Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(sin^2A+sin^2B+sin^2C=2\)
\(\Leftrightarrow sin^2A+\dfrac{1-cos2B}{2}+\dfrac{1-cos2C}{2}=2\)
\(\Leftrightarrow sin^2A-\dfrac{1}{2}\left(cos2B+cos2C\right)=1\)
\(\Leftrightarrow1-cos^2A-cos\left(B+C\right)cos\left(B-C\right)=1\)
\(\Leftrightarrow cos^2A+cos\left(B+C\right)cos\left(B-C\right)=0\)
\(\Leftrightarrow cos^2A-cosA.cos\left(B-C\right)=0\)
\(\Leftrightarrow cosA\left[cosA-cos\left(B-C\right)\right]=0\)
\(\Leftrightarrow cosA.sin\left(\dfrac{A+B-C}{2}\right)sin\left(\dfrac{A+C-B}{2}\right)=0\)
\(\Leftrightarrow cosA.sin\left(90^0-C\right)sin\left(90^0-B\right)=0\)
\(\Leftrightarrow cosA.cosB.cosC=0\)
\(\Leftrightarrow\left[{}\begin{matrix}A=90^0\\B=90^0\\C=90^0\end{matrix}\right.\) hay tam giác ABC vuông
\(sin^2\dfrac{A}{2}=\dfrac{b-c}{2b}\)
\(\Leftrightarrow\dfrac{1-cosA}{2}=\dfrac{b-c}{2b}\)
\(\Leftrightarrow1-\dfrac{b^2+c^2-a^2}{2bc}=\dfrac{b-c}{b}=1-\dfrac{c}{b}\)
\(\Leftrightarrow b^2+c^2-a^2=2c^2\)
\(\Leftrightarrow a^2+c^2=b^2\)
Tam giác vuông tại B
\(sin^3A.sin\left(B-C\right)=sin^2A.sinA.sin\left(B-C\right)\)
\(=sin^2A.sin\left(B+C\right).sin\left(B-C\right)=-\frac{1}{2}sin^2A\left(cos2B-cos2C\right)\)
\(=-\frac{1}{2}sin^2A\left(1-2sin^2B-1+2sin^2C\right)=sin^2A.sin^2B-sin^2A.sin^2C\)
\(\left\{{}\begin{matrix}sinA=\dfrac{a}{2R}\\sinB=\dfrac{b}{2R}\\sinC=\dfrac{c}{2R}\end{matrix}\right.\) \(\Rightarrow sin^2A+sin^2B=\dfrac{a^2+b^2}{4R^2}=\dfrac{9+36}{4R^2}=\dfrac{45}{4R^2}\)
Trong khi đó \(3sin^2C=\dfrac{3.17}{4R^2}=\dfrac{51}{4R^2}\)
Đề bài sai