A.Math Questions

樓上 1c) 計左少個答案

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1a)cosx﹣cos3x = sinx﹣sin3x
-2 cos 2x cosx = 2 cos 2x sinx
2 cos 2x sinx + 2 cos 2x cosx = 0
2 cos 2x (sinx + cosx ) = 0
cos 2x = 0 or sinx + cosx = 0
2x = 90 or 270 or sinx = -cosx
x = 45° or 135° or tan x = -1
or x = 135° or 315°
∴ x = 45°,135°or 315°

b)sin(30﹣x) cos(30+x) = √3/2
{[sin[(30﹣x)+(30+x)]+ sin[(30﹣x)﹣(30+x)]}/2 = √3/2
sin(60) + sin(-2x) = 2sqrt(3)/2
sin(-2x) = 2√3/2 - sin(60)
sin(-2x) = (2√3/2)﹣(√3/2)
sin(-2x) = √3/2
-sin(2x) = √3/2
sin(2x) = -√3/2
2x = 240°, 300°, 600°, 660°
∴x = 120°, 150°, 300°, 330°

c)cos (x + 15°) cos (x﹣15°)﹣sin x sin(x﹣30°) = cos165°
(1/2) (cos 2x + cos 30°) + (1/2) [cos (2x﹣30°)﹣cos 30°] = -cos15°
(1/2) cos 2x + (1/2) cos (2x﹣30°) + cos15° = 0
(1/2) [cos 2x + cos (2x﹣30°)] + cos15°= 0
(1/2) (2) [cos (2x﹣15°)cos 15°] + cos15°= 0
cos 15°[cos (2x﹣15°) + 1] = 0
cos (2x﹣15°) + 1 = 0
cos (2x﹣15°) = - 1
改動過 => 2x﹣15° = 180° or 540°
(由於 0°<=x<=360°, 2x - 15° -15°<=x<=705°)
改動過 => x = 97.5° or 277.5°

2)sinA+sinB+sinC
= 2sin[(A+B)/2]cos[(A﹣B)/2] + sin2(C/2)
= 2sin[90°﹣(C/2)]cos[(A﹣B)/2] + 2sin(C/2)cos(C/2)
= 2cos(C/2)cos[(A﹣B)/2]+2cos[90°﹣(C/2)]cos(C/2)
= 2cos(C/2){cos[(A﹣B)/2]+cos[90°﹣(C/2)]}
= 4cos(C/2) cos{[(A﹣B)/2]+[90°﹣(C/2)] /2}cos{[(A﹣B)/2]﹣[90°﹣(C/2)] /2}
= 4cos(C/2)cos[(A﹣B+180°﹣C)/4] cos[(A﹣B﹣180°+C)/4]
= 4cos(C/2)cos[A+(180°﹣B﹣C)/4] cos[﹣B﹣(180°﹣C﹣A)/4]
= 4cos(C/2)cos[(A+A)/4] cos[(﹣B﹣B)/4]
= 4cos(C/2)cos(A/2)cos(B/2)
= 4cos( A/2)cos(B/2)cos(C/2)

2007-12-29 23:11:49 補充：
sorry 1a) 都係2x = 90° or 270° or 540° or 630° or sinx = -cosx(由於 0° <= x <= 360°, 0° <= 2x <=720°)x = 45° or 135° or 270° or 315° or tan x = -1or x = 135° or 315°∴ x = 45°,135°, 270° or 315°

1a)cosx﹣cos3x = sinx﹣sin3x
-2 cos 2x cosx = 2 cos 2x sinx
2 cos 2x sinx + 2 cos 2x cosx = 0
2 cos 2x (sinx + cosx ) = 0
cos 2x = 0 or sinx + cosx = 0
2x = 90 or 270 or sinx = -cosx
x = 45° or 135° or tan x = -1
or x = 135° or 315°
∴ x = 45°,135°or 315°

b)sin(30﹣x) cos(30+x) = √3/2
{[sin[(30﹣x)+(30+x)]+ sin[(30﹣x)﹣(30+x)]}/2 = √3/2
sin(60) + sin(-2x) = 2sqrt(3)/2
sin(-2x) = 2√3/2 - sin(60)
sin(-2x) = (2√3/2)﹣(√3/2)
sin(-2x) = √3/2
-sin(2x) = √3/2
sin(2x) = -√3/2
2x = 240°, 300°, 600°, 660°
∴x = 120°, 150°, 300°, 330°

c)cos (x + 15°) cos (x﹣15°)﹣sin x sin(x﹣30°) = cos165°
(1/2) (cos 2x + cos 30°) + (1/2) [cos (2x﹣30°)﹣cos 30°] = -cos15°
(1/2) cos 2x + (1/2) cos (2x﹣30°) + cos15° = 0
(1/2) [cos 2x + cos (2x﹣30°)] + cos15°= 0
(1/2) (2) [cos (2x﹣15°)cos 15°] + cos15°= 0
cos 15°[cos (2x﹣15°) + 1] = 0
cos (2x﹣15°) + 1 = 0
cos (2x﹣15°) = - 1
2x﹣15° = 180°
x = 97.5°

2)sinA+sinB+sinC
= 2sin[(A+B)/2]cos[(A﹣B)/2] + sin2(C/2)
= 2sin[90°﹣(C/2)]cos[(A﹣B)/2] + 2sin(C/2)cos(C/2)
= 2cos(C/2)cos[(A﹣B)/2]+2cos[90°﹣(C/2)]cos(C/2)
= 2cos(C/2){cos[(A﹣B)/2]+cos[90°﹣(C/2)]}
= 4cos(C/2) cos{[(A﹣B)/2]+[90°﹣(C/2)] /2}cos{[(A﹣B)/2]﹣[90°﹣(C/2)] /2}
= 4cos(C/2)cos[(A﹣B+180°﹣C)/4] cos[(A﹣B﹣180°+C)/4]
= 4cos(C/2)cos[A+(180°﹣B﹣C)/4] cos[﹣B﹣(180°﹣C﹣A)/4]
= 4cos(C/2)cos[(A+A)/4] cos[(﹣B﹣B)/4]
= 4cos(C/2)cos(A/2)cos(B/2)
= 4cos( A/2)cos(B/2)cos(C/2)