急 Trigonometry sin,cos,tan40

Q1.
cos 210 = cos ( 270 - 60) = - sin 60 = - (sqrt3)/2.
sin 330 = sin (360 - 30) = - sin 30 = -1/2.
sin 120 = sin (180 - 60) = sin 60 = (sqrt3)/2
tan 240 = tan (270 - 30) = 1/tan 30 = sqrt 3.
cos 300 = cos ( 360 - 60) = cos 60 = 1/2.
so the expression = {[- sqrt 3 + 1]^2/4 } / { (sqrt 3)/2 x sqrt 3 + 1/2 x sqrt 3}
= [ 3 + 1 - 2 sqrt 3]/4[3/2 + (sqrt 3)/2]
= (2 - sqrt 3)/(3 + sqrt 3)
= ( 2 - sqrt 3)(3 - sqrt 3)/(9 - 3)
= (6 + 3 - 5 sqrt 3)/6
= (9 - 5 sqrt 3)/6.
Q2.
tan 150 = tan (180 - 30) = - tan 30 = - (sqrt 3)/3
sin 240 = sin ( 180 + 60) = - sin 60 = - (sqrt 3)/2
cos 120 = cos ( 180 - 60) = - cos 60 = -1/2
tan 300 = tan ( 360 - 60) = - tan 60 = - sqrt 3
sin 135 = sin (180 - 45) = sin 45 = (sqrt 2)/2
cos 225 = cos ( 180 + 45) = - cos 45 = - (sqrt 2)/2
so the expression = {[( - sqrt 3)/3 - (sqrt 3)/2 ]^2 - [ -1/2 - sqrt 3]^2}/ [ (sqrt 2)/2 x - (sqrt 2)/2]^2
= [3(1/3 + 1/2)^2] - (1/4 + 3 + sqrt 3)]/(-1/2)^2
= [ 3 ( 25/36) - 1/4 - 3 - sqrt 3]/(1/4)
= 4(25/12 - 1/4 - 3 - sqrt 3)
= 25/3 - 1 - 12 - 4 sqrt 3
= - 14/3 - 4 sqrt 3.

2010-05-04 21:50:43 補充：
cos 210 can be = cos ( 180 + 30) = - cos 30 = - (sqrt 3)2.
tan 240 can be = tan ( 180 + 60) = tan 60 = sqrt 3.