Cho sin α + cos α= 2 phần 5 tính P= sin^3 α+ cos^3 α
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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) Ta có: \(\sin^2\alpha+\cos^2\alpha=1\)
\(\Leftrightarrow\cos^2\alpha=1-\dfrac{9}{25}=\dfrac{16}{25}\)
Ta có: \(A=5\cdot\sin^2\alpha+6\cdot\cos^2\alpha\)
\(=5\left(\sin^2\alpha+\cos^2\alpha\right)+\cos^2\alpha\)
\(=5+\dfrac{16}{25}=\dfrac{141}{25}\)
a: \(\dfrac{\cos\alpha}{1-\sin\alpha}=\dfrac{1+\sin\alpha}{\cos\alpha}\)
\(\Leftrightarrow\cos^2\alpha=1-\sin^2\alpha\)(đúng)
b: Ta có: \(\dfrac{\left(\sin\alpha+\cos\alpha\right)^2-\left(\sin\alpha-\cos\alpha\right)^2}{\sin\alpha\cdot\cos\alpha}\)
\(=\dfrac{4\cdot\sin\alpha\cdot\cos\alpha}{\sin\alpha\cdot\cos\alpha}\)
=4
Ta có : P = sin3 α + cos3 α = ( sinα + cosα) 3 - 3sin α.cosα(sinα + cosα)
Ta có (sin α + cos α) 2 = sin2α + cos2α + 2sinα.cosα = 1 + 24/25 = 49/25.
Vì sin α + cosα > 0 nên ta chọn sinα + cosα = 7/5.
Thay vào P ta được
Có \(sin\alpha+cos\alpha=\dfrac{2}{5}\Leftrightarrow\left(sin\alpha+cos\alpha\right)^2=\dfrac{4}{25}\)
\(\Leftrightarrow sin^2\alpha+2sin\alpha\cdot cos\alpha+cos^2\alpha=\dfrac{4}{25}\)
\(\Leftrightarrow\left(sin^2\alpha+cos^2\alpha\right)+2sin\alpha\cdot cos\alpha=\dfrac{4}{25}\)
\(\Leftrightarrow1+2sin\alpha\cdot cos\alpha=\dfrac{4}{25}\Leftrightarrow sin\alpha\cdot cos\alpha=-\dfrac{21}{50}\)
Ta có:
\(P=sin^3\alpha+cos^3\alpha=\left(sin\alpha+cos\alpha\right)\left(sin^2\alpha-sin\alpha\cdot cos\alpha+cos^2\alpha\right)\)
\(=2\left[1-\left(-\dfrac{21}{50}\right)\right]=\dfrac{71}{25}\)