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NV
20 tháng 4 2020

a/ \(\frac{\pi}{2}\le y\le\pi\Rightarrow cosy< 0\)

\(\Rightarrow cosy=-\sqrt{1-sin^2y}=-\frac{2\sqrt{2}}{3}\)

\(sin2y=2siny.cosy=2.\left(\frac{1}{3}\right).\left(-\frac{2\sqrt{2}}{3}\right)=-\frac{4\sqrt{2}}{9}\)

\(cos\left(\frac{\pi}{3}-y\right)=cos\frac{\pi}{3}cosy+sin\frac{\pi}{3}siny=\frac{\sqrt{3}-2\sqrt{2}}{6}\)

\(tany+5=\frac{siny}{cosy}+5=5-\frac{\sqrt{2}}{4}\)

b/ \(-\frac{\pi}{2}\le a\le9\Rightarrow sina\le0\)

\(\Rightarrow sina=\sqrt{1-cos^2a}=-\frac{4}{5}\)

\(sin2a=2sina.cosa=-\frac{24}{25}\)

\(cos2a=cos^2a-sin^2a=-\frac{7}{25}\)

\(tan2a=\frac{sin2a}{cos2a}=\frac{24}{7}\)

c/ \(\pi\le a\le\frac{3\pi}{2}\Rightarrow\left\{{}\begin{matrix}sina\le0\\cosa\le0\end{matrix}\right.\)

\(\Rightarrow cosa=-\frac{1}{\sqrt{1+tan^2a}}=-\frac{1}{2}\Rightarrow sina=-\frac{\sqrt{3}}{2}\)

\(\Rightarrow sin2a=2sina.cosa=\frac{\sqrt{3}}{2}\)

\(\Rightarrow\left(\sqrt{3}-sin2a\right)sin\frac{2\pi}{3}=\frac{3}{4}\)

NV
19 tháng 6 2020

\(sina.sin\left(\frac{\pi}{3}-a\right)sin\left(\frac{\pi}{3}+a\right)\)

\(=-\frac{1}{2}sina\left[cos\frac{2\pi}{3}-cos2a\right]=-\frac{1}{2}sina\left(-\frac{1}{2}-cos2a\right)\)

\(=\frac{1}{4}sina+\frac{1}{2}sina.cos2a=\frac{1}{4}sina+\frac{1}{4}sin3a-\frac{1}{4}sina\)

\(=\frac{1}{4}sin3a\)

\(sin\frac{\pi}{9}sin\frac{2\pi}{9}sin\frac{4\pi}{9}=sin\frac{\pi}{9}sin\left(\frac{\pi}{3}-\frac{\pi}{9}\right)sin\left(\frac{\pi}{3}+\frac{\pi}{9}\right)=\frac{1}{4}sin\frac{\pi}{3}=\frac{\sqrt{3}}{8}\)

\(cosa.cos\left(\frac{\pi}{3}-a\right)cos\left(\frac{\pi}{3}+a\right)=\frac{1}{2}cosa\left(cos\frac{2\pi}{3}+cos2a\right)\)

\(=\frac{1}{2}cosa\left(cos2a-\frac{1}{2}\right)=\frac{1}{2}cosa.cos2a-\frac{1}{4}cosa\)

\(=\frac{1}{4}cos3a+\frac{1}{4}cosa-\frac{1}{4}cosa=\frac{1}{4}cos3a\)

\(cos\frac{\pi}{18}cos\frac{5\pi}{18}cos\frac{7\pi}{18}=cos\frac{\pi}{18}.cos\left(\frac{\pi}{3}-\frac{\pi}{18}\right).cos\left(\frac{\pi}{3}+\frac{\pi}{18}\right)=\frac{1}{4}cos\frac{\pi}{6}=\frac{\sqrt{3}}{8}\)

NV
8 tháng 6 2020

\(\frac{sina+sin3a+sin2a}{cosa+cos3a+cos2a}=\frac{2sin2a.cosa+sin2a}{2cos2a.cosa+cos2a}=\frac{sin2a\left(2cosa+1\right)}{cos2a\left(2cosa+1\right)}=\frac{sin2a}{cos2a}=tan2a\)

\(cos^2\left(a-\frac{\pi}{4}\right)-sin^2\left(a-\frac{\pi}{4}\right)=cos\left(2a-\frac{\pi}{2}\right)\)

\(=cos\left(\frac{\pi}{2}-2a\right)=sin2a\)