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標題:
AMATH...compound angles (sum and product formulae)
發問:
1. Without using a calculator,prove that cos 兀/11 + cos 3兀/11 + cos 5兀/11 + cos 7兀/11 + cos 9兀/11 =1/22. Slove the equation sin 3x/2 cos 5x/2 - cos x/2 sin 7x/2 = 0 for 0°≦ x ≦ 360° (hint : sin 3x = 3sin x - 4sin^3x)3. Prove the identity (sin x + sin2x + sin3x)/(cos x + cos2x + cos3x) =... 顯示更多 1. Without using a calculator,prove that cos 兀/11 + cos 3兀/11 + cos 5兀/11 + cos 7兀/11 + cos 9兀/11 =1/2 2. Slove the equation sin 3x/2 cos 5x/2 - cos x/2 sin 7x/2 = 0 for 0°≦ x ≦ 360° (hint : sin 3x = 3sin x - 4sin^3x) 3. Prove the identity (sin x + sin2x + sin3x)/(cos x + cos2x + cos3x) = tan2x 4. In△ABC, prove that cos2A + cos2B + cos2C = -1 - 4cosA cosB cosC
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其他解答:
AMATH...compound angles (sum and product formulae)
發問:
1. Without using a calculator,prove that cos 兀/11 + cos 3兀/11 + cos 5兀/11 + cos 7兀/11 + cos 9兀/11 =1/22. Slove the equation sin 3x/2 cos 5x/2 - cos x/2 sin 7x/2 = 0 for 0°≦ x ≦ 360° (hint : sin 3x = 3sin x - 4sin^3x)3. Prove the identity (sin x + sin2x + sin3x)/(cos x + cos2x + cos3x) =... 顯示更多 1. Without using a calculator,prove that cos 兀/11 + cos 3兀/11 + cos 5兀/11 + cos 7兀/11 + cos 9兀/11 =1/2 2. Slove the equation sin 3x/2 cos 5x/2 - cos x/2 sin 7x/2 = 0 for 0°≦ x ≦ 360° (hint : sin 3x = 3sin x - 4sin^3x) 3. Prove the identity (sin x + sin2x + sin3x)/(cos x + cos2x + cos3x) = tan2x 4. In△ABC, prove that cos2A + cos2B + cos2C = -1 - 4cosA cosB cosC
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1. Without using a calculator,prove that cos 兀/11 + cos 3兀/11 + cos 5兀/11 + cos 7兀/11 + cos 9兀/11 =1/2 let y=cosπ/11+cos3π/11+cos5π/11+cos9π/11 2ysinπ/11 =sin2π/11+sin4π/11-sin2π/11+sin6π/11-sin4π/11+sin8π/11-sin6π/11sin10π/11-sin8π/11 【sinAcosB=1/2[sin(A+B)+sin(A-B)]】 =sin10π/11 So 2ysinπ/11=sin10π/11 2ysinπ/11=sin(π-π/11) 2ysinπ/11=sinπ/11 2y=1 y=1/2 Thus cosπ/11+cos3π/11+cos5π/11+cos9π/11=1/2 2. Slove the equation sin 3x/2 cos 5x/2 - cos x/2 sin 7x/2 = 0 for 0°≦ x ≦ 360° (hint : sin 3x = 3sin x - 4sin^3x) sin 3x/2 cos 5x/2=1/2[sin4x-sinx] cos x/2 sin 7x/2=1/2[sin4x+sin3x] Now sin 3x/2 cos 5x/2 - cos x/2 sin 7x/2 = 0 1/2[sin4x-sinx]-1/2[sin4x+sin3x]=0 sin4x-sinx-sin4x-sin3x=0 sinx+sin3x=0 sinx+3sin x - 4sin^3x=0 4sinx(1-sin^2x)=0 sinxcos^2x=0 sinx=0 or cosx=0 when sinx=0 x=0,180,360 when cosx=0 x=90, 270 So x=0,90,180,270,360 3. Prove the identity (sin x + sin2x + sin3x)/(cos x + cos2x + cos3x) = tan2x sin x+sin 2x +sin 3x = sin 2x + (sin 3x +sin x) = sin 2x + {2sin[( x +3x)/2]cos[( 3x - x)/2] = sin 2x + 2(sin 2x )(cos x ) = (sin 2x )(1+2cos x ) cos x+cos 2x +cos 3x = cos 2x + (cos 3x + cos x) = cos 2x + {2cos[( 3x + x)/2]cos[( 3x - x)/2] = cos 2x + 2(cos 2x )(cos x ) = (cos 2x )(1+2cos x ) (sin x+sin 2x +sin 3x )/(cosx+cos 2x+cos 3x ) = (sin 2x )(1+2cos x )/(cos 2x )(1+2cos x ) = sin 2x /cos 2x = tan 2x 4. In△ABC, prove that cos 2A + cos2B + cos 2C = -1 - 4cosA cosB cosC 證明： ∵A、B、C為△ABC的三內角． ∴A+B+C=π，即C=π-(A+B) ∴原式左邊 =cos 2A + cos2B + cos 2C =2cos(A+B)cos(A-B)+2 cos 2C -1 =2cos(A+B)cos(A-B)+2 cos2(A+B)-1 =2cos(A+B)[cos(A+B)+cos(A-B)]-1 =4cos(A+B)cosAcosC-1 =-1-4cosAcosBcosC其他解答:
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