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\(5sin2a-6cosa=0\)
\(\Leftrightarrow sin2a=\dfrac{6}{5}cosa\)
\(\Leftrightarrow2\cdot sina\cdot cosa=\dfrac{6}{5}\cdot cosa\)
\(\Leftrightarrow cosa\left(2sina-\dfrac{6}{5}\right)=0\)
=>cosa=0 hoặc sina=3/5
hay \(a=\dfrac{\Pi}{2}+k\Pi\) hoặc \(\left[{}\begin{matrix}a=arcsin\left(\dfrac{3}{5}\right)+k2\Pi\\a=\Pi-arcsin\left(\dfrac{3}{5}\right)+k2\Pi\end{matrix}\right.\)
mà 0<a<pi/2
nên \(a=arcsin\left(\dfrac{3}{5}\right)\)
\(A=sina+sina+cota=2\cdot sina+cota\)
\(=\dfrac{38}{15}\)
Theo giả thiết ta có 3 góc: \(\alpha;\beta=\alpha+\dfrac{\pi}{3};\gamma=\alpha+\dfrac{2\pi}{3}\).
Ta có:
\(tan\alpha.tan\left(\alpha+\dfrac{\pi}{3}\right)+tan\left(\alpha+\dfrac{\pi}{3}\right).tan\left(\alpha+\dfrac{2\pi}{3}\right)+\)\(tan\left(\alpha+\dfrac{2\pi}{3}\right).tan\alpha\)
\(=tan\alpha\left[tan\left(\alpha+\dfrac{\pi}{3}\right)+tan\left(\alpha+\dfrac{2\pi}{3}\right)\right]\)\(+tan\left(a+\dfrac{\pi}{3}\right)tan\left(\alpha+\dfrac{2\pi}{3}\right)\)
\(=tan\alpha\dfrac{sin\left(2\alpha+\pi\right)}{cos\left(\alpha+\dfrac{\pi}{3}\right)cos\left(\alpha+\dfrac{2\pi}{3}\right)}\)\(+\dfrac{sin\left(\alpha+\dfrac{\pi}{3}\right)sin\left(\alpha+\dfrac{2\pi}{3}\right)}{cos\left(\alpha+\dfrac{\pi}{3}\right)cos\left(\alpha+\dfrac{2\pi}{3}\right)}\)
\(=tan\alpha\dfrac{-sin2\alpha}{cos\left(\alpha+\dfrac{\pi}{3}\right)cos\left(\alpha+\dfrac{2\pi}{3}\right)}\)\(+\dfrac{cos\dfrac{\pi}{3}-cos\left(2\alpha+\pi\right)}{2cos\left(\alpha+\dfrac{\pi}{3}\right)cos\left(\alpha+\dfrac{2\pi}{3}\right)}\)
\(=\dfrac{-2sin^2\alpha}{cos\left(\alpha+\dfrac{\pi}{3}\right)cos\left(\alpha+\dfrac{2\pi}{3}\right)}\)\(+\dfrac{\dfrac{1}{2}+cos2\alpha}{2cos\left(\alpha+\dfrac{\pi}{3}\right)cos\left(\alpha+\dfrac{2\pi}{3}\right)}\)
\(=\dfrac{\dfrac{1}{2}-4sin^2\alpha+cos2\alpha}{2cos\left(\alpha+\dfrac{\pi}{3}\right)cos\left(\alpha+\dfrac{2\pi}{3}\right)}\)
\(=\dfrac{\dfrac{1}{2}-4\left(1-cos^2\alpha\right)+2cos^2\alpha-1}{cos\dfrac{\pi}{3}+cos\left(2\alpha+\pi\right)}\)
\(=\dfrac{6cos^2\alpha-\dfrac{9}{2}}{\dfrac{1}{2}-cos2\alpha}\)
\(=\dfrac{3\left(2cos^2\alpha-\dfrac{3}{2}\right)}{\dfrac{1}{2}-\left(2cos^2\alpha-1\right)}=\dfrac{3\left(2cos^2\alpha-\dfrac{3}{2}\right)}{\dfrac{3}{2}-2cos^2\alpha}=-3\).
\(4cos\alpha.cos\beta cos\gamma=4cos\alpha cos\left(\alpha+\dfrac{\pi}{3}\right)cos\left(\alpha+\dfrac{2\pi}{3}\right)\)
\(=4cos\alpha.\dfrac{1}{2}\left(cos\dfrac{\pi}{3}+cos\left(2\alpha+\pi\right)\right)\)
\(=4cos\alpha.\dfrac{1}{2}\left(\dfrac{1}{2}-cos2\alpha\right)\)
\(=cos\alpha-2cos\alpha.cos2\alpha\)
\(=cos\alpha-\left(cos\alpha+cos3\alpha\right)\)
\(=-cos3\alpha\)
\(=cos\left(\pi+3\alpha\right)\)
\(=cos3\left(\dfrac{\pi}{3}+\alpha\right)\)
\(=cos3\beta\) (đpcm).
\(P=\frac{sina+cosa}{sina-cosa}=\frac{\frac{sina}{sina}+\frac{cosa}{sina}}{\frac{sina}{sina}-\frac{cosa}{sina}}=\frac{1+cota}{1-cota}=\frac{1+2}{1-2}=-3\)
Ta có : \(A=\frac{\tan\alpha}{1+\tan^2\alpha}=\tan\alpha.\cos^2\alpha=\sin\alpha.\cos\alpha=\frac{3}{5}\cos\alpha\left(1\right)\)
\(\cos^2\alpha=1-\sin^2\alpha=1-\left(\frac{3}{5}\right)^2=\frac{16}{25}\) (2)
Vì \(\alpha\in\left(\frac{\pi}{2};\pi\right)\) nên \(\cos\alpha<0\)
Do đó, từ (2) suy ra \(\cos\alpha=-\frac{4}{5}\) (3)
Thế (3) vào (1) ta được \(A=-\frac{12}{25}\)
sin a=1/4
=>sin^2a=1/16
=>cos^2a=15/16
\(B=\dfrac{3\cdot\dfrac{cosa}{sina}-\dfrac{sina}{cosa}}{2\cdot\dfrac{sina}{cosa}+\dfrac{cosa}{sina}}\)
\(=\dfrac{3\cdot cosa^2a-sin^2a}{sina\cdot cosa}:\dfrac{2\cdot sin^2a+cos^2a}{sina\cdot cosa}\)
\(=\dfrac{3\cdot cos^2a-sin^2a}{2\cdot sin^2a+cos^2a}\)
\(=\dfrac{3\cdot\dfrac{15}{16}-\dfrac{1}{16}}{2\cdot\dfrac{1}{16}+\dfrac{15}{16}}=\dfrac{44}{17}\)