tính :A=2a+1.73+2a+1.57
2a.104
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a.
\(tana=\dfrac{sina}{cosa}=\dfrac{1}{15}\Rightarrow sina=\dfrac{cosa}{15}\)
\(\Rightarrow sin2a=2sina.cosa=\dfrac{2cosa}{15}.cosa=\dfrac{2}{15}cos^2a=\dfrac{2}{15}.\dfrac{1}{1+tan^2a}=\dfrac{2}{15}.\dfrac{1}{1+\dfrac{1}{15^2}}=\dfrac{15}{113}\)
b.
\(5^2=\left(3sina+4cosa\right)^2\le\left(3^2+4^2\right)\left(sin^2+cos^2a\right)=25\)
Đẳng thức xảy ra khi và chỉ khi: \(\left\{{}\begin{matrix}\dfrac{sina}{3}=\dfrac{cosa}{4}\\3sina+4cosa=5\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}sina=\dfrac{3}{5}\\cosa=\dfrac{4}{5}\end{matrix}\right.\)
c.
\(\dfrac{1}{tan^2a}+\dfrac{1}{cot^2a}+\dfrac{1}{sin^2a}+\dfrac{1}{cos^2a}=7\)
\(\Leftrightarrow\dfrac{cos^2a}{sin^2a}+\dfrac{sin^2a}{cos^2a}+\dfrac{1}{sin^2a}+\dfrac{1}{cos^2a}=7\)
\(\)\(\Leftrightarrow\dfrac{sin^4a+cos^4a}{sin^2a.cos^2a}+\dfrac{sin^2a+cos^2a}{sin^2a.cos^2a}=7\)
\(\Leftrightarrow\dfrac{\left(sin^2a+cos^2a\right)^2-2sin^2a.cos^2a}{sin^2a.cos^2a}+\dfrac{1}{sin^2a.cos^2a}=7\)
\(\Leftrightarrow\dfrac{2}{sin^2a.cos^2a}=9\)
\(\Leftrightarrow\dfrac{8}{\left(2sina.cosa\right)^2}=9\)
\(\Leftrightarrow\dfrac{8}{sin^22a}=9\)
\(\Leftrightarrow sin^22a=\dfrac{8}{9}\)
\(\sin 2a = \sin \left( {a + a} \right) = \sin a.\cos a + \cos a.\sin a = 2\sin a\cos a\)
\(\begin{array}{l}\cos 2a = \cos \left( {a + a} \right) = \cos a.\cos a - \sin a.\sin a = {\cos ^2}a - {\sin ^2}a\\\tan 2a = \tan \left( {a + a} \right) = \frac{{\tan a + \tan a}}{{1 - \tan a.\tan a}} = \frac{{2\tan a}}{{1 - {{\tan }^2}a}}\end{array}\)
\(\sin 2a = \sin \left( {a + a} \right) = \sin \left( {a + b} \right) = \sin a\cos b + \sin b\cos a = 2\sin a\cos a\)
\(\cos 2a = \cos \left( {a + a} \right) = \cos \left( {a + b} \right) = \cos a\cos b - \sin a\sin b = {\cos ^2}a - {\sin ^2}a = 2{\cos ^2}a - 1\)
\( = 1 - 2{\sin ^2}a\)
\(\tan 2a = \tan \left( {a + a} \right) = \tan \left( {a + b} \right) = \frac{{\tan a + \tan b}}{{1 - \tan a\tan b}} = \frac{{2\tan a}}{{1 - {{\tan }^2}a}}\)
Ta có
\(\frac{2a}{3b}=\frac{3b}{4c}=\frac{4c}{5d}=\frac{5d}{2a}=\frac{2a+3b+4c+5d}{3b+4c+5d+2a}=1.\) (Tính chất dãy tỷ số bằng nhau)
\(\Rightarrow\frac{2a}{3b}+\frac{3b}{4c}+\frac{4c}{5d}+\frac{5d}{2a}=4.1=4\)
\(\dfrac{1}{tan^2a}+\dfrac{1}{cot^2a}+\dfrac{1}{sin^2a}+\dfrac{1}{cos^2a}=7\)
=>\(\dfrac{sin^2a+1}{cos^2a}+\dfrac{cos^2a+1}{sin^2a}=7\)
=>\(\dfrac{sin^4a+sin^2a+cos^4a+cos^2a}{sin^2a\cdot cos^2a}=7\)
=>\(sin^4a+cos^4a+1=7\cdot sin^2a\cdot cos^2a\)
=>\(\left(sin^2a+cos^2a\right)^2-2\cdot sin^2a\cdot cos^2a+1=7\cdot sin^2a\cdot cos^2a\)
=>\(2=9\cdot sin^2a\cdot cos^2a\)
=>\(8=9\cdot sin^22a\)
=>16=9(1-cos4a)
=>1-cos4a=16/9
=>cos4a=-7/9
1:
a: \(\left(x+y+z\right)^2=x^2+y^2+z^2+2xy+2zx+2yz\)
b: \(\left(x-y+z\right)^2=x^2+y^2+z^2-2xy+2xz-2yz\)
c: \(\left(x-y-z\right)^2=x^2+y^2+z^2-2xy-2xz+2yz\)
Vì 2a/3b=3b/4c=4c/5d=5d/2a nên suy ra 2a=3b=4c=5d ( Theo công thức dãy tỉ số bằng nhau)
=> 2a/3b=3b/4c=4c/5d=5d/2a=1
=>C=1+1+1+1=4
Vậy C=4
\(A=\frac{2^{a+1}.73+2^{a+1}.57}{2^a.104}\)
\(=\frac{2^{a+1}.\left(73+57\right)}{2^a.104}\)
\(=\frac{2^a.2.130}{2^a.104}\)
\(=\frac{1.1.130}{1.52}\)
\(=\frac{5}{2}\)
a là số mấy vậy bạn