Advertisements (Quảng cáo)
Bài 7. Chứng minh các đồng nhất thức.
a) \({{1 – \cos x + \cos 2x} \over {\sin 2x – {\mathop{\rm s}\nolimits} {\rm{in x}}}} = \cot x\)
b) \({{{\mathop{\rm sinx}\nolimits} + sin{x \over 2}} \over {1 + \cos x + \cos {x \over 2}}} = \tan {x \over 2}\)
c) \({{2\cos 2x – \sin 4x} \over {2\cos 2x + \sin 4x}} = {\tan ^2}({\pi \over 4} – x)\)
d) \(\tan x – \tan y = {{\sin (x – y)} \over {\cos x.cosy}}\)
a)
\({{1 – \cos x + \cos 2x} \over {\sin 2x – {\mathop{\rm s}\nolimits} {\rm{in x}}}} = {{1 + \cos 2x – \cos x} \over {2\sin x\cos x – {\mathop{\rm sinx}\nolimits} }} = {{\cos x(2\cos x – 1)} \over {{\mathop{\rm s}\nolimits} {\rm{inx}}(2\cos x – 1)}} = \cot x\)
b)
\( {{{\mathop{\rm sinx}\nolimits} + sin{x \over 2}} \over {1 + \cos x + \cos {x \over 2}}}\)
\(= {{2\sin {x \over 2}\cos {x \over 2} + \sin {x \over 2}} \over {2{{\cos }^2}{x \over 2} + \cos {x \over 2}}}\)
\(= {{\sin {x \over 2}(2\cos {x \over 2} + 1)} \over {\cos {x \over 2}(2\cos {x \over 2} + 1)}}\)=
\(=\tan {x \over 2} \ \)
c)
Advertisements (Quảng cáo)
\({{2\cos 2x – \sin 4x} \over {2\cos 2x + \sin 4x}}\)
\(= {{2\cos 2x – 2\sin2 x\cos 2x} \over {2\cos 2x + 2\sin 2x\cos 2x}}\)
\(= {{1 – \sin 2x} \over {1 + \sin 2x}}\)
\(= {{1 – \cos ({\pi \over 2} – 2x)} \over {1 + \cos ({\pi \over 2} – 2x)}}\)
\(= {{2{{\sin }^2}({\pi \over 4} – x)} \over {2{{\cos }^2}({\pi \over 4} – x)}}\)
\(= {\tan ^2}({\pi \over 4} – x) \)
d)
\(\tan x – \tan y\)
\(= {{{\mathop{\rm sinx}\nolimits} } \over {{\mathop{\rm cosx}\nolimits} }} – {{\sin y} \over {\cos y}}\)
\(= {{\sin {\rm{x}}\cos y – \cos x\sin y} \over {\cos x\cos y}}\)
\(= {{\sin (x – y)} \over {\cos x\cos y}}\)
Mục lục môn Toán 10