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CHỨNG MINH BẰNG QUI NẠP:
[TEX]1+cosx+cos2x+...+cosnx= \frac{sin((n+1)x/2)cos(nx/2)}{sin(x/2)}(1)[/TEX]
+ n=1 ta có :
[TEX]\blue{1+cosx= \frac{sinxcos{\frac{x}{2}}}{sin{\frac{x}{2}}}\\ \Leftrightarrow 1+cosx= 2cos^2\frac{x}{2} \\ \Leftrightarrow cosx= 2cos^2\frac{x}{2} -1 ( luon dung) [/TEX]
+ Giả sử (1) đúng với [TEX]\blue{n=k, \forall k\in N* . [/TEX]
Khi đó
[TEX]\blue{1+cosx+cos2x+...+cosk x + cos(k+1)x= \frac{sin{\frac{(k+1)x}{2}}cos{\frac{kx}{2}}}{sin{\frac{x}{2}}}+ cos(k+1)x \\ = \frac{sin{\frac{(k+1)x}{2}}cos{\frac{kx}{2}} + cos(k+1)xsin{\frac{x}{2}}}{sin{\frac{x}{2}}} \\ = \frac{sin{\frac{(k+1)x}{2}}cos{\frac{kx}{2}} + (1-2sin^2{\frac{(k+1)x}{2})sin{\frac{x}{2}}}}{sin{\fr{x}{2}}} \\ = \frac{sin{\frac{(k+1)x}{2}}cos{\frac{kx}{2}} -2sin^2{\frac{(k+1)x}{2}sin{\frac{x}{2}} +sin{\frac{x}{2}}} }{sin{\frac{x}{2}}} \\ = \frac{sin{\frac{(k+1)x}{2}}(cos{\frac{kx}{2}} -2sin{\frac{(k+1)x}{2}}sin{\frac{x}{2}})+sin{\frac{x}{2}}}{sin{\frac{x}{2}}} \\ = \frac{sin{\frac{(k+1)x}{2}}(cos{\frac{kx}{2}}+ cos{\frac{(k+2)x}{2}}-cos{\frac{kx}{2}})+sin{\frac{x}{2}}}{sin{\frac{x}{2}}} \\ = \frac{\frac{1}{2}(sin{\frac{(2k+3)x}{2}}-sin{\frac{x}{2}}) + sin{\frac{x}{2}}}{sin{\frac{x}{2}}} \\ = \frac{sin{\frac{(k+2)x}{2}}cos{\frac{(k+1)x}{2}}}{sin{\frac{x}{2}}} (dpcm)[/TEX]
=> (1) đúng với [TEX]\blue{\forall k\in N* [/TEX]
[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn