How do you find the exact value of #cos((11pi)/3)#?

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Answer 1 · 2018-04-25T03:56:29

#1/2#

Given: #cos ((11 pi)/3)#

The cosine function has a period of #2 pi#. This means the values of the cosine repeat every period.

See if #(11 pi)/3# is #> 2pi#

#(2 pi)/1 = (2 pi)/1 * 3/3 = (6 pi)/3#

#(11 pi)/3 = (6 pi)/3 + (5 pi)/3 = 2pi + (5 pi)/3#

Yes #(11 pi)/3# is #> 2pi#

The equivalent cosine value of #(11 pi)/3# is #(5 pi)/3#

On a trigonometric circle, the #cos ((5 pi)/3) = 1/2#

Answer 2 · 2018-04-25T03:56:30

# cos( {11 pi} / 3 ) = 1 / 2 #

I'm not sure why we have an entire subject dedicated to just two right triangles (30,60,90 and 45,45,90) but we sure seem to.

# \cos( {11 pi}/3 ) #

#= cos({11 pi}/3 - 4pi )#

# = cos( -pi/3) #

# = cos(pi/3) #

# = cos 60^circ #

# = 1/2 #