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\(\cos a=\dfrac{-12}{13}\)
\(\sin b=\dfrac{4}{5}\)
\(\sin\left(a+b\right)=\sin a\cos b+\sin b\cos a\)
\(=\dfrac{5}{13}\cdot\dfrac{3}{5}+\dfrac{4}{5}\cdot\dfrac{-12}{13}=\dfrac{-45}{65}=\dfrac{-9}{13}\)
\(\dfrac{3\pi}{2}< a< 2\pi\Rightarrow sina< 0\)
\(\Rightarrow sina=-\sqrt{1-cos^2a}=-\sqrt{1-\left(\dfrac{3}{5}\right)^2}=-\dfrac{4}{5}\)
\(\Rightarrow sin2a=2sina.cosa=2.\left(-\dfrac{4}{5}\right).\left(\dfrac{3}{5}\right)=-\dfrac{24}{25}\)
Câu sau có nhầm đề ko nhỉ?
\(sin\left(\pi-\dfrac{\pi}{3}\right)=sin\left(\dfrac{2\pi}{3}\right)=\dfrac{\sqrt{3}}{2}\)
\(A=\frac{1}{2}+\frac{1}{2}cos2x+\frac{1}{2}+\frac{1}{2}cos\left(2x+\frac{4\pi}{3}\right)+\frac{1}{2}+\frac{1}{2}cos\left(2x-\frac{4\pi}{3}\right)\)
\(=\frac{3}{2}+\frac{1}{2}cos2x+cos2x.cos\frac{4\pi}{3}\)
\(=\frac{3}{2}+\frac{1}{2}cos2x-\frac{1}{2}cos2x=\frac{3}{2}\)
\(B=\frac{1}{2}-\frac{1}{2}cos2x+\frac{1}{2}-\frac{1}{2}cos\left(2x+\frac{4\pi}{3}\right)+\frac{1}{2}-\frac{1}{2}cos\left(2x-\frac{4\pi}{3}\right)\)
\(=\frac{3}{2}-\frac{1}{2}cos2x-cos2x.cos\frac{4\pi}{3}\)
\(=\frac{3}{2}-\frac{1}{2}cos2x+\frac{1}{2}cos2x=\frac{3}{2}\)
\(A=cos^2a+cos^2b+2cosa.cosb+sin^2a+sin^2b+2sina.sinb\)
\(=cos^2a+sin^2a+cos^2b+sin^2b+2\left(cosa.cosb+sina.sinb\right)\)
\(=2+2cos\left(a-b\right)=2+2cos\frac{\pi}{3}=3\)
\(\left(cosa+sina\right)^2=\frac{36}{25}\Leftrightarrow1+2sina.cosa=\frac{36}{25}\)
\(\Rightarrow sin2a=\frac{36}{25}-1=\frac{11}{25}\)
\(cos2a=cos^2a-sin^2a=\left(cosa-sina\right)\left(cosa+sina\right)>0\)
\(\Rightarrow cos2a=\sqrt{1-sin^22a}=\frac{6\sqrt{14}}{25}\)
1.
\(sinA+sinB-sinC=2sin\dfrac{A+B}{2}.cos\dfrac{A-B}{2}-sin\left(A+B\right)\)
\(=2sin\dfrac{A+B}{2}.cos\dfrac{A-B}{2}-2sin\dfrac{A+B}{2}.cos\dfrac{A+B}{2}\)
\(=2sin\dfrac{A+B}{2}.\left(cos\dfrac{A-B}{2}-cos\dfrac{A+B}{2}\right)\)
\(=2sin\dfrac{A+B}{2}.2sin\dfrac{A}{2}.sin\dfrac{B}{2}\)
\(=4sin\dfrac{A}{2}.sin\dfrac{B}{2}.cos\dfrac{C}{2}\)
Sao t lại đc như này v, ai check hộ phát
Vì \(\dfrac{\pi}{2}< \alpha< \pi\) \(\Rightarrow\) cos \(\alpha\) < 0
\(\Rightarrow\) cos \(\alpha\) = \(-\sqrt{1-sin^2\alpha}\) = \(-\dfrac{2\sqrt{2}}{3}\)
\(\Rightarrow\) tan \(\alpha\) = \(\dfrac{sin\alpha}{cos\alpha}=\dfrac{-\sqrt{2}}{4}\)
\(\Rightarrow\) cot \(\alpha\) = \(\dfrac{1}{tan\alpha}\) = \(-2\sqrt{2}\)
Chúc bn học tốt!
Ta có sin 2 α + c os 2 α = 1 ⇒ 1 9 + c os 2 α = 1 ⇒ c os 2 α = 8 9
Đáp án A