Câu 2:
a: \(\dfrac{pi}{2}< x< pi\)
nên sin x>0 và cos x<0
\(sin2x=2\cdot sinx\cdot cosx=2\cdot\dfrac{1}{4}\cdot\dfrac{\sqrt{15}}{4}=\dfrac{\sqrt{15}}{8}\)
\(cos2x=1-2sin^2x=1-2\cdot\dfrac{1}{16}=1-\dfrac{1}{8}=\dfrac{7}{8}\)
b: \(cos\left(a+\dfrac{pi}{6}\right)\)
\(=cosa\cdot\dfrac{cospi}{6}+sina\cdot sin\left(\dfrac{pi}{6}\right)\)
\(=\dfrac{\sqrt{3}}{2}cosa+\dfrac{1}{2}sina\)
\(=\dfrac{\sqrt{3}}{2}\cdot\dfrac{\sqrt{15}}{4}+\dfrac{1}{2}\cdot\dfrac{1}{4}=\dfrac{3\sqrt{5}+1}{8}\)
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