How do you use the unit circle to find the exact value for #cos ((7pi)/3)#?
2 Answers
Go
The pattern goes
Or, you can use the additive identities of
#\mathbf(cos(u + v) = cosucosv - sinusinv)#
Using
#color(blue)(cos((7pi)/3))#
#cos((6pi)/3 + pi/3) = cos(2pi + pi/3)#
#= cos2picos(pi/3) - sin2pisin(pi/3)#
#= cos0cos(60^o) - sin0sin(60^o)#
#= 1*cos(60^o) - 0*sin(60^o)#
#= cos(60^o) = color(blue)(1/2)#
1/2
Explanation:
Another way.
cos ((7pi)/3) = cos (pi/3 + 2pi) = cos (pi/3), or cos 60^@
Call M the extremity of arc (pi/3), Call O the origin, and A the origin of all arcs. The triangle MAO is equilateral, since its 3 angles all equal to
Therefor,