chứng minh 1-2sin^2x/1+sin2x=1-tanx/1+tanx

0 bình luận về “chứng minh 1-2sin^2x/1+sin2x=1-tanx/1+tanx”

1. $\dfrac{1-2\sin^2x}{1+\sin2x}$

   $=\dfrac{(1-\sin^2x-\sin^2x}{\sin^2x+2\sin x\cos x+\cos^2x}$

   $=\dfrac{\cos^2x-\sin^2x}{(\sin x+\cos x)^2}$

   $=\dfrac{(\cos x-\sin x)(\cos x+\sin x)}{(\cos x+\sin x)^2}$

   $=\dfrac{\cos x-\sin x}{\cos x+\sin x}$

   $=\dfrac{1-\tan x}{1+\tan x}$

   Bình luận

2. Đáp án:

   dpcm

   Giải thích các bước giải:

   \(\begin{array}{l}
   \dfrac{{1 – 2{{\sin }^2}x}}{{1 + \sin 2x}} = \dfrac{{1 – {{\sin }^2}x – {{\sin }^2}x}}{{{{\sin }^2}x + {{\cos }^2}x + 2\sin x.\cos x}}\\
    = \dfrac{{{{\cos }^2}x – {{\sin }^2}x}}{{{{\left( {\cos x + \sin x} \right)}^2}}}\\
    = \dfrac{{\left( {\cos x – \sin x} \right)\left( {\cos x + \sin x} \right)}}{{{{\left( {\cos x + \sin x} \right)}^2}}}\\
    = \dfrac{{\cos x – \sin x}}{{\cos x + \sin x}}\left( 1 \right)\\
   \dfrac{{1 – \tan x}}{{1 + \tan x}} = \left( {1 – \dfrac{{\sin x}}{{\cos x}}} \right):\left( {1 + \dfrac{{\sin x}}{{\cos x}}} \right)\\
    = \left( {\dfrac{{\cos x – \sin x}}{{\cos x}}} \right):\left( {\dfrac{{\cos x + \sin x}}{{\cos x}}} \right)\\
    = \dfrac{{\cos x – \sin x}}{{\cos x}}.\dfrac{{\cos x}}{{\cos x + \sin x}}\\
    = \dfrac{{\cos x – \sin x}}{{\cos x + \sin x}}\left( 2 \right)
   \end{array}\)

   Từ (1) và (2) ⇒ dpcm

    

   Bình luận