Sine/cosine question - basic

[lucy189]

I am reviewing some geometry and am stumped by something very basic. I just can't see it. In an isosceles triangle PRQ, with sides of 1, 1, and (sqrt 2), where the hypotenuse is PR, why is sin P expressed as (sqrt 2)/2, rather than 1/(sqrt 2)?

Thanks for your help.

 

[pka]

You have an isosceles right triangle.
By definition the sine is length of the opposite divided by the length of the hypotenuse.
Now your question is about simple algebra:

\(\displaystyle \L
\frac{1}{{\sqrt 2 }} = \frac{1}{{\sqrt 2 }}\frac{{\sqrt 2 }}{{\sqrt 2 }} = \frac{{\sqrt 2 }}{2}\).

 

[soroban]

Hello, Lucy!

Ever hear of "rationalizing the denominator"?


Traditionally, it is not a good idea to leave radicals in the denominator.

Back in 1950 B.C. (before calculators), we had to use long division on \(\displaystyle \frac{1}{\sqrt{2}}\)

You can imagine the set up: \(\displaystyle \:1.414213562\:\overline{)\:1.00000000...}\)

while \(\displaystyle \frac{\sqrt{2}}{2}\) looks like this: \(\displaystyle \;2\;\overline{)\;1.414213562...}\) . . . much easier!


Have you ever had to add: \(\displaystyle \:\frac{1}{2}\,+\,\frac{1}{\sqrt{2}}\) ? . . . What's the common denominator?

If we write it like this: \(\displaystyle \:\frac{1}{2}\,+\,\frac{\sqrt{2}}{2}\) . . . it's easy! . . . We already have a common denominator.

\(\displaystyle \;\;\)Answer: \(\displaystyle \:\frac{1\,+\,\sqrt{2}}{2}\)