Video hướng dẫn giải - bài 7 trang 156 sgk đại số 10

LG a

\(\displaystyle {{1 - \cos x + \cos 2x} \over {\sin 2x - {\mathop{\rm s}\nolimits} {\rm{in x}}}} = \cot x\)

Phương pháp giải:

Sử dụng các công thức:

\(\begin{array}{l}
\cos 2\alpha = 2{\cos ^2}\alpha - 1\\
\sin 2\alpha = 2\sin \alpha \cos \alpha
\end{array}\)

Lời giải chi tiết:

\(\begin{array}{l}
\dfrac{{1 - \cos x + \cos 2x}}{{\sin 2x - \sin x}}\\
= \dfrac{{1 - \cos x + 2{{\cos }^2}x - 1}}{{2\sin x\cos x - \sin x}}\\
= \dfrac{{2{{\cos }^2}x - \cos x}}{{2\sin x\cos x - \sin x}}\\
= \dfrac{{\cos x\left( {2\cos x - 1} \right)}}{{\sin x\left( {2\cos x - 1} \right)}}\\
= \dfrac{{\cos x}}{{\sin x}}\\
= \cot x
\end{array}\)

LG b

\(\displaystyle {{{\mathop{\rm \sin x}\nolimits} + \sin{x \over 2}} \over {1 + \cos x + \cos {x \over 2}}} = \tan {x \over 2}\)

Phương pháp giải:

Sử dụng các công thức:

\(\begin{array}{l}
\cos 2\alpha = 2{\cos ^2}\alpha - 1\\
\sin 2\alpha = 2\sin \alpha \cos \alpha
\end{array}\)

Lời giải chi tiết:

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

LG c

\(\displaystyle {{2\cos 2x - \sin 4x} \over {2\cos 2x + \sin 4x}} = {\tan ^2}({\pi \over 4} - x)\)

Phương pháp giải:

Sử dụng các công thức:

\(\begin{array}{l}
\sin 2\alpha = 2\sin \alpha \cos \alpha
\end{array}\)

\(\sin \alpha = \cos \left( {\dfrac{\pi }{2} - \alpha } \right)\)

\(\cos 2\alpha = 2{\cos ^2}\alpha - 1 = 1 - 2{\sin ^2}\alpha \)

Lời giải chi tiết:

\(\displaystyle \, \, {{2\cos 2x - \sin 4x} \over {2\cos 2x + \sin 4x}}\)

\(\displaystyle = {{2\cos 2x - 2\sin2 x\cos 2x} \over {2\cos 2x + 2\sin 2x\cos 2x}}\)

\(\displaystyle = \dfrac{{2\cos 2x\left( {1 - \sin 2x} \right)}}{{2\cos 2x\left( {1 + \sin 2x} \right)}}\)

\(\displaystyle = {{1 - \sin 2x} \over {1 + \sin 2x}}\)
\(\displaystyle = {{1 - \cos ({\pi \over 2} - 2x)} \over {1 + \cos ({\pi \over 2} - 2x)}}\)

\(\displaystyle \begin{array}{l}
= \dfrac{{1 - \cos \left[ {2.\left( {\dfrac{\pi }{4} - x} \right)} \right]}}{{1 + \cos \left[ {2.\left( {\dfrac{\pi }{4} - x} \right)} \right]}}\\
= \dfrac{{1 - \left[ {1 - 2{{\sin }^2}\left( {\dfrac{\pi }{4} - x} \right)} \right]}}{{1 + \left[ {2{{\cos }^2}\left( {\dfrac{\pi }{4} - x} \right) - 1} \right]}}
\end{array}\)

\(\displaystyle = {{2{{\sin }^2}({\pi \over 4} - x)} \over {2{{\cos }^2}({\pi \over 4} - x)}}\)
\(\displaystyle = {\tan ^2}({\pi \over 4} - x) \)

Cách khác:

\(\displaystyle VT= {{2\cos 2x - \sin 4x} \over {2\cos 2x + \sin 4x}}\)

\(\displaystyle = {{2\cos 2x - 2\sin2 x\cos 2x} \over {2\cos 2x + 2\sin 2x\cos 2x}}\)

\(\displaystyle = \dfrac{{2\cos 2x\left( {1 - \sin 2x} \right)}}{{2\cos 2x\left( {1 + \sin 2x} \right)}}\)

\(\displaystyle = {{1 - \sin 2x} \over {1 + \sin 2x}}\)

\(\begin{array}{l}
VP = {\tan ^2}\left( {\frac{\pi }{4} - x} \right)\\
= \dfrac{{{{\sin }^2}\left( {\frac{\pi }{4} - x} \right)}}{{{{\cos }^2}\left( {\frac{\pi }{4} - x} \right)}}\\
= \dfrac{{\frac{{1 - \cos \left[ {2\left( {\frac{\pi }{4} - x} \right)} \right]}}{2}}}{{\frac{{1 + \cos \left[ {2\left( {\frac{\pi }{4} - x} \right)} \right]}}{2}}}\\
= \dfrac{{2.\frac{{1 - \cos \left[ {2\left( {\frac{\pi }{4} - x} \right)} \right]}}{2}}}{{2.\frac{{1 + \cos \left[ {2\left( {\frac{\pi }{4} - x} \right)} \right]}}{2}}}\\
= \frac{{1 - \cos \left( {\frac{\pi }{2} - 2x} \right)}}{{1 + \cos \left( {\frac{\pi }{2} - 2x} \right)}}\\
= \frac{{1 - \sin 2x}}{{1 + \sin 2x}}
\end{array}\)

Vậy VT=VP hay ta có đpcm.

LG d

\(\displaystyle \tan x - \tan y = {{\sin (x - y)} \over {\cos x.cosy}}\)

Phương pháp giải:

Sử dụng các công thức:

\(\begin{array}{l}
\tan \alpha = \dfrac{{\sin \alpha }}{{\cos \alpha }}\\
\sin a\cos b - \sin b\cos a = \sin \left( {a - b} \right)
\end{array}\)

Lời giải chi tiết:

\(\displaystyle d) \tan x - \tan y\)

\(\displaystyle = {{{\mathop{\rm sinx}\nolimits} } \over {{\mathop{\rm cosx}\nolimits} }} - {{\sin y} \over {\cos y}}\)

\(\displaystyle = {{\sin {\rm{x}}\cos y - \cos x\sin y} \over {\cos x\cos y}}\)

\(\displaystyle = {{\sin (x - y)} \over {\cos x\cos y}}.\)