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\(VT=\frac{1-\cos x}{\sin x}=\frac{\sin x\left(1-\cos x\right)}{\sin^2x}=\frac{\sin x\left(1-\cos x\right)}{1-\cos^2x}=\frac{\sin x\left(1-\cos x\right)}{\left(1-\cos x\right)\left(1+\cos x\right)}=\frac{\sin x}{1+\cos x}\)
Nếu ko CM theo cách này thì bn nhân chéo lên nó sẽ ra điều luôn đúng => ĐT đc CM
Lời giải:
Ta có:
\(\frac{\cot ^2a-\cos ^2}{\cot ^2a}+\frac{\sin a\cos a}{\cot a}=1-\frac{\cos ^2a}{\cot ^2a}+\frac{\sin a\cos a}{\cot a}\)
\(=1-\frac{\cos ^2a}{\frac{\cos ^2a}{\sin ^2a}}+\frac{\sin a\cos a}{\frac{\cos a}{\sin a}}=1-\sin ^2a+\sin ^2a=1\)
Ta có đpcm.
Có \(\sin^2x+\cos^2x=1\Rightarrow2\sin^2x=1-\cos^2x+\sin^2x\)
\(\Rightarrow1+\sin^2x=2\sin^2x+\cos^2x\)
\(\Rightarrow VT=\frac{2\sin^2x+\cos^2x}{\cos^2x}=2\tan^2x+1\)
\(1-\frac{sin^3x}{sinx+cosx}-\frac{cos^3x}{sinx+cosx}=1-\frac{sin^3x+cos^3x}{sinx+cosx}\)
\(=1-\frac{\left(sinx+cosx\right)\left(sin^2x+cos^2x-sinx.cosx\right)}{sinx+cosx}=1-\left(1-sinxcosx\right)\)
\(=sinx.cosx\)
\(VP=\frac{2\sin^2x-1}{\sin^4x}=\frac{\sin^2x+\sin^2x-1}{\sin^4x}=\frac{\sin^2x-\cos^2x}{\sin^4x}\)
\(=\frac{\left(\sin^2x-\cos^2x\right).1}{\sin^4x}=\frac{\left(\sin^2x-\cos^2x\right)\left(\sin^2x+\cos^2x\right)}{\sin^4x}=\frac{\sin^4x-\cos^4x}{\sin^4x}\)
\(=1-\cot^4x\)=VT
\(VT=\sin^2\alpha.\frac{\sin\alpha}{\cos\alpha}+\cos^2\alpha.\frac{\cos\alpha}{\sin\alpha}+2\sin\alpha.\cos\alpha\)
\(=\frac{\sin^4\alpha+\cos^4\alpha+2\sin^2\alpha.\cos^2\alpha}{\sin\alpha.\cos\alpha}=\frac{\left(\sin^2\alpha+\cos^2\alpha\right)^2}{\sin\alpha.\cos\alpha}=\frac{1}{\sin\alpha.\cos\alpha}\)
\(=\frac{\sin^2\alpha+\cos^2\alpha}{\sin\alpha.\cos\alpha}=\tan\alpha+\cot\alpha=VP\)
P/s: đổi \(\alpha\) thành x nha! Làm gần hết bài ms nhớ ra ! :D
Giả sử có \(\Delta ABC\) có \(A=90^o;AH\) là đường cao
Có \(\sin\widehat{B}=\frac{AC}{BC};\cos\widehat{B}=\frac{AB}{BC};\tan\widehat{B}=\frac{AC}{AB};\cot\widehat{B}=\frac{AB}{AC}\)
\(\frac{\cot^2\widehat{B}-\cos^2\widehat{B}}{\cot^2\widehat{B}}+\frac{\sin\widehat{B}.\cos\widehat{B}}{\cot\widehat{B}}=\frac{\frac{AB^2}{AC^2}-\frac{AB^2}{BC^2}}{\frac{AB^2}{AC^2}}+\frac{\frac{AC}{BC}.\frac{AB}{BC}}{\frac{AB}{AC}}\)
\(=\frac{\frac{AB^2}{AC^2}}{\frac{AB^2}{AC^2}}-\frac{\frac{AB^2}{BC^2}}{\frac{AB^2}{AC^2}}+\frac{\frac{AC.AB}{BC^2}}{\frac{AB}{AC}}=1-\frac{AC^2}{BC^2}+\frac{AC^2}{BC^2}=1\)
\(\sin^6x+\cos^6x\\ =\left(\sin^2x\right)^3+\left(\cos^2x\right)^3\\ =\left(\sin^2x+\cos^2x\right)^3-3\sin^2x\cos^2x\left(\sin^2x+\cos^2x\right)\\ =1-3\sin^2x\cos^2x\left(đpcm\right)\)
\(sin^6x+cos^6x\)
=\(\left(sin^2x+cos^2x\right)\left(sin^4x-sin^2x.cos^2x+cos^4x\right)\)
=\(sin^4x-sin^2x.cos^2x+cos^4x\)
=\(\left(1-2sin^2x.cos^2x\right)-sin^2x.cos^2x\)
=\(1-3sin^2x.cos^2x\)(đpcm)
➞\(sin^6x+cos^6x\)=\(1-3sin^2x.cos^2x\)