How do you find the exact values of the sine, cosine, and tangent of the angle #(-13pi)/12#?

Question

1964 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-01-11T03:11:38

#sin (-(13pi)/12) = 1/2sqrt(2-sqrt(3))#

#cos (-(13pi)/12) = -1/2sqrt(2+sqrt(3))#

#tan(-(13pi)/12) = -sqrt((2-sqrt(3))/(2+sqrt(3)))#

First we note that:

#(13pi)/12 = pi+pi/12#

Remebering that:

#sin(-x) = -sinx#
#cos(-x) = cosx#
#tan(x) = sinx/cosx#

We have:

#sin (-(13pi)/12) = -sin ((13pi)/12) = - sin(pi+pi/12) = -sinpicos(pi/12) - cospisin(pi/12) = sin(pi/12)#

#cos (-(13pi)/12) = cos ((13pi)/12) = cos(pi+pi/12) = cospicos(pi/12) - sinpisin(pi/12) = -cos(pi/12)#

#tan(-(13pi)/12) = sin (-(13pi)/12) / cos (-(13pi)/12) =- sin(pi/12)/cos(pi/12)#

Now we can find the value of #sin(pi/12)# and #cos(pi/12)# from the value of:

#cos(pi/6) = sqrt(3)/2#

#sin(pi/12) = sin(1/2*pi/6)= sqrt((1-cos(pi/6))/2) = sqrt(( 1- sqrt(3)/2)/2)=sqrt((2-sqrt(3))/4)=1/2sqrt(2-sqrt(3))#

#cos(pi/12) = sin(1/2*pi/6)= sqrt((1+cos(pi/6))/2)=1/2sqrt(2+sqrt(3))#

Finally:

#sin (-(13pi)/12) = 1/2sqrt(2-sqrt(3))#

#cos (-(13pi)/12) = -1/2sqrt(2+sqrt(3))#

#tan(-(13pi)/12) = -sqrt((2-sqrt(3))/(2+sqrt(3)))#