use half angle indentities and the fact that sin(pi/6)=1/2 to find cos(-pi/12) and sin (-pi/12)?

Answer 1

sin(pi/6) = (1/2)

cos(-pi/12) = cos(pi/12) = cos((pi/6)/2)
cos((pi/6)/2) = sqrt((1 + cos(pi/6))/2) =
sqrt((1 + (sqrt(3)/2))/2) = sqrt(((2 + sqrt(3))/2)/2) =
sqrt((2 + sqrt(3))/4) = (1/2)sqrt(2 + sqrt(3))

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

ANS :
cos(-pi/12) = (1/2)sqrt(2 - sqrt(3))
sin(-pi/12) = (-1/2)sqrt(2 - sqrt(3))

Answer 2

Half-angle identities:

sin (x / 2) = sqrt[ (1 - cos x) / 2 ]
cos (x / 2) = sqrt[ (1 + cos x) / 2 ]

cos(pi/6) = sqrt[3] / 2
[You can find this using either the 30-60-90 right triangle, or, as suggested, the fact that sin^2 x + cos^2 x = 1 allows you to find cosine if you know sine].

sin (pi / 12) = sqrt[ (1 - cos (pi / 6) ) / 2 ]
cos (pi / 12) = sqrt[ (1 + cos (pi / 6) ) / 2 ]

sin (pi / 12) = sqrt[ (1 - sqrt[3] / 2) ) / 2 ]
cos (pi / 12) = sqrt[ (1 + sqrt[3] / 2) ) / 2 ]

But they ask for the sine and cosine of -pi/12:

sin(-x) = -sin (x) --> sin (-pi / 12) = - sin (pi / 12)
cos(-x) = cos (x) --> cos (-pi / 12) = cos (pi / 12)

sin (-pi / 12) = - sqrt[ (1 - sqrt[3] / 2) ) / 2 ]
cos (-pi / 12) = sqrt[ (1 + sqrt[3] / 2) ) / 2 ]


[By the way, you don't need to memorize the half-angle identities. Just derive them from the double angle identities for cosine:

cos 2x = 1 - 2sin^2 x
cos 2x = 2cos^2 x - 1

Solve for sin x and cos x in each equation and then substitute x/2 for x.]

Answer 3

i never memorized those...i just filled the space with as many numbers, symbols, and scribbles as possible so to only lose 1 out of 10 points...it works.