數學 三角函數

Q1:
原式=> 16^(1+sin^2 x)+ 16^(1-sin^2 x)= 40
令 y= 16^(sin^2 x), 則 16y+ 16/y= 40=> y= 2, 1/2
y= 16^(sin^2 x)= 2 or 1/2 => sin^2 x= 1/4 or -1/4(不合)
=> sin x= 1/2 or -1/2 => x= nπ +/- π/6 , n in Z
Q2:
sin^2 (28π/3) = [ - sin(π/3) ]^2= 3/4
原式=> 5^(sin2x + cosx -2 sinx - 1) = 1 => sin(2x) + cosx - 2sinx -1 =0
2sinx cosx + cosx - (2sinx + 1) =0
(2sinx + 1)(cosx - 1)=0
sinx = -1/2 or cosx = 1
so, x= -π/6+ 2nπ, -5π/6+ 2nπ, or 2nπ, n in Z

As follows:

圖片參考：http://f.imagehost.org/0118/ScreenHunter_02_May_04_09_32.gif


圖片參考：http://f.imagehost.org/0343/ScreenHunter_03_May_04_09_33.gif


2009-05-06 21:05:03 補充：
sorry我轉左弧度。