Giúp t câu lượng giác

L

lovelycat_handoi95

[TEX]tanx+cosx-cos^2x=sinx[1+tanxtan(\frac{x}{2})][/TEX]


[TEX]DK:cosx \not=0\ ; cos\frac{x}{2}\not=0\[/TEX]

Có
[TEX]1+ tanx.tan\frac{x}{2} = \frac{1}{cosx}[/TEX]

[TEX] \Leftrightarrow tanx+cosx-cos^2x=sinx\frac{1}{cosx}\\ \Leftrightarrow tanx+cosx-cos^2x=tanx\\ \Leftrightarrow cos^2x-cosx=0 \\ \Leftrightarrow \left[cosx=0(L)\\cosx=1(T/M)[/TEX]
 
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H

heartrock_159

[TEX]tanx + cosx - cos^2 x = sinx(1+tanx.tan{\frac{x}{2})[/TEX] (a)

Đk:....

(a)[TEX] \Leftrightarrow tanx + cosx - cos^2 x = tanx[/TEX]

[TEX]\Leftrightarrow cosx(1-cosx) = 0[/TEX]

[TEX]\Leftrightarrow \[ cosx = 0 \\ cosx = 1[/TEX]

[TEX]\Leftrightarrow \[ x = k2{\pi}[/TEX]
 
N

n0vem13er

cám ơn bạn, cho t hỏi tiếp bài này với:
sin(x/2)sin(x) - cos(x/2)sin^2(x) + 1 = cos^2(pi/4 - x/2)
 
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