F.4 AMATHS..

2008-04-06 8:54 pm
Prove the following indentities.

(sinA-sin3A+sin5A-sin7A)/(cosA-cos3A+cos5A-cos7A) = -cot4A

回答 (2)

2008-04-06 9:11 pm
✔ 最佳答案
L.H.S=(sinA-sin3A+sin5A-sin7A)/(cosA-cos3A+cos5A-cos7A)
=(sinA-sin7A+sin5A-sin3A)/(cosA-cos7A+cos5-cos3A)
=(-2cos4Asin3A+2cos4AsinA)/(2sin4Asin3A-2sin4AsinA)
=2cos4A(-sin3A+sinA) / 2sin4A(sin3A-sinA)
=2cos4A(-sin3A+sinA) / -2sin4A(-sin3A+sinA)
=-cot4A
=R.H.S
2008-04-06 9:39 pm
(sinA-sin3A+sin5A-sin7A)/(cosA-cos3A+cos5A-cos7A) = -cot4A

LHS : (sinA-sin3A+sin5A-sin7A)/(cosA-cos3A+cos5A-cos7A)
=sin(2A-A)-sin(2A+A)+sin(6A-A)-sin(6A+a)/cos(2A-A)-cos(2A+A)+cos(6A-A)-cos(6A+a)
= (sin2AcosA-cos2AsinA-sin2AcosA-cos2AsinA+sin6AcosA-cos6AsinA-sin6AcosA-cos6AsinA) / (cos2AcosA+sin2AsinA-cos2AcosA+sin2AsinA+cos6AcosA+sin6AsinA-cos6AcosA+sin6AsinA)
= (-2cos2AsinA-2cos6AsinA) / (2sin2AsinA+2sin6AsinA)
= - ( cos2A+cos6A ) / ( sin2A+sin6A )
=- ( cos(4A-2A) + cos(4A+2A) )/ (sin(4A-2A)+sin(4A+2A))
= - ( cos4Acos2A+sin4Asin2A+cos4Acos2A-sin4Asin2A ) / ( sin4Acos2A-cos4Asin2A+sin4Acos2A+cos4Asin2A )
= -( cos4A / sin4A)
= - cot4A =RHS


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