Cho cota = 3. Khi đó 3 sin a - 2 c o s a 12 sin 3 a + 4 cos 3 a có giá trị bằng
A. - 1 4
B. - 5 4
C. 3 4
D. 1 4
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\(sin^2\alpha+cos^2\alpha=1\\ \Rightarrow cos\alpha=\sqrt{1-sin^2\alpha}\\ \Rightarrow cos\alpha=\dfrac{2\sqrt{2}}{3}\)
Mà \(90^0< \alpha< 180^0\)
\(\Rightarrow cos\alpha=-\dfrac{2\sqrt{2}}{3}\)
\(cot\alpha=3\Leftrightarrow\dfrac{cos\alpha}{sin\alpha}=3\Leftrightarrow cos\alpha=3sin\alpha\)
Khi đó:
\(\dfrac{3sin\alpha-2cos\alpha}{12sin^3\alpha+4cos^3\alpha}=\dfrac{3sin\alpha-6sin\alpha}{12sin^3\alpha+108sin^3\alpha}=-\dfrac{3sin\alpha}{120sin^3\alpha}=-\dfrac{1}{40sin^2\alpha}\)
\(sina=\frac{3}{5}\Rightarrow sin^2a=\frac{9}{25}\) ; \(cos^2a=1-\frac{9}{25}=\frac{16}{25}\)
\(A=\frac{cota+tana}{cota-tana}=\frac{sina.cosa\left(cota+tana\right)}{sina.cosa\left(cota-tana\right)}=\frac{cos^2a+sin^2a}{cos^2a-sin^2a}=\frac{1}{cos^2a-sin^2a}=\frac{1}{\frac{16}{25}-\frac{9}{25}}=\frac{25}{7}\)
\(B=\frac{sin^2a-cos^2a}{sin^2a-3cos^2a}=\frac{\frac{sin^2a}{sin^2a}-\frac{cos^2a}{sin^2a}}{\frac{sin^2a}{sin^2a}-\frac{3cos^2a}{sin^2a}}=\frac{1-cot^2a}{1-3cot^2a}=\frac{1-\left(-\frac{1}{3}\right)^2}{1-3\left(-\frac{1}{3}\right)^2}=\)
\(C_1=sin^2a+cos^2a+cos^2a=1+cos^2a=1+\frac{1}{1+tan^2a}=1+\frac{1}{1+\left(-2\right)^2}\)
\(C_2=\left(sin^2a+cos^2a\right)\left(sin^2a-cos^2a\right)=sin^2a-cos^2a=1-2cos^2a\)
\(=1-\frac{2}{1+tan^2a}=1-\frac{2}{1+\left(-2\right)^2}\)
Bài 2:
Sửa đề: \(\sin\alpha=\dfrac{3}{5}\)
Ta có: \(\sin^2\alpha+\cos^2\alpha=1\)
\(\Leftrightarrow\cos^2\alpha=1-\left(\dfrac{3}{5}\right)^2=1-\dfrac{9}{25}=\dfrac{16}{25}\)
\(\Leftrightarrow\cos\alpha=\dfrac{4}{5}\)
Ta có: \(\tan\alpha=\dfrac{\sin\alpha}{\cos\alpha}\)
\(=\dfrac{3}{5}:\dfrac{4}{5}=\dfrac{3}{5}\cdot\dfrac{5}{4}=\dfrac{3}{4}\)
Ta có: \(\cot\alpha=\dfrac{1}{\tan\alpha}\)
\(=\dfrac{1}{\dfrac{3}{4}}=\dfrac{4}{3}\)
Ta có \sin 5\alpha -2\sin \alpha \left({\cos} 4\alpha +\cos 2\alpha \right)=\sin 5\alpha -2\sin \alpha .\cos 4\alpha -2\sin \alpha .\cos 2\alphasin5α−2sinα(cos4α+cos2α)=sin5α−2sinα.cos4α−2sinα.cos2α
=\sin 5\alpha -\left(\sin 5\alpha -\sin 3\alpha \right)-\left(\sin 3\alpha -\sin \alpha \right)=sin5α−(sin5α−sin3α)−(sin3α−sinα)
=\sin \alpha .=sinα.
Vậy \sin 5\alpha -2\sin \alpha \left({\cos} 4\alpha +\cos 2\alpha \right)=\sin \alphasin5α−2sinα(cos4α+cos2α)=sinα
áp dụng công thức sin2a+cos2a=1
A= sin2a +cos2a-2sina.cosa-sin2a-cos2a+2sina.cosa = 0
B=(sỉn2a+cos2a)2 =12 =1
C= cos2a(cos2a+sin2a)+ sin2a=cos2a+sin2a=1
D=sin2a(sin2p+cos2p)+cos2a=sin2a+cos2a=1
E= (sin2a+cos2a)(sin4a-sin2a.cos2a+cos4a)+3sin2a.cos2a
=sin4a+2sin2a.cos2a+ cos4a=(sin2a+cos2a)2=1
a) Ta có : sin\(^2\)12o=cos278o=> sin212o+sin278o=1.
tương tự => A=3
b) tương tự câu (a) ta có: cos215o=sin275o ( do 15+75=90 nha bạn ) => cos215o+cos275o=1. Tương tự => B=0
Chọn A.