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Bài 1:
Ta có:
\(A=\sin ^6a+\cos ^6a+3\sin ^2a\cos ^2a\)
\(=(\sin ^2a)^3+(\cos ^2a)^3+3\sin ^2a\cos ^2a\)
\(=(\sin ^2a+\cos ^2a)(\sin ^4a-\sin ^2a\cos ^2a+\cos ^4a)+3\sin ^2a\cos ^2a\)
\(=\sin ^4a-\sin ^2a\cos ^2a+\cos ^4a+3\sin ^2a\cos ^2a\)
\(=\sin ^4a+2\sin ^2a\cos ^2a+\cos ^4a\)
\(=(\sin ^2a+\cos ^2a)^2=1^2=1\)
Lời giải:
Xét tam giác $ABC$. Gọi cạnh $AB, AC$ là $a,b$ và góc \(\widehat{BAC}=\alpha\)
Kẻ đường cao $BH$ của tam giác $ABC$
Khi đó:
\(S=\frac{BH.AC}{2}\)
Mặt khác, theo công thức lượng giác:
\(\frac{BH}{AB}=\sin \widehat{BAC}=\sin \alpha\Rightarrow BH=\sin \alpha.AB\)
Do đó: \(S=\frac{BH.AC}{2}=\frac{\sin \alpha.AB.AC}{2}=\frac{\sin \alpha.a.b}{2}\) (đpcm)
\(\sin^4\alpha+\cos^4\alpha=\left(\sin^2\alpha+\cos^2\alpha\right)^2-2\sin^2\alpha.\cos^2\alpha=1-2.\frac{1}{4^2}=\frac{7}{8}\)
Ta có:
\(\hept{\begin{cases}3sina+cosa=2\\sin^2a+cos^2a=1\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}cosa=2-3sina\left(1\right)\\sin^2a+\left(2-3sina\right)^2=1\left(2\right)\end{cases}}\)
\(\left(2\right)\Leftrightarrow10sin^2a-12sina+3=0\)
\(\Leftrightarrow\orbr{\begin{cases}sina=\frac{3}{5}+\frac{\sqrt{6}}{10}\\sina=\frac{3}{5}-\frac{\sqrt{6}}{10}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}cosa=\frac{1}{5}-\frac{3.\sqrt{6}}{10}\left(l\right)\\cosa=\frac{1}{5}+\frac{3.\sqrt{6}}{10}\end{cases}}\)
Thế vô tính tiếp
\(A=sin^6\alpha+cos^6\alpha+3sin^2\alpha-cos^2\alpha\)
\(=\left(sin^2\alpha\right)^3+\left(cos^2\alpha\right)^3+3sin^2\alpha-cos^2\alpha\)
\(=\left(sin^2\alpha+cos^2\alpha\right)\left(sin^4\alpha+cos^4\alpha-sin^2\alpha.cos^2\alpha\right)+3sin^2\alpha-cos^2\alpha\)
\(=sin^4\alpha+cos^4\alpha-sin^2\alpha.cos^2\alpha+3sin^2\alpha-cos^2\alpha\)
\(=\left(sin^2\alpha+cos^2\alpha\right)^2-2sin^2\alpha.cos^2\alpha-sin^2\alpha.cos^2\alpha+3sin^2\alpha-cos^2\alpha\)
\(1-3sin^2\alpha.cos^2\alpha+3sin^2\alpha-cos^2\alpha=3sin^2\alpha\left(1-cos^2\alpha\right)+\left(1-cos^2\alpha\right)\)
\(=\left(3sin^2\alpha+1\right).sin^2\alpha=0\)
Ta có:
\(sin^2a+cos^2a=1\Leftrightarrow sin^2a+\left(\frac{1}{3}\right)^2=1\Leftrightarrow sin^2a=\frac{8}{9}\Rightarrow sina=\frac{2\sqrt{2}}{3}.\)
\(B=\frac{sin\alpha-3cosa}{sina+2cosa}=\frac{\frac{2\sqrt{2}}{3}-3.\frac{1}{3}}{\frac{2\sqrt{2}}{3}+2.\frac{1}{3}}=\frac{7-5\sqrt{2}}{2}\)
\(A=\dfrac{\dfrac{3sina}{cosa}-\dfrac{5cosa}{cosa}}{\dfrac{5sina}{cosa}+\dfrac{8cosa}{cosa}}=\dfrac{3tana-5}{5tana+8}=\dfrac{3.\left(\dfrac{5}{7}\right)-5}{5.\left(\dfrac{5}{7}\right)+8}=...\)