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Ta có :\(sin^2a+cos^2a=1\)
Thay số: \(\left(\frac{2}{3}\right)^2\)\(+cos^2a=1\)\(\Rightarrow cos^2a=\frac{5}{9}\)
A=\(2sin^2a+5cos^2a\)\(\Rightarrow2.\frac{4}{9}+5.\frac{5}{9}\)\(\Rightarrow A=\frac{11}{3}\)
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
\(sina=\sqrt{1-cos^2a}=\dfrac{2}{\sqrt{5}}\)
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) Ta có: \(\sin^2a^o=\cos^2\left(90^o-a^o\right)\)
Biểu thức trên
\(=\left(\sin^21^o+\sin^o89\right)+\left(\sin^22^o+\sin^288^o\right)+...+\left(\sin^244^o+\sin^246^o\right)+\sin^245^o\)
\(=\left(\sin^21^o+\cos^21^o\right)+\left(\sin^22^o+\cos^22^o\right)+...+\left(\sin^244^o+\cos^246^o\right)+\sin^245^o\)
\(=1+1+..+1+\sin^245^o=44+\frac{1}{2}=\frac{89}{2}\)
b)
Ta có: \(\sin^2x+\cos^2x=1\)
\(0^o< x< 90^o\)
=> \(0< \sin x;\cos x< 1\)
Ta có: \(\frac{\sin^2x+\cos^2x}{\text{}\text{}\sin x.\cos x}=\frac{1}{\frac{12}{25}}=\frac{25}{12}\Leftrightarrow\frac{\sin x}{\cos x}+\frac{\cos x}{\sin x}=\frac{25}{12}\)
\(\Leftrightarrow\tan x+\frac{1}{\tan x}=\frac{25}{12}\Leftrightarrow\tan^2x-\frac{25}{12}\tan x+1=0\)
Đặt t =tan x => có phương trình bậc 2 ẩn t => Giải đen ta => ra đc t => ra đc tan t
\(\Leftrightarrow\orbr{\begin{cases}\tan x=\frac{3}{4}\\\tan x=\frac{4}{3}\end{cases}}\)
Ta có \(\sin^2a+\cos^2a=1\)
\(\Rightarrow0.6^2+\cos^2a=1\)
\(\Rightarrow\cos^2a=0.64\)
Mà sin ,cos,tan đều bằng thương các cạnh tam giác nên sẽ lớn hơn 0
Vậy \(\cos a=0.8\)
Từ đó A=7.6
\(A=2\left(sin^2a+cos^2a\right)+3cos^2a=2+3\cdot cos^2a\)
mặt khác: \(sina=\dfrac{2}{3}\Leftrightarrow a=sin^{-1}\left(\dfrac{2}{3}\right)\)
thay vào A , ta được:
\(A=2+3\cdot sin^{-1}\left(\dfrac{2}{3}\right)=....\) (số xấu quá!)
A=2(sin2a + cos2a) +3 cos2a=2+ 3 cos2a
ta có sin2a+cos2a=1
(2/3)2 + cos2a =1
cosa=\(\dfrac{\sqrt{5}}{3}\)
A=....
\(A=2\cdot sin^2a+5\cdot\left(1-sin^2a\right)\)
\(=-3\cdot sin^2a+5\)
\(=-3\cdot\dfrac{4}{9}+5\)
\(=5-\dfrac{4}{3}=\dfrac{11}{3}\)
Ta có:
\(sin^2a+cos^2a=1\)
\(\Rightarrow cos^2a=1-sin^2a\)
Thay vào A ta có:
\(A=2\cdot sin^2a+5\cdot\left(1-sin^2a\right)\)
\(A=2\cdot sin^2a+5-5sin^2a\)
\(A=-3\cdot sin^2a+5\)
Mà: \(sina=\dfrac{2}{3}\Rightarrow sin^2a=\dfrac{4}{9}\)
\(A=-3\cdot\dfrac{4}{9}+5\)
\(A=-\dfrac{4}{3}+5\)
\(A=\dfrac{11}{3}\)