{3}}{2}$ is simpler to use than $\sqrt{\frac{3}{4}}$. I obtained $\sqrt{\frac{3}{4}}$ from $\sqrt{1^2 - (\frac{1}{2})^2}$

• $\sqrt{\frac{a}{b}} = \frac{\sqrt{a}}{\sqrt{b}}$, for $a,\,b > 0$. Jul 25, 2013 at 10:20
• square both sides. In both cases, you get 3/4. Therefore these positive expressions are themselves equal. Jul 25, 2013 at 10:23
• it's easier to use because you only have one square root left Jul 25, 2013 at 10:24
• I cannot agree more with @DanielFischer . Jul 25, 2013 at 10:35
• @DanielFischer That is a amazing. I never knew that. Thank you.
– Sam
Jul 25, 2013 at 10:37

A brief word on why $\sqrt{p/q} = \sqrt{p}/\sqrt{q}$:

The square root of $x$ is defined to be the positive number that, when squared, gives $x$. So, to prove that $a = \sqrt x$ I just have to prove that $a > 0$ and $a^2 = x$.

Well, $\sqrt p$ and $\sqrt q$ are both positive, so $\frac{\sqrt p}{\sqrt q}$ is going to be positive too. Moreover, $\left(\frac{\sqrt p}{\sqrt q}\right)^2 = \frac{\sqrt p^2}{\sqrt q^2} = \frac{p}{q}$ so $\sqrt{p}/\sqrt{q}$ satisfies all the conditions necessary to be the square root of $p/q$.

If you're aware of how $\sqrt{x}$ can be written as $x^{1/2}$ and $1/x$ can be written as $x^{-1}$, you might also consider $\sqrt{a/b} = \sqrt{ab^{-1}} = (ab^{-1})^{1/2} = a^{1/2}(b^{-1})^{1/2} = a^{1/2}b^{-1/2} = a^{1/2}/b^{1/2}$

With regards to how $\sqrt{3/4}$ is "simpler to use" than $\sqrt{3}/2$, well, it isn't really. But the key is to realise that the former can be written in terms of $\sqrt{3}$, so that if you have something else that can be written in terms of $\sqrt{3}$, you might be able to cancel terms or simplify or something else interesting. The key is to be consistent about what you use: either $\sqrt{3/4}$ everywhere or $\sqrt{3}$ everywhere, so your similar terms really do look the same and you can compare them easily.

$$\sqrt{\frac{3}{4}}=\sqrt{3\cdot4^{-1}}=\sqrt{3\cdot2^{-2}}=\sqrt{3}\cdot2^{-1}=\frac{\sqrt{3}}{2}$$

As to which to use, normally people tend to like to have as little surds as possible.

use this simple properties that $a=\sqrt {a^2}$ where $a>0$ and $\dfrac {\sqrt p}{\sqrt q}=\sqrt{\dfrac {p}{q}}$ where $p,q>0$

$$\dfrac {\sqrt 3}{2}\implies \dfrac {\sqrt 3}{\sqrt 4}\implies\sqrt {\dfrac {3}{4}}$$

for your second question that why $\dfrac {\sqrt 3}{2}$ is simpler to use than other one,I want to say that in every expression we try to remove or simplify surds cause they make calculation troublesome as they go complex.so just try to keep things simple. One of my math teacher used to say that $\sqrt {}$ sign is like a snake who can get down a horse so try to get rid of this sign ASAP.