Hi,

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

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

Michael

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- Aug 31st 2005, 03:04 AMmichael5139really 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 AMticbol
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

2pi rad = 360 deg

Divide both sides by 2,

pi rad = 180 deg

Divide both sides by 3,

pi/3 rad = 180/3 deg

pi/3 rad = 60 deg