Cho \(\cos \left( a \right) = \frac{1}{4}\). Tính \(\cos \left( {\frac{{3a}}{2}} \right)\cos \left( {\frac{a}{2}} \right)\)
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A.
\(\frac{{ - 5}}{{16}}\)
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B.
\(\frac{5}{{16}}\)
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C.
\(\frac{{ - 5}}{6}\)
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D.
\( - 5\)
Sử dụng công thức \(\cos \left( a \right)\cos \left( b \right) = \frac{1}{2}\left[ {\cos \left( {a + b} \right) + \cos \left( {a - b} \right)} \right]\);
\(\cos \left( {2a} \right) = 2{\cos ^2}\left( a \right) - 1\)
Ta có \(\cos \left( {\frac{{3a}}{2}} \right)\cos \left( {\frac{a}{2}} \right) = \frac{1}{2}\left[ {\cos \left( {\frac{{3a}}{2} + \frac{a}{2}} \right) + \cos \left( {\frac{{3a}}{2} - \frac{a}{2}} \right)} \right]\)
\( = \frac{1}{2}\left[ {\cos \left( {2a} \right) + \cos \left( a \right)} \right] = \frac{1}{2}\left[ {2{{\cos }^2}\left( a \right) - 1 + \cos \left( a \right)} \right] = \frac{1}{2}\left( {2.\frac{1}{{{4^2}}} - 1 + \frac{1}{4}} \right) = \frac{{ - 5}}{{16}}\)
Đáp án : A
