a+b+c : dựa vào cái hệ thức \(\sin^2\alpha+\cos^2\alpha=1\)
a) Ta có : \(\left(\sin x+\cos x\right)^2\)
\(=\sin^2x+2.\sin x.\cos x+\cos^2x\)
\(=1+2.\sin x.\cos x\left(đpcm\right)\)
b) Ta có : \(\left(\sin x+\cos x\right)^2+\left(\sin x-\cos x\right)^2\)
\(=\sin^2x+2.\sin x.\cos x+\cos^2x+\sin^2x-2.\sin x.\cos x+\cos^2x\)
\(=\sin^2x+\cos^2x+\sin^2x+\cos^2x\)
\(=2\left(\sin^2x+\cos^2x\right)\)
\(=2\times1=2\left(đpcm\right)\)
c) Ta có : \(\sin^4x+\cos^4x\)
\(=\left(\sin^2x\right)^2+\left(\cos^2x\right)^2\)
\(=\left(\sin^2x+\cos^2x\right)^2-2.\sin^2x.\cos^2x\)
\(=1-2.\sin^2x.\cos^2x\left(đpcm\right)\)
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