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ta có : \(sin^2\alpha+cos^2\alpha=1\Leftrightarrow sin^2\alpha+\dfrac{9}{16}=1\Leftrightarrow sin^2\alpha=\dfrac{7}{16}\)
\(\Leftrightarrow sin\alpha=\pm\dfrac{\sqrt{7}}{4}\)
với \(sin\alpha=\dfrac{\sqrt{7}}{4}\)\(\Rightarrow tan\alpha=\dfrac{sin\alpha}{cos\alpha}=\dfrac{\dfrac{\sqrt{7}}{4}}{\dfrac{3}{4}}=\dfrac{\sqrt{7}}{3}\) \(\Rightarrow cot=\dfrac{3}{\sqrt{7}}\)
với \(sin\alpha=\dfrac{-\sqrt{7}}{4}\)\(\Rightarrow tan\alpha=\dfrac{sin\alpha}{cos\alpha}=\dfrac{\dfrac{-\sqrt{7}}{4}}{\dfrac{3}{4}}=\dfrac{-\sqrt{7}}{3}\) \(\Rightarrow cot=\dfrac{-3}{\sqrt{7}}\)
vậy \(sin\alpha=\pm\dfrac{\sqrt{7}}{4}\) ; \(tan\alpha=\pm\dfrac{\sqrt{7}}{3}\) ; \(cot=\pm\dfrac{3}{\sqrt{7}}\)
cos an pha =căn(1-sin2anpha)=\(\sqrt{1-\left(\dfrac{7}{25}\right)^2}\)=\(\dfrac{24}{25}\)
cot anpha =cos anpha :sin anpha =\(\dfrac{24}{25}\):\(\dfrac{7}{25}\) =\(\dfrac{24}{7}\)
a: 
b: \(B=3-sin^290^0+2\cdot cos^260^0-3\cdot tan^245^0\)
\(=3-1+2\cdot\left(\dfrac{1}{2}\right)^2-3\cdot1^2\)
\(=2-3+2\cdot\dfrac{1}{4}=-1+\dfrac{1}{2}=-\dfrac{1}{2}\)
c: \(C=sin^245^0-2\cdot sin^250^0+3\cdot cos^245^0-2\cdot sin^240^0+4\cdot tan55\cdot tan35\)
\(=\left(\dfrac{\sqrt{2}}{2}\right)^2+3\cdot\left(\dfrac{\sqrt{2}}{2}\right)^2-2\cdot\left(sin^250^0+sin^240^0\right)+4\)
\(=\dfrac{1}{2}+3\cdot\dfrac{1}{2}-2+4\)
\(=2-2+4=4\)
\(90^0< a< 180^0\)
=>\(cosa< 0\)
\(sin^2a+cos^2a=1\)
=>\(cos^2a+\dfrac{9}{25}=1\)
=>\(cos^2a=1-\dfrac{9}{25}=\dfrac{16}{25}\)
mà cosa<0
nên \(cosa=-\dfrac{4}{5}\)
\(tana=\dfrac{sina}{cosa}=\dfrac{3}{5}:\dfrac{-4}{5}=-\dfrac{3}{4}\)
\(A=2\cdot cos^2a-5\cdot tan^2a\)
\(=2\cdot\left(-\dfrac{4}{5}\right)^2-5\cdot\left(-\dfrac{3}{4}\right)^2\)
\(=2\cdot\dfrac{16}{25}-5\cdot\dfrac{9}{16}\)
\(=\dfrac{32}{25}-\dfrac{45}{16}=\dfrac{-613}{400}\)
Ta có: \(1+tan^2a=\frac{1}{cos^2a}\)
=>\(\frac{1}{cos^2a}=1+3^2=10\)
=>\(cos^2a=\frac{1}{10}\)
TA có: \(\sin^2a+cos^2a=1\)
=>\(\sin^2a=1-\frac{1}{10}=\frac{9}{10}\)
\(\sin^4a-cos^4a=\left(\sin^2a-cos^2a\right)\left(\sin^2a+cos^2a\right)\)
\(=\sin^2a-cos^2a=\frac{9}{10}-\frac{1}{10}=\frac{8}{10}=\frac45\)
\(A=\frac{3\cdot\sin^2a+5}{\sin^4a-cos^4a}\)
\(=\left(3\cdot\frac{9}{10}+5\right):\frac45=\left(\frac{27}{10}+\frac{50}{10}\right)\cdot\frac54=\frac{77}{10}\cdot\frac54=\frac{77}{8}\)
=9,625
\(B=cos^2x.cot^2x+cos^2x-cot^2x+2\left(sin^2x+cos^2x\right)\)
\(=cos^2x\left(cot^2x+1\right)-cot^2x+2\)
\(=\frac{cos^2x}{sin^2x}-cot^2x+1=cot^2x-cot^2x+1=1\)
\(M=cos^4x-sin^4x+cos^4x+sin^2x.cos^2x+3sin^2x\)
\(=\left(cos^2x-sin^2x\right)\left(cos^2x+sin^2x\right)+cos^2x\left(cos^2x+sin^2x\right)+3sin^2x\)
\(=cos^2x-sin^2x+cos^2x+3sin^2x\)
\(=2\left(sin^2x+cos^2x\right)=2\)
P= \(1-cos^2x+2cos^2x=1+cos^2x\)
Ta có:
\(0\le cos^2x\le1\)
=> \(1\le P\le2\)
min P=1 <=> \(cos^2x=0\Leftrightarrow cosx=0\Leftrightarrow x=\frac{\pi}{2}+k\pi\)
b) \(\sin x+\cos x=\dfrac{3}{2}\)
\(\left(\sin x+\cos x\right)^2=\dfrac{1}{4}\)
\(\sin^2x+\cos^2x+2\sin x\cos x=\dfrac{1}{4}\)
\(2\sin x\cos x=-\dfrac{3}{4}=\sin2x\)
\(cosa=-\sqrt{1-\dfrac{16}{25}}=-\dfrac{3}{5}\)
\(M=\dfrac{3\cdot\dfrac{4}{5}+2\cdot\dfrac{-3}{5}}{6+16\cdot\left(-\dfrac{3}{5}:\dfrac{4}{5}\right)^2}=\dfrac{\dfrac{6}{5}}{6+16\cdot\dfrac{9}{16}}=\dfrac{\dfrac{6}{5}}{6+9}=\dfrac{6}{5}:15=\dfrac{6}{75}=\dfrac{2}{25}\)
cos a ở đâu vậy?