A.Math Questions－yanahv7的部落格

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A.Math Questions

發問:

1) Solve the following equations for 0°
最佳解答:

樓上 1c) 計左少個答案引用返個答案先 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 = 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°
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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 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)63D0B758E2D502CC