B= ( 3sina + 4cosa)^2 + (4sina - 3cos^2)^2
Biết góc nhọn a. Tính B?
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a.
\(tana=\dfrac{sina}{cosa}=\dfrac{1}{15}\Rightarrow sina=\dfrac{cosa}{15}\)
\(\Rightarrow sin2a=2sina.cosa=\dfrac{2cosa}{15}.cosa=\dfrac{2}{15}cos^2a=\dfrac{2}{15}.\dfrac{1}{1+tan^2a}=\dfrac{2}{15}.\dfrac{1}{1+\dfrac{1}{15^2}}=\dfrac{15}{113}\)
b.
\(5^2=\left(3sina+4cosa\right)^2\le\left(3^2+4^2\right)\left(sin^2+cos^2a\right)=25\)
Đẳng thức xảy ra khi và chỉ khi: \(\left\{{}\begin{matrix}\dfrac{sina}{3}=\dfrac{cosa}{4}\\3sina+4cosa=5\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}sina=\dfrac{3}{5}\\cosa=\dfrac{4}{5}\end{matrix}\right.\)
c.
\(\dfrac{1}{tan^2a}+\dfrac{1}{cot^2a}+\dfrac{1}{sin^2a}+\dfrac{1}{cos^2a}=7\)
\(\Leftrightarrow\dfrac{cos^2a}{sin^2a}+\dfrac{sin^2a}{cos^2a}+\dfrac{1}{sin^2a}+\dfrac{1}{cos^2a}=7\)
\(\)\(\Leftrightarrow\dfrac{sin^4a+cos^4a}{sin^2a.cos^2a}+\dfrac{sin^2a+cos^2a}{sin^2a.cos^2a}=7\)
\(\Leftrightarrow\dfrac{\left(sin^2a+cos^2a\right)^2-2sin^2a.cos^2a}{sin^2a.cos^2a}+\dfrac{1}{sin^2a.cos^2a}=7\)
\(\Leftrightarrow\dfrac{2}{sin^2a.cos^2a}=9\)
\(\Leftrightarrow\dfrac{8}{\left(2sina.cosa\right)^2}=9\)
\(\Leftrightarrow\dfrac{8}{sin^22a}=9\)
\(\Leftrightarrow sin^22a=\dfrac{8}{9}\)
\(3sina-\sqrt{3}\cdot cosa=0\)
=>\(3\cdot sina=\sqrt{3}\cdot cosa\)
=>\(\dfrac{sina}{cosa}=\dfrac{\sqrt{3}}{3}=\dfrac{1}{\sqrt{3}}\)
=>\(tana=\dfrac{1}{\sqrt{3}}\)
=>\(a=30^0\)
\(tan\left(a+b\right)=\dfrac{tana+tanb}{1-tana.tanb}=1\)
\(\Rightarrow a+b=45^0\)
b: \(=-cos\left(3\cdot10\right)=-cos30=-\dfrac{\sqrt{3}}{2}\)
c: \(=\dfrac{1}{2}\cdot\left(2-4\cdot\dfrac{2+\sqrt{3}}{4}\right)\)
=-căn 3/2
a)\(VT=sinA+sinB+sinC=2sin\frac{A+B}{2}.cos\frac{A-B}{2}+2sin\frac{C}{2}.cos\frac{C}{2}\)
\(=2cos\frac{C}{2}\left(cos\frac{A-B}{2}+cos\frac{A+B}{2}\right)=4cos\frac{C}{2}.cos\frac{A}{2}.cos\frac{B}{2}\)(đpcm)
\(B=\left(3sina+4cosa\right)^2+\left(4sina-3cosa\right)^2\)
\(=9sin^2a+24sina.cosa+16cos^2a+16sin^2a-24sina.cosa+9cos^2a\)
\(=33sin^2a+33cos^2a=33\)