1, Chứng minh
 $sin^{2}(a+b)-sin^{2}a-sin^{2}b=2sina.sinb.cos(a+b)$
  2, Cho $m.sin(a+b)=cos(a-b)$
Chứng minh biểu thức sau k phụ thuộc vào biến
  A=$\frac{1}{1-m.sin2a}+\frac{1}{1-m.sin2b}$
1)
$VT=(sina.cosb+sinb.cosa)^2-sin^2a-sin^2b$
$=sin^2a.cos^2b+sin^2b.cos^2a-sin^2a-sin^2b+2sina.sinb.cosa.cosb$
$=sin^2a(cos^2b-1)+sin^2b(cos^2a-1)+2sina.sinb.cosa.cosb$
$=2sina.sinb.cosa.cosb-2sin^2a.sin^2b$
$=2sina.sinb.cos(a+b)=VF$
nhóm $(sin^2a.cos^2b-sin^2a)$ và $(sin^2b.cos^2a-sin^2b)$ –  never give up 09-05-14 08:27 AM
bạn có thể giải thích cho mk từ dòng 2 xuống dòng 3 như thế nào k . mk k hiểu –  Dân Nguyễn 07-05-14 02:02 PM
A = $\frac{1}{1 - msin2a}$ + $\frac{1}{1 - msin2b}$
   = $\frac{2 - msin2b-msin2a}{(1-msin2a)(1-msin2b}$
   = $\frac{2-2msin(a+b)cos(a-b)}{1+m^{2}sin2asinb-msin2a-msin2b}$
   = $\frac{2-2cos^{2}(a-b)}{1+\frac{1}{2}m^{2}cos2(a-b)+\frac{1}{2}m^{2}cos2(a+b)-2msin(a+b)cos(a-b)}$
   = $\frac{2-2cos^{2}(a-b)}{1+\frac{1}{2}m^{2}2cos^{2}(a-b)-\frac{1}{2}m^{2}+\frac{1}{2}m^{2.}2sin^{2}(a+b)-\frac{1}{2}m^{2}-2cos^{2}(a-b)}$
   = $\frac{2[1-cos^{2}(a-b)]}{[1-cos^{2}(a-b)]+m^{2}[cos^{2}(a-b)-1]}$
   = $\frac{2}{1-m^{2}}$
um,kcj bạn –  girlkute_9x_nd98 07-05-14 10:30 PM
bạn giải thích cho mình cái chỗ biến dổi ra "1/2m^2.2sin^2(a b)-1/2m^2" –  luckyboy_kg1998 07-05-14 04:56 PM
à k . đề bài cho là k phụ thuộc vào a, b . Thank nhiều nhé –  Dân Nguyễn 07-05-14 01:55 PM
vẫn phụ thuộc vào biến bạn ạ –  girlkute_9x_nd98 07-05-14 11:10 AM

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