If you know the values of sin(π3) and cos(π3), you can write that
tan(π3)=sin(π3)cos(π3)=√3212=√32(21)=√3
Alternatively, you could think of this as tan(60˚), and then draw a 30˚−60˚−90˚ triangle:
tan(60˚) will be equal to oppositeadjacent in reference to the 60˚ angle, so we see that opposite=√3 and adjacent=1. Hence,
tan(60˚)=oppositeadjacent=√31=√3
We can also examine the unit circle at π3:
If we know the point (12,√32), we can determine tangent if we think about tangent as the slope of the line in the unit circle. Since the line originates at (0,0), its slope is
tan(π3)=√32−012−0=√3
This idea of slope=ΔyΔx is analogous to tangent because the sine values correlate to the y values of the ordered pair, and cosine with x, so remembering that tan(x)=sin(x)cos(x) and that tangent is slope should be fairly intuitive.
