show that 2 [ cos A + cos (A+2B) + cos (A+4B) + cos (A+6B) + cos (A+8B) ]sinA = sin (A+9B) - sin (A-B)
更新1:
呀 係wor 錯左... 係sinB....
F.4 Amaths please....
2007-02-21 3:45 am
using the identity 2 cos x sin y = sin (x+y) - sin (x-y)
show that 2 [ cos A + cos (A+2B) + cos (A+4B) + cos (A+6B) + cos (A+8B) ]sinA = sin (A+9B) - sin (A-B)
show that 2 [ cos A + cos (A+2B) + cos (A+4B) + cos (A+6B) + cos (A+8B) ]sinA = sin (A+9B) - sin (A-B)
回答 (2)
2007-02-23 12:34 am
✔ 最佳答案
I think there may be some error in the question.i.e. to prove 2[cos A + cos(A+2B) + cos(A+4B) + cos(A+6B) + cos(A+8B)]sinB = sin(A+9B) - sin(A-B)
Proof:
sin (A+9B) - sin(A-B)
= sin (A+8B + B) - sin (A+8B - B) + sin (A+8B - B) - sin (A-B)
= 2cos (A+8B)sin B + sin (A+7B) - sin (A-B)
= 2cos (A+8B)sin B + sin (A+6B + B) - sin (A+6B - B) + sin (A+6B - B) - sin (A-B)
= 2cos (A+8B)sin B + 2cos (A+6B)sin B + sin (A+5B) - sin (A-B)
= ... = 2cos (A+8B)sin B + 2cos (A+6B)sin B + 2cos (A+4B)sin B + 2cos (A+2B)sin B + sin (A+B) - sin (A-B)
= 2cos (A+8B)sin B + 2cos (A+6B)sin B + 2cos (A+4B)sin B + 2cos (A+2B)sin B + 2cos A sin B
= 2[cos (A+8B) + cos (A+6B) + cos (A+4B) + cos (A+2B) + cos A] sin B
2007-02-21 4:23 am
2 [ cos A + cos (A+2B) + cos (A+4B) + cos (A+6B) + cos (A+8B) ]sinA
=[sin2A+sin0]+[sin(2A+2B)+sin(-2B)]+[sin(2A+4B)+sin(-4B)]+[sin(2A+6B)+sin(-6B)]+[sin(2A+8B)+sin(-8B)]
=sin2A+sin2(A+B)+sin2(A+2B)+sin2(A+3B)+sin2(A+4B)-sin2B-sin4B-sin6B-sin8B=?
=[sin2A+sin0]+[sin(2A+2B)+sin(-2B)]+[sin(2A+4B)+sin(-4B)]+[sin(2A+6B)+sin(-6B)]+[sin(2A+8B)+sin(-8B)]
=sin2A+sin2(A+B)+sin2(A+2B)+sin2(A+3B)+sin2(A+4B)-sin2B-sin4B-sin6B-sin8B=?
收錄日期: 2021-05-01 10:22:10
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