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\(\pi< a< \frac{3\pi}{2}\Rightarrow sina< 0\)
\(\Rightarrow sina=-\sqrt{1-cos^2a}=-\frac{12}{13}\)
\(sin2a=2sina.cosa=\frac{120}{169}\)
\(cos2a=2cos^2a-1=-\frac{119}{169}\)
\(tan2a=\frac{sin2a}{cos2a}=-\frac{120}{119}\)
\(\dfrac{3\pi}{2}< a< 2\pi\Rightarrow sina< 0\)
\(\Rightarrow sina=-\sqrt{1-cos^2a}=-\sqrt{1-\left(\dfrac{3}{5}\right)^2}=-\dfrac{4}{5}\)
\(\Rightarrow sin2a=2sina.cosa=2.\left(-\dfrac{4}{5}\right).\left(\dfrac{3}{5}\right)=-\dfrac{24}{25}\)
Câu sau có nhầm đề ko nhỉ?
\(sin\left(\pi-\dfrac{\pi}{3}\right)=sin\left(\dfrac{2\pi}{3}\right)=\dfrac{\sqrt{3}}{2}\)
\(F=cos\left(\frac{\pi}{4}+a\right)\cdot cos\left(\frac{\pi}{4}-a\right)\)
\(=\frac12\cdot\left\lbrack cos\left(\frac{\pi}{4}+a-\frac{\pi}{4}+a\right)+cos\left(\frac{\pi}{4}+a+\frac{\pi}{4}+a\right)\right\rbrack\)
\(=\frac12\cdot\left\lbrack cos\left(2a\right)+cos\left(\frac{\pi}{2}\right)\right\rbrack=\frac12\cdot cos2a\)
\(G=\sin\left(\frac{\pi}{3}+a\right)\cdot cos\left(\frac{\pi}{3}-a\right)\)
\(=\frac12\cdot\left\lbrack\sin\left(\frac{\pi}{3}+a+\frac{\pi}{3}-a\right)+\sin\left(\frac{\pi}{3}+a-\frac{\pi}{3}+a\right)\right\rbrack\)
\(=\frac12\cdot\left\lbrack\sin\left(\frac23\pi\right)+\sin2a\right\rbrack=\frac12\cdot\left\lbrack\frac12+\sin2a\right\rbrack\)
\(H=cos\left(\frac{\pi}{2}-a\right)\cdot\sin\left(\frac{\pi}{2}+a\right)\)
\(=\frac12\cdot\left\lbrack\sin\left(\frac{\pi}{2}+a+\frac{\pi}{2}-a\right)+\sin\left(\frac{\pi}{2}+a-\frac{\pi}{2}+a\right)\right\rbrack\)
\(=\frac12\cdot\left\lbrack\sin\left(\pi\right)+\sin2a\right\rbrack=\frac12\left\lbrack2\cdot\sin a\cdot cosa\right\rbrack=\sin a\cdot cosa\)
\(I=\sin\left(\frac{\pi}{4}+a\right)-cos\left(\frac{\pi}{4}-a\right)\)
\(=\sin\left(\frac{\pi}{4}+a\right)-\sin\left(\frac{\pi}{2}-\frac{\pi}{4}+a\right)=\sin\left(\frac{\pi}{4}+a\right)-\sin\left(\frac{\pi}{4}+a\right)\)
=0
\(K=cos\left(\frac{\pi}{6}-x\right)-\sin\left(\frac{\pi}{3}+x\right)\)
\(=\sin\left(\frac{\pi}{2}-\frac{\pi}{6}+x\right)-\sin\left(\frac{\pi}{3}+x\right)=\sin\left(\frac{\pi}{3}+x\right)-\sin\left(\frac{\pi}{3}+x\right)\)
=0
\(\cos a=\dfrac{-12}{13}\)
\(\sin b=\dfrac{4}{5}\)
\(\sin\left(a+b\right)=\sin a\cos b+\sin b\cos a\)
\(=\dfrac{5}{13}\cdot\dfrac{3}{5}+\dfrac{4}{5}\cdot\dfrac{-12}{13}=\dfrac{-45}{65}=\dfrac{-9}{13}\)
\(sinx+cosx=\sqrt{2}\left(\frac{\sqrt{2}}{2}sinx+\frac{\sqrt{2}}{2}cosx\right)=\sqrt{2}\left(sinx.cos\frac{\pi}{4}+cosx.sin\frac{\pi}{4}\right)=\sqrt{2}sin\left(x+\frac{\pi}{4}\right)\)
\(=\sqrt{2}cos\left(\frac{\pi}{2}-\left(x+\frac{\pi}{4}\right)\right)=\sqrt{2}cos\left(\frac{\pi}{4}-x\right)=\sqrt{2}cos\left(x-\frac{\pi}{4}\right)\)
\(sinx-cosx=\sqrt{2}\left(\frac{\sqrt{2}}{2}sinx-\frac{\sqrt{2}}{2}cosx\right)=\sqrt{2}\left(sinx.cos\frac{\pi}{4}-cosx.sin\frac{\pi}{4}\right)=\sqrt{2}sin\left(x-\frac{\pi}{4}\right)\)
\(=-\sqrt{2}sin\left(\frac{\pi}{4}-x\right)=-\sqrt{2}cos\left(\frac{\pi}{2}-\left(\frac{\pi}{4}-x\right)\right)=-\sqrt{2}cos\left(x+\frac{\pi}{4}\right)\)
\(sin^4x-cos^4x=\left(sin^2x-cos^2x\right)\left(sin^2x+cos^2x\right)+sin2x\)
\(=sin^2x-cos^2x+sin2x=sin2x-cos2x\)
\(=\sqrt{2}sin\left(2x-\frac{\pi}{4}\right)\)
Bạn ghi ko đúng đề