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\(\frac{\left(sin^2x\right)^2-\left(cos^2x\right)^2}{2sinxcosx}\)=\(\frac{\left(sin^2x+cos^2x\right).\left(sin^2x-cos^2x\right)}{2sinxcosx}\)=\(\frac{1.\left(sin^2x-cos^2x\right)}{2sinxcosx}\)=\(\frac{sin^2x-cos^2x}{sin2x}\)=\(\frac{\frac{1-cos2x}{2}-\frac{1+cos2x}{2}}{sin2x}\)=\(\frac{1-1-cos2x-cos2x}{2}.\frac{1}{sin2x}\)=\(\frac{-2cos2x}{2sin2x}=\frac{-cos2x}{sin2x}=-cot2x\left(đpcm\right)\)
\(3\overrightarrow{AP}-2\overrightarrow{AC}=\overrightarrow{0}\)
\(VT=3\left(\overrightarrow{AD}+\overrightarrow{DP}\right)-2\left(\overrightarrow{AD}+\overrightarrow{DC}\right)\)
\(=3\overrightarrow{AD}+3\overrightarrow{DP}-2\overrightarrow{AD}-2\overrightarrow{DC}\)
\(=\overrightarrow{AD}+3\overrightarrow{DP}-2\overrightarrow{DC}\)
\(=\overrightarrow{AD}+3\left(\overrightarrow{DC}+\overrightarrow{CP}\right)-2\overrightarrow{DC}\)
\(=\overrightarrow{AD}+3\overrightarrow{DC}+3\overrightarrow{CP}-2\overrightarrow{DC}\)
\(=\widehat{AD}+\overrightarrow{DC}+3.\dfrac{2}{3}\overrightarrow{CO}\)
\(=\overrightarrow{AD}+\overrightarrow{DC}+2.\dfrac{1}{2}\overrightarrow{CA}\)
\(=\overrightarrow{AD}+\overrightarrow{DC}+\overrightarrow{CA}\)
\(=\overrightarrow{AC}+\overrightarrow{CA}\)
\(=\overrightarrow{AA}=\overrightarrow{0}=VP\) (điều phải chứng minh)
\(\frac{sin4x-sin2x}{1-cos2x+cos4x}=\frac{2sin2x.cos2x-sin2x}{1-cos2x+2cos^22x-1}=\frac{sin2x\left(2cos2x-1\right)}{cos2x\left(2cos2x-1\right)}=\frac{sin2x}{cos2x}=tan2x\)
\(\Rightarrow\) đề sai
b/
\(\frac{1-cos4x}{sin4x}=\frac{1-\left(1-2sin^22x\right)}{2sin2x.cos2x}=\frac{2sin^22x}{2sin2x.cos2x}=\frac{sin2x}{cos2x}=tan2x\)
Đề sai tiếp lần 2
\(sin^6a+cos^6a=\left(sin^2a\right)^3+\left(cos^2a\right)^3\)
\(=\left(sin^2a+cos^2a\right)^3-3sin^2a.cos^2a\left(sin^2a+cos^2a\right)\)
\(=1-3sin^2a.cos^2a\)
VT = \(\frac{1-cosx+cos2x}{sin2x-sin}\)
= \(\frac{1-cosx+2cos^2x-1}{2sinx.cosx-sinx}\)
= \(\frac{cosx\left(2cosx-1\right)}{sinx\left(2cosx-1\right)}\)
= cotx = VP ( đpcm )
mình cảm ơn ạ