Câu 1: [tex]Sin^4x=\frac{a}{8}-\frac{1}{2}cos2x+\frac{b}{8}.cos4x[/tex], với [tex]a,b\in\mathbb{Q}[/tex] khi đó tổng a+b bằng ?
Câu 2: ta có [tex]Sin^8x+cos^8x=\frac{a}{64}+\frac{b}{16}cos4x+\frac{c}{16}[/tex] với [tex]a,b\in \mathbb{Q}[/tex] khi đó a-5b+c bằng ?
Câu 1:
Có: [tex]A=sin^4x=(sin^2x)^2=(1-cos^2x)^2[/tex]
[tex]=1-2cos^2x+cos^4x(*)[/tex]
Mà:
- [tex]cos2x=2cos^2x-1\rightarrow 2cos^2x=cos2x+1[/tex]
- [tex]cos4x=2cos^22x-1=2(2cos^2x-1)^2-1=8cos^4x-8cos^2x+1[/tex] [tex]\rightarrow cos^4x=\frac{-1}{8}cos4x-cos^2x+\frac{1}{8}[/tex]
Thay vào (*):
[tex]\Rightarrow A=1-2cos^2x-\frac{1}{8}cos4x-cos^2x+\frac{1}{8}[/tex]
[tex]\Leftrightarrow A=\frac{3}{8}-\frac{1}{2}cos2x+\frac{1}{8}cos4x[/tex]
[tex]\Rightarrow \left\{\begin{matrix} a=3\\ b=1 \end{matrix}\right.\Rightarrow a+b=4[/tex] :v
Câu 2:
Đề bài đúng phải là: [tex]Sin^8x+cos^8x=\frac{a}{64}+\frac{b}{16}cos4x+\frac{c}{16}cos8x[/tex] chứ nhỉ?
$\sin ^{8}x+\cos ^{8}x= (sin^{4}x+cos^{4}x)^{2}-2sin^{4}xcos^{4}x $
$= (1-\frac{1}{2}sin^{2}2x)^{2}-2sin^{4}xcos^{4}x $
$= 1-4sin^{2}xcos^{2}x + 2sin^{4}xcos^{4}x$
$= 1-sin^{2}2x + 2(\frac{1}{4}sin^{2}2x)^{2} $
$= 1- \frac{1-cos4x}{2} + \frac{1}{8}(\frac{1-cos4x }{2})^{2}$
$ = \frac{1+cos4x}{2} + \frac{1}{8}(\frac{1-2cos4x + cos^{2}4x }{4})$
$= \frac{1+cos4x}{2} + \frac{1-2cos4x}{32}+ \frac{1+cos8x}{64}$
$=\frac{1}{64}cos 8x + \frac{7}{16}cos 4x + \frac{35}{64}$
[tex]\Rightarrow \left\{\begin{matrix} a=35\\ b=7 \\ c=1 \end{matrix}\right.\rightarrow a-5b+c=1[/tex]
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