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# Trig help

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How do you convert any trigonometric value, ie, sin60 into rationalized form from it's standard decimal form?

Guest Mar 19, 2017

#1
+7155
+12

The easiest way is to look at a unit circle.

But not all angles have a rational sin or cosine.

The only ones that are on the unit circle (that I know of!) are 0º, 30º, 45º, 60º, 90º, and those same ones "flipped over" into the second, third, and fourth quadrants.

If you don't have a unit circle you can find the values by just drawing a triangle and using what you know about trig and the pythagorean theorem to find the values.

To use your example, this is how sin60º can be found:

First just draw out what you know sin 60º is. That's the black triangle.

And since there are 180º in every triangle, the remaining angle is 30º.

Next draw a triangle identical to it flipped over.

All the angles are 60º! This makes an equilateral triangle.

Since it is equilateral, the bottom line is also 1.

The black horizontal line is half that: 1/2.

(which is the cosine of 60º, but we are looking for sin.)

Now just go back to the good ol' pythagorean theorem

to find the vertical leg of that right triangle, aka sin60º.

(1/2)2 + (sin60º)2 = 12

sin60º = √(1 - 1/4)

sin60º = √(3/4)

sin60º = √(3) / 2

hectictar  Mar 19, 2017
#1
+7155
+12

The easiest way is to look at a unit circle.

But not all angles have a rational sin or cosine.

The only ones that are on the unit circle (that I know of!) are 0º, 30º, 45º, 60º, 90º, and those same ones "flipped over" into the second, third, and fourth quadrants.

If you don't have a unit circle you can find the values by just drawing a triangle and using what you know about trig and the pythagorean theorem to find the values.

To use your example, this is how sin60º can be found:

First just draw out what you know sin 60º is. That's the black triangle.

And since there are 180º in every triangle, the remaining angle is 30º.

Next draw a triangle identical to it flipped over.

All the angles are 60º! This makes an equilateral triangle.

Since it is equilateral, the bottom line is also 1.

The black horizontal line is half that: 1/2.

(which is the cosine of 60º, but we are looking for sin.)

Now just go back to the good ol' pythagorean theorem

to find the vertical leg of that right triangle, aka sin60º.

(1/2)2 + (sin60º)2 = 12

sin60º = √(1 - 1/4)

sin60º = √(3/4)

sin60º = √(3) / 2

hectictar  Mar 19, 2017