How do you use the unit circle to find the exact value for #cos ((7pi)/3)#?

2 Answers
Jun 12, 2015

#cos((7pi)/3)# is just #cos(2pi + pi/3)#. Since #cos(2pi) = cos 0#, #cos 2pi = 1#.

Go #pi/3# (#60^o#) past that, and you'll have #cos((7pi)/3) = cos(420^o) = cos(60^o)#. You'd go down two #30^o# steps from the value of 1, which becomes #1 -> sqrt3/2 -> 1/2#.

The pattern goes #1, sqrt3/2, 1/2, 0, -1/2, -sqrt3/2, -1, -sqrt3/2, -1/2, 0, 1/2, sqrt3/2, 1# at every #30^o#.

Or, you can use the additive identities of #cos#.

#\mathbf(cos(u + v) = cosucosv - sinusinv)#

Using #u + v = (7pi)/3#, we get:

#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)#

Mar 5, 2016

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 #60^@#.
#cos pi/3#, or cos 60^@, is equal to half of the radius OA = 1.
Therefor, #cos (pi/3) = 1/2#