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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