Here's an alternative way - check out this fancy pic of the unit circle:
To use this, you just need to know that because sin = opposite / hypotenuse, in the unit circle sin = y/1. [The hypotenuse is the radius of the unit circle, which is 1]
(Similarly, cos = x/1 and tan = y/x)
E.g. sin(0), look at the point (1,0). sin(0) = y = 0.
sin(pi/2) = y = 1
sin(pi) = y = 0
sin(3pi/2) = y = -1.
^ For these boundary values, when I was first starting out, the unit circle was how I remembered the values. Eventually I was so familiar with the graphs that it was faster to use these instead.
I found the pi/4 value easy to remember, because it is half-way between 0 degrees and 90 degrees (pi/2). This is where the x-value is the same as the y-value, so the cosine value is the same as the sine value.
From the Pythagorean identity, we know that x^2 + y^2 = cos(pi/2)^2 + sin(pi/2)^2 = 1 (the hypotenuse).
Therefore x = y = sqrt(2) /2, which is the exact value for both cos(pi/2) and sin(pi/2)
Notice that the Pythagorean identity works for all of the values. You can use it as a simple check to see if your triangle is correct, if you choose to use that method. Personally I found that using the triangle for pi/3 and pi/6 was the fastest.
 \cos \left (\frac{3 \pi}{7} \right ) \cos \left (\frac{5 \pi}{7} \right ) = -\frac{1}{8} )
