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DD
24 tháng 6 2021

a) \(sin^2x+cos^2x=1\Leftrightarrow cos^2x=1-sin^2x=1-\frac{3}{4}=\frac{1}{4}\)

\(\Leftrightarrow\orbr{\begin{cases}cosx=\frac{1}{2}\\cosx=-\frac{1}{2}\end{cases}}\)

\(cosx=\frac{1}{2}\)

\(tanx=\frac{sinx}{cosx}=\frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}}=\sqrt{3}\)

\(tanx.cotx=1\Rightarrow cotx=\frac{1}{tanx}=\frac{1}{\sqrt{3}}=\frac{\sqrt{3}}{3}\)

\(cosx=\frac{-1}{2}\)

\(tanx=\frac{sinx}{cosx}=\frac{\frac{\sqrt{3}}{2}}{\frac{-1}{2}}=-\sqrt{3}\)

\(tanx.cotx=1\Rightarrow cotx=\frac{1}{tanx}=\frac{1}{-\sqrt{3}}=\frac{-\sqrt{3}}{3}\)

b) Bạn làm tương tự câu a) nha. 

9 tháng 10 2016

a) sin anpha = 2/3 => góc anpha = 42o 

cos 42o = 0,743

tan 42o =  0,9

cot  42o = 1/tan 42o = 1/0,9 = 1,111

b) tan anpha + cot anpha = 3

<=> tan anpha + 1/tan anpha = 3

<=> tananpha = 2

<=> tan anpha = \(\sqrt{2}\)

=> góc anpha =  55

Ta có: a = sin 55o . cos 55o

<=> a = 0,469

DD
22 tháng 6 2021

a) \(cos^4x-sin^4x=\left(cos^2x+sin^2x\right)\left(cos^2x-sin^2x\right)=cos^2x-sin^2x\)

b) \(\frac{1}{1+tanx}+\frac{1}{1+cotx}=\frac{1}{1+tanx}+\frac{tanxcotx}{tanxcotx+cotx}=\frac{1}{1+tanx}+\frac{tanx}{tanx+1}\)

\(=\frac{1+tanx}{1+tanx}=1\)

c) Ta có: \(1+tan^2x=1+\frac{sin^2x}{cos^2x}=\frac{cos^2x+sin^2x}{cos^2x}=\frac{1}{cos^2x}\)

\(\Rightarrow\frac{1}{1+tan^2x}=cos^2x\)

Tương tự \(\frac{1}{1+tan^2y}=cos^2y\)

\(\Rightarrow cos^2x-cos^2y=\frac{1}{1+tan^2x}-\frac{1}{1+tan^2y}\)

\(cos^2x-cos^2y=\left(1-sin^2x\right)-\left(1-sin^2y\right)=sin^2y-sin^2x\)

d) \(\frac{1+sin^2x}{1-sin^2x}=\frac{cos^2x+sin^2x+sin^2x}{cos^2x+sin^2x-sin^2x}=\frac{cos^2x+2sin^2x}{cos^2x}=1+2\left(\frac{sinx}{cosx}\right)^2=1+2tan^2x\)

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

1: \(=\dfrac{cotx+1+tanx+1}{\left(tanx+1\right)\left(cotx+1\right)}\)

\(=\dfrac{\dfrac{1}{cotx}+cotx+2}{2+tanx+cotx}\)

\(=1\)

2: \(VT=\dfrac{cos^2x+cosxsinx+sin^2x-sinx\cdot cosx}{sin^2x-cos^2x}\)

\(=\dfrac{1}{sin^2x-cos^2x}\)

\(VP=\dfrac{1+cot^2x}{1-cot^2x}=\left(1+\dfrac{cos^2x}{sin^2x}\right):\left(1-\dfrac{cos^2x}{sin^2x}\right)\)

\(=\dfrac{1}{sin^2x}:\dfrac{sin^2x-cos^2x}{sin^2x}=\dfrac{1}{sin^2x-cos^2x}\)

=>VT=VP

cot x=2

=>tan x=1/cotx=1/2

\(1+tan^2x=\dfrac{1}{cos^2x}\)

=>\(\dfrac{1}{cos^2x}=1+\dfrac{1}{4}=\dfrac{5}{4}\)

=>cos^2x=4/5

=>cosx=2/căn 5

\(sinx=\sqrt{1-\dfrac{4}{5}}=\sqrt{\dfrac{1}{5}}=\dfrac{1}{\sqrt{5}}\)

a) \(4sinx-1=1\Leftrightarrow4sinx=2\Leftrightarrow sinx=\dfrac{2}{4}=\dfrac{1}{2}\)

\(\Leftrightarrow x=30^o\)

b) \(2\sqrt{3}-3tanx=\sqrt{3}\Leftrightarrow3tanx=2\sqrt{3}-\sqrt{3}=\sqrt{3}\Leftrightarrow tanx=\dfrac{\sqrt{3}}{3}\)

\(\Leftrightarrow x=30^o\)

c) \(7sinx-3cos\left(90^o-x\right)=2,5\Leftrightarrow7sinx-3sinx=2,5\Leftrightarrow4sinx=2,5\Leftrightarrow sinx=\dfrac{5}{8}\Leftrightarrow x=30^o41'\)

d)\(\left(2sin-\sqrt{2}\right)\left(4cos-5\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}2sin-\sqrt{2}=0\\4cos-5=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}2sin=\sqrt{2}\\4cos=5\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}sin=\dfrac{\sqrt{2}}{2}\\cos=\dfrac{5}{4}\left(loai\right)\end{matrix}\right.\)\(\Rightarrow x=45^o\)

 

Xin lỗi nãy đang làm thì bấm gửi, quên còn câu e, f nữa:"(

e) \(\dfrac{1}{cos^2x}-tanx=1\Leftrightarrow1+tan^2x-tanx-1=0\Leftrightarrow tan^2x-tanx=0\Leftrightarrow tanx\left(tanx-1\right)=0\Rightarrow tanx-1=0\Leftrightarrow tanx=1\Leftrightarrow x=45^o\)

f) \(cos^2x-3sin^2x=0,19\Leftrightarrow1-sin^2x-3sin^2x=0,19\Leftrightarrow1-4sin^2x=0,19\Leftrightarrow4sin^2x=0,81\Leftrightarrow sin^2x=\dfrac{81}{400}\Leftrightarrow sinx=\dfrac{9}{20}\Leftrightarrow x=26^o44'\)