標題:
A.maths:sine cosine
發問:
suppose acosx+bsinx+c=0 an equation of x has 2 roots m and n in the range of x is between 0 to 180 (a) show that a(cosm-cosn)+b(sinm-sinn)=0 (b)hence show that tan[(m+n)/2 = b/a
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suppose acosx+bsinx+c=0 an equation of x has 2 roots m and n in the range of x is between 0 to 180 (a) show that a(cosm-cosn)+b(sinm- sinn)=0 (b)hence show that tan[(m+n)/2 = b/a sol (a) substitute m and n into the equation acosm+bsinm+c=0...(1) acosn+bsinn+c=0...(2) (1)-(2) a(cosm-cosn)+b(sinm- sinn)=0 (b) Since a(cosm-cosn)+b(sinm- sinn)=0 a(cosm-cosn)=-b(sinm- sinn) (cosm-cosn)/(sinm- sinn)=-b/a {-2sin[(m+n)/2]sin[(m-n)/2]}/{2cos[(m+n)/2]sin[(m-n)/2]}=-b/a -sin[(m+n)/2]/cos[(m+n)/2]=-b/a tan[(m+n)/2] = b/a 2007-04-09 22:08:43 補充： sum to product formula(cosm-cosn)=-2sin[(m+n)/2]sin[(m-n)/2](sinm- sinn)=2cos[(m+n)/2]sin[(m-n)/2]
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