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\(A=sin^210+sin^220+sin^230+sin^280+sin^270+sin^260=sin^210+sin^220+sin^230+cos^210+cos^220+cos^230=1+1+1=3\)\(B=\left(1+tan^2\alpha\right)\left(1-sin^2\alpha\right)+\left(1+cot^2\alpha\right)\left(1-cos^2\alpha\right)=\dfrac{1}{cos^2\alpha}.cos^2\alpha+\dfrac{1}{sin^2\alpha}.sin^2\alpha=1+1=2\)
tui rất thích lượng giác:
a) = s2 + 2s.c +c2 +s2- 2s.c + c2 =1+1=2
b) = s.c(s/c + c/s) = s.c(s2 + c2) / s.c = 1
.............................bài nào cx dễ
( k có việc j khó, chỉ sợ lòng k bền....)
ta có : \(A=\dfrac{sin^3\alpha+cos^3\alpha}{2sin\alpha.cos^2\alpha+cos^2\alpha.sin^2\alpha}\)
\(\Leftrightarrow A=\dfrac{\dfrac{sin^3\alpha}{cos^3\alpha}+\dfrac{cos^3\alpha}{cos^3\alpha}}{\dfrac{2sin\alpha.cos^2\alpha}{cos^3\alpha}+\dfrac{cos\alpha.sin^2\alpha}{cos^3\alpha}}=\dfrac{tan^3\alpha+1}{2tan\alpha+tan^2\alpha}\)
\(\Leftrightarrow A=\dfrac{\left(\dfrac{3}{4}\right)^3+1}{2\left(\dfrac{3}{4}\right)+\left(\dfrac{3}{4}\right)^2}=\dfrac{91}{132}\)
Biểu thức\(=\tan^2\alpha.\cos^2\alpha+\tan^2\alpha.\cos^2\alpha.\cot^2\alpha\)\(=\frac{sin^2\alpha}{\cos^2\alpha}.\cos^2\alpha+\cos^2\alpha=\sin^2\alpha+\cos^2\alpha=1\)
a) ta có : \(sin^2\alpha+cos^2\alpha=1\Leftrightarrow sin^2\alpha=1-cos^2\alpha\)
\(\Leftrightarrow sin^2\alpha=\left(1-cos\alpha\right)\left(1+cos\alpha\right)\Leftrightarrow\dfrac{sin\alpha}{1+cos\alpha}=\dfrac{1-cos\alpha}{sin\alpha}\left(đpcm\right)\)
b) ta có : \(tan^2\alpha-sin^2\alpha=sin^2\alpha\left(\dfrac{1}{cos^2\alpha}-1\right)=sin^2\alpha\left(\dfrac{1-cos^2\alpha}{cos^2\alpha}\right)\)
\(=sin^2\alpha.\dfrac{sin^2\alpha}{cos^2\alpha}=sin^2\alpha.tan^2\alpha\left(đpcm\right)\)
Sao ko chuyển về cái kia nó dễ hiểu hơn :v AHihi
\(sin^2a=\left(1-cosa\right)\left(1+cosa\right)\Leftrightarrow sin^2a=1-cos^2a\Leftrightarrow sin^2a+cos^2a=1\)