maths - trigo2 (proof)－nzphddr的部落格

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maths - trigo2 (proof)

發問:

Please show all the necessary steps. 圖片參考：http://imgcld.yimg.com/8/n/HA00656648/o/701107270103213873433680.jpg 更新: Yes, there are some mistakes 1) It should be sin(A+B)sin(B-A) instead of sin(A+B)sin(A-B) 2) It should be cos^2 A - cos^2 B - cos^2 C= -1 instead of cos^2 A - cos^2 B + cos^2 C= -1 更新 2: Question(iii), Why sin (π/2 + x - y) sin (π/2 - (x + y)) = cos -(x - y) cos(x + y) should it be cos(x - y) cos(x + y)

最佳解答:

Typing error ?? i) sin(A+B) sin(A-B) should be - sin(A+B) sin(A-B) ?? ii) b) cos2A - cos2B + cos2C = - 1 should be cos2A - cos2B + cos2C = 1 ?? 2011-07-27 23:51:06 補充： i)cos2A - cos2B= (cosA - cosB) (cosA + cosB)= [ - 2sin (A+B)/2 sin(A-B)/2 ] [ 2cos (A+B)/2 cos(A-B)/2 ]= - 2 sin (A+B)/2 cos (A+B)/2 * 2 sin(A-B)/2 cos(A-B)/2= - sin(A+B) * sin(A-B) = sin(A+B) sin(B-A) ii)a)cos2A - cos2B + sin2C= sin2C - sin(A+B) sin(A-B)= sin2C - sin(π - C) sin(A-B)= sin2C - sinC sin(A-B)= sinC (sinC - sin(A-B))= sinC (sin(A+B) - sin(A-B))= sinC ( sinAcosB + cosAsinB - (sinAcosB - cosAsinB) )= sinC (2 cosAsinB )= 2 cosA sinB sinCb)cos2A - cos2B - cos2C = - 1cos2A - cos2B + (1 - cos2C) = 0cos2A - cos2B + sin2C = 0By the result of a) :2 cosA sinB sinC = 0cosA = 0 or sinB = 0 (rejected since sinB≠0 for 0 < B < π) or sinC = 0(rejected)A = π/2Therefore △ABC is a right-angled △. iii)cos2x - sin2y= cos2x - cos2(π/2 - y)By the result of i) ,= - sin (x + π/2 - y) sin (x - π/2 + y)= sin (π/2 + x - y) sin (π/2 - (x + y))= cos -(x - y) cos(x + y)= cos (x + y) cos (x - y) 2011-07-28 13:34:33 補充： Yes! No need to write cos -(x - y) cos(x + y). but cos -(x-y) = cos(x-y) is no problem.

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