a:Sửa đề: \(I=\frac{cos\left(-288^0\right)\cdot\cot72^0}{\tan\left(-162^0\right)\cdot\sin108^0}-\tan18^0\)

Ta có: \(I=\frac{cos\left(-288^0\right)\cdot\cot72^0}{\tan\left(-162^0\right)\cdot\sin108^0}-\tan18^0\)

\(=\frac{cos\left(72^0-360^0\right)\cdot\cot72^0}{\tan\left(18^0-180^0\right)\cdot\sin108^0}-\tan18^0\)

\(=\frac{cos72^0\cdot\cot72^0}{\tan18^0\cdot\sin108^0}-\tan18^0\)

\(=\frac{cos72^0\cdot\frac{cos72^0}{\sin72^0}}{\tan18^0\cdot\sin108^0}-\tan18^0=\frac{cos^272^0}{\sin72^0\cdot\tan18^0\cdot\sin\left(180^0-108^0\right)}-\tan18^0\)

\(=\frac{cos^272^0}{\sin72^0\cdot\tan18^0\cdot\sin72^0}-\tan18^0=\frac{cos^272^0}{\sin^272^0\cdot\tan18^0}-\tan18^0\)

\(=\frac{cos^272^0}{\sin^272^0\cdot\cot72^0}-\tan18^0=\frac{cos^272^0}{\sin^272^0\cdot\frac{cos72^0}{\sin72^0}}-\tan18^0\)

\(=\frac{cos72^0}{\sin72^0}-\tan18^0=\cot72^0-\tan18^0=0\)

b: Ta có: \(J=2\cdot\sin\left(790^0+x\right)+cos\left(1260^0-x\right)+\tan\left(630^0+x\right)\cdot\tan\left(1260^0-x\right)\)

\(=2\cdot\sin\left(720^0+70^0+x\right)+cos\left(1080^0+180^0-x\right)+\tan\left(720^0+x-90^0\right)\cdot\tan\left(1080^0+180^0-x\right)\)

\(=2\cdot\sin\left(70^0+x\right)+cos\left(180^0-x\right)+\tan\left(x-90^0\right)\cdot\tan\left(180^0-x\right)\)

\(=2\cdot\sin\left(70^0+x\right)-cosx-\tan\left(90^0-x\right)\cdot\left(-\tan x\right)\)

\(=2\cdot\sin\left(70^0+x\right)-cosx+\tan\left(90^0-x\right)\cdot\tan x\)

\(=2\cdot\sin\left(70^0+x\right)-cosx+\cot x\cdot\tan x=2\cdot\sin\left(x+70^0\right)-cosx+1\)

mai đăng lại bài này nhé t làm cho h đi ngủ

ừ

\(cos3x=-cos\left(x-120^0\right)\)

\(\Leftrightarrow cos3x=cos\left(x+60^0\right)\)

\(\Rightarrow\left[{}\begin{matrix}3x=x+60^0+k360^0\\3x=-x-60^0+k360^0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=30^0+k180^0\\x=-15^0+k90^0\end{matrix}\right.\)

\(\Leftrightarrow sin\left(2x-90^0\right)=cos2x\)

\(\Leftrightarrow-cos2x=cos2x\)

\(\Rightarrow cos2x=0\Rightarrow2x=90^0+k180^0\)

\(\Rightarrow x=45^0+k90^0\)

\(cos^2x+sin^2x+2sinx.cosx=1+cos4x\)

\(\Leftrightarrow1+sin2x=1+cos4x\)

\(\Leftrightarrow cos4x=sin2x=cos\left(\frac{\pi}{2}-2x\right)\)

\(\Rightarrow\left[{}\begin{matrix}4x=\frac{\pi}{2}-2x+k2\pi\\4x=2x-\frac{\pi}{2}+k2\pi\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{12}+\frac{k\pi}{3}\\x=-\frac{\pi}{4}+k\pi\end{matrix}\right.\)