b) \(cosa=-\dfrac{3}{5}\left(\pi< a< \dfrac{3\pi}{2}\right)\)
\(sin^2a+cos^2a=1\Rightarrow sin^2a=1-cos^2a=1-\dfrac{9}{25}=\dfrac{16}{25}\)
\(\Rightarrow sina=-\dfrac{4}{5}\) vì \(\pi< a< \dfrac{3\pi}{2}\)
\(cos2a=cos^2a-sin^2a=\dfrac{9}{25}-\dfrac{16}{25}=-\dfrac{7}{25}\)
\(sin2a=2sinacosa=2.\left(-\dfrac{4}{5}\right).\left(-\dfrac{3}{5}\right)=\dfrac{24}{25}\)
\(cosa=2cos^2\dfrac{a}{2}-1=1-2sin^2\dfrac{a}{2}\)
\(\Rightarrow\left\{{}\begin{matrix}sin^2\dfrac{a}{2}=\dfrac{1}{2}\left(1-cosa\right)=\dfrac{1}{2}\left(1+\dfrac{3}{5}\right)=\dfrac{4}{5}\\cos^2\dfrac{a}{2}=\dfrac{1}{2}\left(1+cosa\right)=\dfrac{1}{2}\left(1-\dfrac{3}{5}\right)=\dfrac{1}{5}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}sin\dfrac{a}{2}=-\dfrac{2}{\sqrt[]{5}}=-\dfrac{2\sqrt[]{5}}{5}\\cosa=\dfrac{1}{\sqrt[]{5}}=\dfrac{\sqrt[]{5}}{5}\end{matrix}\right.\) \(\left(\pi< a< \dfrac{3\pi}{2}\Rightarrow\dfrac{\pi}{2}< \dfrac{a}{2}< \dfrac{3\pi}{4}\right)\)
c.
\(cos\left(a+b\right)cos\left(a-b\right)=\dfrac{1}{2}\left[cos\left(a+b+a-b\right)+cos\left(a+b-a+b\right)\right]\)
\(=\dfrac{1}{2}\left(cos2a+cos2b\right)=\dfrac{1}{2}\left(2cos^2a-1+1-2sin^2b\right)\)
\(=cos^2a-sin^2b\)
\(=1-sin^2a-\left(1-cos^2b\right)=cos^2b-sin^2a\)
d.
\(sin4x.sin10x-sin11x.sin3x-sin7x.sinx\)
\(=\dfrac{1}{2}\left(cos6x-cos14x\right)-\dfrac{1}{2}\left(cos8x-cos14x\right)-\dfrac{1}{2}\left(cos6x-cos8x\right)\)
\(=\dfrac{1}{2}\left(cos6x-cos6x-cos14x+cos14x-cos8x+cos8x\right)\)
\(=\dfrac{1}{2}.0=0\)
e.
\(sinA+sinB+sinC=2sin\left(\dfrac{A+B}{2}\right)cos\left(\dfrac{A-B}{2}\right)+2sin\left(\dfrac{C}{2}\right)cos\left(\dfrac{C}{2}\right)\)
\(=2sin\left(\dfrac{\pi}{2}-\dfrac{C}{2}\right)cos\left(\dfrac{A-B}{2}\right)+2cos\left(\dfrac{\pi}{2}-\dfrac{C}{2}\right)cos\left(\dfrac{C}{2}\right)\)
\(=2cos\left(\dfrac{C}{2}\right)cos\left(\dfrac{A-B}{2}\right)+2cos\left(\dfrac{A+B}{2}\right)cos\left(\dfrac{C}{2}\right)\)
\(=2cos\left(\dfrac{C}{2}\right)\left[cos\left(\dfrac{A-B}{2}\right)+cos\left(\dfrac{A+B}{2}\right)\right]\)
\(=4cos\left(\dfrac{C}{2}\right)cos\left(\dfrac{A}{2}\right)cos\left(\dfrac{B}{2}\right)\)
a.
\(sin\left(a+\dfrac{\pi}{6}\right)-cos\left(a-\dfrac{\pi}{3}\right)=sin\left(a+\dfrac{\pi}{6}\right)-cos\left[a-\left(\dfrac{\pi}{2}-\dfrac{\pi}{6}\right)\right]\)
\(=sin\left(a+\dfrac{\pi}{6}\right)-cos\left(a+\dfrac{\pi}{6}-\dfrac{\pi}{2}\right)\)
Áp dụng: \(cos\left(x-\dfrac{\pi}{2}\right)=sinx\)
\(=sin\left(a+\dfrac{\pi}{6}\right)-sin\left(a+\dfrac{\pi}{6}\right)=0\) (với mọi a luôn)
b.
\(\pi< a< \dfrac{3\pi}{2}\Rightarrow sina< 0\Rightarrow sina=-\sqrt{1-cos^2a}=-\dfrac{4}{5}\)
\(cos2a=2cos^2a-1=2.\left(-\dfrac{3}{5}\right)^2-1=-\dfrac{7}{25}\)
\(sin2a=2sina.cosa=2.\left(-\dfrac{4}{5}\right).\left(-\dfrac{3}{5}\right)=\dfrac{24}{25}\)
\(\pi< a< \dfrac{3\pi}{2}\Rightarrow\dfrac{\pi}{2}< \dfrac{a}{2}< \dfrac{3\pi}{4}< \pi\) \(\Rightarrow cos\left(\dfrac{a}{2}\right)< 0\)
\(cosa=2cos^2\left(\dfrac{a}{2}\right)-1\Rightarrow cos\left(\dfrac{a}{2}\right)=-\sqrt{\dfrac{1+cosa}{2}}=-\dfrac{\sqrt{5}}{5}\)
\(sina=2sin\left(\dfrac{a}{2}\right).cos\left(\dfrac{a}{2}\right)\Rightarrow sin\left(\dfrac{a}{2}\right)=\dfrac{sina}{2cos\left(\dfrac{a}{2}\right)}=\dfrac{2\sqrt{5}}{5}\)
