Why does sin 2(30°) = sin 2(60°)

sin 2(30º) = sin 60º

sin 2(60º) = sin 120º

Think about the angles

120º is a 60º reference angle in quad II

sin (x) is positive in both quad I and quad II.

Sin(30) = 1/2
Sin^2(30) = (1/2)^2 = 1/4

Sin(60) = sqrt(3) / 2
Sin^2(60) = 3/4

So Sin^2(30) does NOT equal Sin^2(60)

However,
if you mean
Sin(2 x 30) = Sin(60) = sqrt(3)/2
Sin(2 x 60) = Sin(120) = sqrt(3)/2

If you trace the #Sine curve from '0' to ;180; you will notice the curve rises to 1 @ 90 and then falls to0 @ 180
Inspecting the 'y' - vertical axis you will notice that the 'y' axis values are the same at 'sqrt(3)/ 2 at these two angular values.

The question becomes why does sin60 equal sin120?

Draw a unit circle with center at (0,0). Draw a radius in quadrant I making a 60 degree angle with the positive x axis. Sin 60 equals the length of the vertical line to the x axis, divided by the hypotenuse which is 1. Draw a radius in quadrant II making a 120 degree angle with the positive x axis. Draw a vertical line from that radius to the negative x axis. Sin 120 equals the length of that vertical line , divided by the hypotenuse which is 1. But that second vertical line is the vertical line of a triangle whose angle supplementary to the 120 angle is a 60 degree angle. The two sines are equal.

So you are asking why sin(60°) = sin(120°)

sin(180°-x) = sin180°cosx - cos180°sinx
sin(180°-x) = 0cosx - (-1)sinx
sin(180°-x) = sinx . . . this is also clear from looking at the graph of sin
Substitute x = 60°.
sin(120°) = sin(60°)

sin(60) = sin(120)

Note that from the unit circle, sin(60) = √3/2, which is 60 degrees above the positive x-axis.

Note that 120 is 60 degrees above the negative x-axis and will yield the same result since sine is positive in this case as well.

complimentary angles around 180° are always of the same sin and of equal absolute value in the cosine but of opposite signs, this is because they are complimentary in twin quadrants.