luong gjac.giup em gjaj bai nay vs.kho qua

N

nguyenbahiep1

chung mjnh dang thuc
1:(sin^2 3a/sin^2a) -(cos^2 3a/cos2a) =8cos2a
2:6+2cos4a/1-cos4a =tan^2a+ cot^2a
giup em vs kho qua em k lam dk

[TEX] \frac{sin^2 3a}{sin^2 a} - \frac{cos^2 3a}{cos^2 a} = (\frac{sin3a}{sina} - \frac{cos3a}{cosa})(\frac{sin3a}{sina} +\frac{cos3a}{cosa}) \\ (\frac{sin3a.cosa - sinacos3a}{cosa.sina})(\frac{sin3a.cosa + sinacos3a}{cosa.sina}) = \frac{sin2a.sin4a}{(sina.cosa)^2} \\ \frac{2sina.cosa.4.sina.cosa.cos2a}{(sinacosa)^2} = 8cos2a[/TEX]
 
  • Like
Reactions: nguyenkhanhtruong
Top Bottom