How do you evaluate sine, cosine, tangent of #-pi/6# without using a calculator?

2 Answers
Nov 18, 2017

#sin(-pi/6)=-1/2#
#cos(-pi/6)=sqrt2/2#
#tan(-pi/6)=-sqrt3/3#

Explanation:

Recall that:
#sin(-x)=-sin(x)#
#cos(-x)=cos(x)#
#tan(-x)=-tan(x)#

and,
#pi/6#/#(30^@)# is a special angle.

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Hence,
#sin(-pi/6)=-sin(pi/6)#
#color(white)(xxxxx....)=-1/2#

#cos(-pi/6)=cos(pi/6)#
#color(white)(xxxxxx..)=sqrt2/2#

#tan(-pi/6)=-tan(pi/6)#
#color(white)(xxxxxx..)=-sqrt3/3#

Nov 18, 2017

#sin(-pi/6)=-1/2#
#cos(-pi/6)=sqrt(3)/2#
#tan(-pi/6)=-1/sqrt(3)=-sqrt(3)/3#

Explanation:

Draw an equilateral triangle with unit lengths. Each angle is #pi/3#.

Now, bisect one of the angles. This gives two right triangles.

GeoGebra

Look at the right triangle to the right. The hypotenuse is #1#. The angle marked is #pi/6#. The side opposite the angle is #1/2#. The side adjacent the angle can be found using Pythagoras' Theorem #sqrt(1^2-(1/2)^2)=sqrt(3)/2#.

Then, #sin(pi/6)=(1/2)/1=1/2# (opposite over hypotenuse); #cos(pi/6)=(sqrt(3)/2)/1=sqrt(3)/2# (adjacent over hypotenuse); #tan(pi/6)=(1/2)/(sqrt(3)/2)=1/sqrt(3)#, or #sqrt(3)/3# (opposite over adjacent).

Recall the following identities (or use a unit circle):
#sin(-x)=-sin(x)# (since #sin# is an odd function)
#cos(-x)=cos(x)# (since #cos# is an even function)
#tan(-x)=-tan(x)# (since #tan# is an odd function)

Thus,
#sin(-pi/6)=-1/2#
#cos(-pi/6)=sqrt(3)/2#
#tan(-pi/6)=-1/sqrt(3)=-sqrt(3)/3#