Bài 7 trang 155 SGK Đại số 10

LG a

$1 - \sin x$;

Phương pháp giải:

Áp dụng các công thức: 

$\begin{array}{l} + )\;\;\sin a + \sin b = 2\sin \dfrac{{a + b}}{2}\cos \dfrac{{a - b}}{2}.\\ + )\;\;\sin a - \sin b = 2\cos \dfrac{{a + b}}{2}\sin \dfrac{{a - b}}{2}.\\ + )\;\;\cos a + \cos b = 2\cos \dfrac{{a + b}}{2}\cos \dfrac{{a - b}}{2}.\\ + )\;\;\cos a - \cos b = - 2\sin \dfrac{{a + b}}{2}\sin \dfrac{{a - b}}{2}. \end{array}$

Lời giải chi tiết:

$1 - \sin x = \sin \dfrac{\pi }{2} - \sin x$

$= 2\cos \dfrac{\dfrac{\pi }{2}+x}{2}\sin \dfrac{\dfrac{\pi}{2}-x}{2}$

$= 2 \cos \left ( \dfrac{\pi }{4} +\dfrac{x}{2}\right )\sin\left ( \dfrac{\pi }{4} -\dfrac{x}{2}\right )$

Cách khác:

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

LG b

$1 + \sin x$;

Lời giải chi tiết:

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

$= 2\sin \left ( \dfrac{\pi }{4} +\dfrac{x}{2}\right )\cos \left ( \dfrac{\pi }{4} -\dfrac{x}{2}\right )$

Cách khác:

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

LG c

$1 + 2\cos x$;

Lời giải chi tiết:

$1 + 2\cos x = 2( \dfrac{1}{2} + \cos x)$

$= 2(\cos \dfrac{\pi}{3} + \cos x)$

$= 4\cos \left ( \dfrac{\pi }{6} +\dfrac{x}{2}\right )\cos \left ( \dfrac{\pi }{6} -\dfrac{x}{2}\right )$

Cách khác:

$\begin{array}{l} 1 + 2\cos x = 1 + 2\left( {2{{\cos }^2}\dfrac{x}{2} - 1} \right)\\ = 4{\cos ^2}\dfrac{x}{2} - 1 = {\left( {2\cos \dfrac{x}{2}} \right)^2} - 1\\ = \left( {2\cos \dfrac{x}{2} - 1} \right)\left( {2\cos \dfrac{x}{2} + 1} \right) \end{array}$

LG d

$1 - 2\sin x$

Lời giải chi tiết:

$1 - 2\sin x = 2( \dfrac{1}{2} - \sin x)$

$= 2(\sin \dfrac{\pi}{6} - \sin x)$

$= 4\cos \left ( \dfrac{\pi }{12} +\dfrac{x}{2}\right )\sin \left ( \dfrac{\pi }{12} -\dfrac{x}{2}\right )$

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