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1:
a: sin a=căn 3/2
\(cosa=\sqrt{1-sin^2a}=\sqrt{1-\dfrac{3}{4}}=\sqrt{\dfrac{1}{4}}=\dfrac{1}{2}\)
\(tana=\dfrac{\sqrt{3}}{2}:\dfrac{1}{2}=\sqrt{3}\)
cot a=1/tan a=1/căn 3
b: \(tana=2\)
=>cot a=1/tan a=1/2
\(1+tan^2a=\dfrac{1}{cos^2a}\)
=>\(\dfrac{1}{cos^2a}=5\)
=>cos^2a=1/5
=>cosa=1/căn 5
\(sina=\sqrt{1-cos^2a}=\sqrt{\dfrac{4}{5}}=\dfrac{2}{\sqrt{5}}\)
c: \(cosa=\sqrt{1-\left(\dfrac{5}{13}\right)^2}=\dfrac{12}{13}\)
tan a=5/13:12/13=5/12
cot a=1:5/12=12/5
a) \(\dfrac{2sina+3cosa}{3sina-4cosa}=\dfrac{9}{5}\)
b) \(\dfrac{sina.cosa}{sin^2a-sina.cosa+cos^2a}=0\)
\(a.\dfrac{2\sin\alpha+3\cos\alpha}{3\sin\alpha-4\cos\alpha}=\dfrac{2\left(3cos\alpha\right)+3cos\alpha}{3\left(3cos\alpha\right)-4cos\alpha}=\dfrac{9cos\alpha}{5cos\alpha}=\dfrac{9}{5}\)
\(b.\dfrac{sin\alpha cos\alpha}{sin^2\alpha-sin\alpha cos\alpha+cos^2\alpha}=\dfrac{3cos^2\alpha}{9cos^2\alpha-3cos^2\alpha+cos^2\alpha}=\dfrac{3cos^2\alpha}{7cos^2\alpha}=\dfrac{3}{7}\)
a) Ta có: \(\sin^2\alpha+\cos^2\alpha=1\)
\(\Leftrightarrow\cos^2\alpha=1-\dfrac{9}{25}=\dfrac{16}{25}\)
Ta có: \(A=5\cdot\sin^2\alpha+6\cdot\cos^2\alpha\)
\(=5\left(\sin^2\alpha+\cos^2\alpha\right)+\cos^2\alpha\)
\(=5+\dfrac{16}{25}=\dfrac{141}{25}\)
\(ADCT:\sin^2\alpha+\cos^2\alpha=1\)
\(A=\left(\sin^242^0+\sin^248^0\right)+\left(\sin^243^0+\sin^247^0\right)+\left(\sin^244^0+\sin^246^0\right)+\sin45^0\)
\(A=\left(\sin^242^0+\cos^242^0\right)+\left(\sin^243^0+\cos^243^0\right)+\left(\sin^244^0+\cos^244^0\right)+\frac{\sqrt{2}}{2}\)
\(A=1+1+1+\frac{\sqrt{2}}{2}=\frac{6+\sqrt{2}}{2}\)
Câu b lm tương tự
Lời giải:
\(M=\frac{\frac{\sin a}{\cos a}+1}{\frac{\sin a}{\cos a}-1}=\frac{\tan a+1}{\tan a-1}=\frac{\frac{3}{5}+1}{\frac{3}{5}-1}=-4\)
\(N = \frac{\frac{\sin a\cos a}{\cos ^2a}}{\frac{\sin ^2a-\cos ^2a}{\cos ^2a}}=\frac{\frac{\sin a}{\cos a}}{(\frac{\sin a}{\cos a})^2-1}=\frac{\tan a}{\tan ^2a-1}=\frac{\frac{3}{5}}{\frac{3^2}{5^2}-1}=\frac{-15}{16}\)
a: sin a=2/3
=>cos^2a=1-(2/3)^2=5/9
=>\(cosa=\dfrac{\sqrt{5}}{3}\)
\(tana=\dfrac{2}{3}:\dfrac{\sqrt{5}}{3}=\dfrac{2}{\sqrt{5}}\)
\(cota=1:\dfrac{2}{\sqrt{5}}=\dfrac{\sqrt{5}}{2}\)
b: cos a=1/5
=>sin^2a=1-(1/5)^2=24/25
=>\(sina=\dfrac{2\sqrt{6}}{5}\)
\(tana=\dfrac{2\sqrt{6}}{5}:\dfrac{1}{5}=2\sqrt{6}\)
\(cota=\dfrac{1}{2\sqrt{6}}=\dfrac{\sqrt{6}}{12}\)
c: cot a=1/tana=1/2
\(1+tan^2a=\dfrac{1}{cos^2a}\)
=>1/cos^2a=1+4=5
=>cos^2a=1/5
=>cosa=1/căn 5
\(sina=\sqrt{1-cos^2a}=\dfrac{2}{\sqrt{5}}\)
(cos36-sin36)(cos37-sin38)(sin48-sin48)=0
thanks