三題三角函數問題 謝謝

Question

麻煩要有步驟唷謝謝
1, sin^4(X)+cos^4(X)=1/2sin(2X) 解方程式 答案給 拍的任意倍數+四分之拍

2. 請問把 log[(cosX-1)^2] 化成 4log| sinX/2 | +2log2 的步驟

3. 2X^3-3X^2-X+P=0的三根為1, tan a , tan b 求 sin2(a+b)+cos2(a+b)的值 謝謝唷

Answer 1

1.
sin4x + cos4x=1/2 sin(2x)
(sin2x + cos2x)2-2sin2x cos2x=1/2 sin(2x)
1-1/2(2sinx cosx)2 = 1/2 sin(2x) , 設y=sin(2x)
1-y2/2=y/2 => y2 +y-2=0 , y=1, -2(不合)
y=sin(2x)=1 => 2x=(2n-1)π/2 => x=(2n-1)π/4 , n為任意整數
2.
(cosx-1)2=[1-2sin2(x/2) -1]2 = 4sin4(x/2)
=>log[(cosx-1)2]=log[4sin4(x/2)]=log4+4log|sin(x/2)|=4log|sin(x/2)|+2log2
3.根與係數
1 +tana +tanb=3/2 =>tana +tanb=1/2
tana +tanb +tana*tanb=-1/2 => tana*tanb=-1
=>tan(a +b)=(tana +tanb)/(1-tana*tanb)=1/4
sin2(a+b)+cos2(a+b)=2tan(a+b)/[1+tan2(a+b)]+[1-tan2(a+b)]/[1+tan2(a+b)]
= (1/2)/(1+1/16)+(1-1/16)/(1+1/16)=23/17