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A.maths~急
1.given that sinA+sinB=a , cosA+cosB=b find that sin(A+B) and cos(A+B) with a and b.Show the step clearly. 2.find the value cos(180/11)+cos(180x3/11)+cos(5x180/11)+cos(7x180/11)+cos(9x180/11) without calculator. 更新: I can not see the answer 2.. 更新 2: I want to ask you one more question... Find cos(180/7)cos(2x180/7)cos(4x180/7) without calculator. Thx~
最佳解答:
1) a^2 + b^2 = sin^2 A + 2 sin A sin B + sin^2 B + cos^2 A + 2 cos A cos B + cos^2 B = 2 + 2 cos a cos B+ 2 sin A sin B = 2 + 2 cos ( A - B ) So, cos ( A - B ) = ( a^2 + b^2 - 2 ) / 2 ab = ( sin A + sin B )( cos A + cos B ) = sinAcosA + sinAcosB + sinBcosA + sinBcosB = sin(A+B) + (1/2)(sin2A) + (1/2)(sin2B) = sin(A+B) + (1/2)(2)[sin(A+B)+cos(A-B)] ab = sin(A+B)+sin(A+B)+cos(A-B) ab = 2sin(A+B)+( a^2 + b^2 - 2 ) / 2 2sin(A+B) = ab- ( a^2 + b^2 - 2 ) / 2 2sin(A+B)= (-1/2)(a^2-2ab+b^2)+1 = (-1/2)(a-b)^2+1 = (-1/4)(a-b)^2+1/2 =[2-(a-b)^2]/4 cos(A+B)=sqr[14+(a-b)^2]/4 2) Please go to http://upload.u-state.com/images/1196174134.JPG 2007-11-28 16:47:50 補充： Sorry for the technical probelms, plx go tohttp://i238.photobucket.com/albums/ff245/chocolate328154/Maths004.jpg 2007-11-28 23:37:29 補充： Plx go to http://i238.photobucket.com/albums/ff245/chocolate328154/ScreenHunter_071.jpgfor the extra Q. 2007-11-29 13:25:05 補充： To the below one, dun copy my answer!!!
其他解答:
1) a^2 + b^2 = sin^2 A + 2 sin A sin B + sin^2 B + cos^2 A + 2 cos A cos B + cos^2 B = 2 + 2 cos a cos B+ 2 sin A sin B = 2 + 2 cos ( A - B )
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發問:1.given that sinA+sinB=a , cosA+cosB=b find that sin(A+B) and cos(A+B) with a and b.Show the step clearly. 2.find the value cos(180/11)+cos(180x3/11)+cos(5x180/11)+cos(7x180/11)+cos(9x180/11) without calculator. 更新: I can not see the answer 2.. 更新 2: I want to ask you one more question... Find cos(180/7)cos(2x180/7)cos(4x180/7) without calculator. Thx~
最佳解答:
1) a^2 + b^2 = sin^2 A + 2 sin A sin B + sin^2 B + cos^2 A + 2 cos A cos B + cos^2 B = 2 + 2 cos a cos B+ 2 sin A sin B = 2 + 2 cos ( A - B ) So, cos ( A - B ) = ( a^2 + b^2 - 2 ) / 2 ab = ( sin A + sin B )( cos A + cos B ) = sinAcosA + sinAcosB + sinBcosA + sinBcosB = sin(A+B) + (1/2)(sin2A) + (1/2)(sin2B) = sin(A+B) + (1/2)(2)[sin(A+B)+cos(A-B)] ab = sin(A+B)+sin(A+B)+cos(A-B) ab = 2sin(A+B)+( a^2 + b^2 - 2 ) / 2 2sin(A+B) = ab- ( a^2 + b^2 - 2 ) / 2 2sin(A+B)= (-1/2)(a^2-2ab+b^2)+1 = (-1/2)(a-b)^2+1 = (-1/4)(a-b)^2+1/2 =[2-(a-b)^2]/4 cos(A+B)=sqr[14+(a-b)^2]/4 2) Please go to http://upload.u-state.com/images/1196174134.JPG 2007-11-28 16:47:50 補充： Sorry for the technical probelms, plx go tohttp://i238.photobucket.com/albums/ff245/chocolate328154/Maths004.jpg 2007-11-28 23:37:29 補充： Plx go to http://i238.photobucket.com/albums/ff245/chocolate328154/ScreenHunter_071.jpgfor the extra Q. 2007-11-29 13:25:05 補充： To the below one, dun copy my answer!!!
其他解答:
1) a^2 + b^2 = sin^2 A + 2 sin A sin B + sin^2 B + cos^2 A + 2 cos A cos B + cos^2 B = 2 + 2 cos a cos B+ 2 sin A sin B = 2 + 2 cos ( A - B )
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