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a)\(\left(\sin x+\cos x\right)^2=\sin^2x+\cos^2x+2\sin x\cdot\cos x\)
\(=1+2\cdot\frac{1}{2}=1+1=2\)
\(\Rightarrow\sin x+\cos x=\sqrt{2}\)
b)\(\sin^4x+\cos^4x=\left(\sin^2x+\cos^2x\right)^2-2\sin^2x\cdot\cos^2x\)
\(=1^2-2\cdot\frac{1}{2}^2=1-\frac{1}{2}=\frac{1}{2}\)
c)\(\left|\sin x-\cos x\right|^2=\left(\sin x-\cos x\right)^2=\sin^2x+\cos^2x-2\sin x\cdot\cos x=1-2\cdot\frac{1}{2}=1-1=0\)
\(\left|\sin x+\cos x\right|=0\)
\(\left(sinx+cosx\right)^2=\frac{25}{16}\Rightarrow sin^2x+cos^2x+2sinxcosx=\frac{25}{16}\)
\(\Rightarrow2sinxcosx=\frac{25}{16}-1=\frac{9}{16}\Rightarrow A=\frac{9}{32}\)
\(B^2=\left(sinx-cosx\right)^2=1-2sinx.cosx=1-\frac{9}{16}=\frac{7}{16}\Rightarrow B=\pm\frac{\sqrt{7}}{4}\)
\(C=\left(sinx+cosx\right)\left(sinx-cosx\right)=\frac{5}{4}.\left(\pm\frac{\sqrt{7}}{4}\right)=\pm\frac{5\sqrt{7}}{16}\)
ta co \(sin^2a+cos^2a=1\Rightarrow cosa=0.36\)
\(\frac{sina}{cosa}=tana\Rightarrow tana=\frac{20}{9}\)
\(tana\cdot cotga=1\Rightarrow cotga=\frac{9}{20}\)
câu b tương tự nha cau c \(\frac{sina+cosa}{sina-cosa}=\) bn
\(\sin^6x+\cos^6x\\ =\left(\sin^2x\right)^3+\left(\cos^2x\right)^3\\ =\left(\sin^2x+\cos^2x\right)^3-3\sin^2x\cos^2x\left(\sin^2x+\cos^2x\right)\\ =1-3\sin^2x\cos^2x\left(đpcm\right)\)
\(sin^6x+cos^6x\)
=\(\left(sin^2x+cos^2x\right)\left(sin^4x-sin^2x.cos^2x+cos^4x\right)\)
=\(sin^4x-sin^2x.cos^2x+cos^4x\)
=\(\left(1-2sin^2x.cos^2x\right)-sin^2x.cos^2x\)
=\(1-3sin^2x.cos^2x\)(đpcm)
➞\(sin^6x+cos^6x\)=\(1-3sin^2x.cos^2x\)
Điều kiện góc cần tính là góc nhọn
\(1+tan^2x=1+\frac{sin^2x}{cos^2x}=\frac{sin^2x+cos^2x}{cos^2x}=\frac{1}{cos^2x}\)
\(\Rightarrow cos^2x=\frac{1}{1+tan^2x}\Rightarrow cosx=\frac{1}{\sqrt{1+tan^2x}}=\frac{4}{5}\)
\(\Rightarrow sinx=\sqrt{1-cos^2x}=\sqrt{1-\frac{16}{25}}=\frac{3}{5}\)
\(\Rightarrow A=\frac{7}{5}\)