# really easy trig question

• Aug 31st 2005, 03:04 AM
michael5139
really easy trig question
Hi,

I used to know this, now I can't remember.

Why is sin (pi/3) (rad) = (root 3)/2

Michael
• Aug 31st 2005, 03:37 AM
ticbol
Were you familiar with angles in degrees before? Like 60 degrees is pi/3 radians? Like 30 degrees is pi/6 radians?
If you were, then, maybe you still recall that:
"In a 30-60-90-degree right triangle, the leg opposite the 30 degrees is half in measure of the hypotenuse."

So the 30-60-90 right triangle has these:
hypotenuse = 2 units
leg opposite the 30-deg angle = 1 unit

the other leg, or the leg opposite the 60-deg angle is then sqrt(3) units.
why? because by Pythagorean theoem,
(one leg)^2 +(the other leg)^2 = (hypotenuse)^2
x^2 +1^2 = 2^2
x^2 +1 = 4
x^2 = 4 -1
x^2 = 3
x = sqrt(3)

Then, maybe you still know that:
sine = (opposite side)/(hypotenuse)

So if the angle is the 60 degrees,
sin(60deg) = (sqrt(3)) / 2

Then, since pi/3 rad = 60 deg, it follows that
sin(pi/3) = (sqrt(3)) / 2

------------
Why is pi/3 rad = 60 degree?

1 revolution = 360 deg
1 revolution = 2pi rad also
So,
1 rev = 1 rev