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\(A=\sin^6\alpha+cos^6\alpha+3\sin^2\alpha\cos^2\alpha\left(\sin^2\alpha+\cos^2\alpha\right).\)vì\(\sin^2\alpha+\cos^2\alpha=1\)
\(=\left(\sin^2\alpha+\cos^2\alpha\right)^3=1^3=1\)
\(B=2\left(\cos^2\alpha+\sin^2\alpha\right)=2.1=2\)
\(C=\frac{-4\cos\alpha\sin\alpha}{\sin\alpha\cos\alpha}=-4\)
a) Ta có : sin\(^2\)12o=cos278o=> sin212o+sin278o=1.
tương tự => A=3
b) tương tự câu (a) ta có: cos215o=sin275o ( do 15+75=90 nha bạn ) => cos215o+cos275o=1. Tương tự => B=0
sữa đề chút nha :
+) ta có : \(A=\dfrac{1+2sin\alpha.cos\alpha}{cos^2\alpha-sin^2\alpha}=\dfrac{\left(sin\alpha+cos\alpha\right)^2}{\left(sin\alpha+cos\alpha\right)\left(cos\alpha-sin\alpha\right)}=\dfrac{sin\alpha+cos\alpha}{cos\alpha-sin\alpha}\)
+) ta có :
\(B=sin^6\alpha+cos^6\alpha+3sin^2\alpha.cos^2\alpha\)
\(=\left(sin^2\alpha+cos^2\alpha\right)^3-3sin^2\alpha.cos^2\alpha\left(sin^2\alpha+cos^2\alpha\right)+3sin^2\alpha.cos^2\alpha\)
\(=1-3sin^2\alpha.cos^2\alpha+3sin^2\alpha.cos^2\alpha=1\)
\(1+tan^2a=1+\frac{sin^2a}{cos^2a}=\frac{cos^2a+sin^2a}{cos^2a}=\frac{1}{cos^2a}\)
\(1+cot^2a=1+\frac{cos^2a}{sin^2a}=\frac{sin^2a+cos^2a}{sin^2a}=\frac{1}{sin^2a}\)
\(cot^2a-cos^2a=\frac{cos^2a}{sin^2a}-cos^2a=cos^2a\left(\frac{1}{sin^2a}-1\right)=cos^2a\left(\frac{1-sin^2a}{sin^2a}\right)\)
\(=cos^2a\left(\frac{cos^2a}{sin^2a}\right)=cos^2a.cot^2a\)
\(\frac{1+cosa}{sina}=\frac{sina\left(1+cosa\right)}{sin^2a}=\frac{sina\left(1+cosa\right)}{1-cos^2a}=\frac{sina\left(1+cosa\right)}{\left(1-cosa\right)\left(1+cosa\right)}=\frac{sina}{1-cosa}\)
1.Ta có :
\(\cot41=\tan49\) ; \(\cot46=\tan44\)
sắp xếp :\(\tan27< \tan44< \tan47< \tan49\)\(\Rightarrow\tan27< \cot46< \tan47< \cot41\)
2.ta có
\(\cos28=\sin62;\cos41=\sin49\)
\(A=\cos^228+\cos^241+\cos^262+\cos^249\)
\(\Rightarrow A=\sin^262+\cos^262+\sin^249+\cos^249\)
\(\Rightarrow A=1+1=2\)