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\(A=sin42^0-cos48^0=cos\left(90^0-42^0\right)-cos48^0=cos48^0-cos48^0=0\)
\(B=cot56^0-tan34^0=tan\left(90^0-56^0\right)-tan34^0=tan34^0-tan34^0=0\)
\(C=sin30^0-cot50^0-cos60^0+tan40^0\)
\(=cos\left(90^0-30^0\right)-tan\left(90^0-50^0\right)-cos60^0+tan40^0\)
\(=cos60^0-tan40^0-cos60^0+tan40^0=0\)
\(A=\sin42^0-\cos48^0=\sin42^0-\sin42^0=0\)
\(B=\cot56^0-\tan34^0=\tan34^0-\tan34^0=0\)
\(cos\left(4x+60^o\right)-5cos\left(2x+30^o\right)+4=0\)
\(\Leftrightarrow2cos^2\left(2x+30^o\right)-5cos\left(2x+30^o\right)+3=0\)
\(\Leftrightarrow\left[cos\left(2x+30^o\right)-1\right]\left[2cos\left(2x+30^o\right)-3\right]=0\)
\(\Leftrightarrow cos\left(2x+30^o\right)=1\)
\(\Leftrightarrow2x+30^o=k.360^o\)
\(\Leftrightarrow x=-15^o+k.180^o\)
ĐKXĐ: \(cos\left(x-30^0\right)\ne0\Leftrightarrow x\ne120^0+k180^0\)
\(tan\left(x-30^0\right)cos\left(2x-150^0\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}tan\left(x-30^0\right)=0\\cos\left(2x-150^0\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-30^0=k180^0\\2x-150^0=90^0+k180^0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=30^0+k180^0\\x=120^0+k90^0\end{matrix}\right.\)
\(\Rightarrow x=30^0+k180^0\)
đáp án
A=Sin 42o - cos 48o =cos(90o - 42o) - cos 48o= cos48o - cos48o=0
hok tốt
B=cos56o-tan34o=tan(90o - 56o) - tan34o=tan34o - tan34o=0
a) \(A=2sin30^o+3cos45^o-sin60^0\)
\(\Leftrightarrow A=2.\dfrac{1}{2}+3.\dfrac{\sqrt[]{2}}{2}-\dfrac{\sqrt[]{3}}{2}\)
\(\Leftrightarrow A=1+\dfrac{3\sqrt[]{2}}{2}-\dfrac{\sqrt[]{3}}{2}\)
\(\Leftrightarrow A=1+\dfrac{\sqrt[]{3}\left(\sqrt[]{6}-1\right)}{2}\)
b) \(B=3cos30^o+3sin45^o-cos45^o\)
\(\Leftrightarrow B=3\dfrac{\sqrt[]{3}}{2}+3\dfrac{\sqrt[]{2}}{2}-\dfrac{\sqrt[]{2}}{2}\)
\(\Leftrightarrow B=\dfrac{3\sqrt[]{3}}{2}+\dfrac{2\sqrt[]{2}}{2}\)
\(\Leftrightarrow B=\dfrac{3\sqrt[]{3}}{2}+\sqrt[]{2}\)
\(tử:=\dfrac{1}{2}\left[sin\left(60^o-x+30^o-x\right)+sin\left(60^o-x-30^2+x\right)\right]+\dfrac{1}{2}\left[sin\left(30^o-x+60^o-x\right)+sin\left(30^o-x-60^o+x\right)\right]\)
\(=\dfrac{1}{2}\left[2sin\left(\dfrac{\pi}{2}-2x\right)+sin\left(\dfrac{\pi}{6}\right)+sin\left(-\dfrac{\pi}{6}\right)\right]=\dfrac{1}{2}.\left[2sin\left(\dfrac{\pi}{2}-2x\right)+0\right]=sin\left(\dfrac{\pi}{2}-2x\right)=cos2x\)
\(VT=\dfrac{cos2x}{sin4x}=\dfrac{cos2x}{2sin2x.cos2x}=\dfrac{1}{2sin2x}=\dfrac{1}{4sinx.cosx}=\dfrac{\dfrac{1}{cos^2x}}{\dfrac{4sinx.cosx}{cos^2x}}=\dfrac{1+tan^2x}{\dfrac{4sĩnx}{cosx}}=\dfrac{1+tan^2x}{4tanx}=VP\)
