How would you find the exact value of the six trigonometric function of −π/3?

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Answer 1 · 2017-01-29T03:05:43

Use the identities:
#cos(x)=cos(-x)#
#sin(x)=+-sqrt(1-cos^2(x))#
#tan(x)=sin(x)/cos(x)#
#cot(x)=1/tan(x)#
#sec(x)=1/cos(x)#
#csc(x)=1/sin(x)#

It is well known that #cos(pi/3) = 1/2#

Using the identity, #cos(x) = cos(-x)#, then:

#cos(-pi/3) = 1/2#

Using the identity #sin(x) = +-sqrt(1-cos^2(x))#:

#sin(-pi/3) = +-sqrt(1-cos^2(-pi/3))#

Substitute #(1/2)^2# for #cos^2(-pi/3)#

#sin(-pi/3) = +-sqrt(1-(1/2)^2)#

#sin(-pi/3) = +-sqrt(3)/2#

We know that the sine function is negative in this quadrant, therefore, drop the +:

#sin(-pi/3) = -sqrt(3)/2#

Use the identity #tan(x) = sin(x)/cos(x)#

#tan(-pi/3) = (-sqrt(3)/2)/(1/2)#

#tan(-pi/3) = -sqrt(3)#

Use the identity #cot(x) = 1/tan(x)#:

#cot(x) = 1/-sqrt(3) = -sqrt3/3#

Use the identity #sec(x) = 1/cos(x)#

#sec(-pi/3) = 1/cos(pi/3) = 1/(1/2) = 2#

Use the identity #csc(x) = 1/sin(x)#

#csc(-pi/3) = 1/sin(pi/3) = 1/(-sqrt(3)/2) = -2sqrt(3)/3#