Cho cot=3căn8.Tính
A=(sin-cos):(2sin+cos)
B=(1+sin^2):(2+sincos)
K
Khách
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ND
Nguyễn Đức Trí
VIP
12 tháng 9 2023
1) \(cot\alpha=\sqrt[]{5}\Rightarrow tan\alpha=\dfrac{1}{\sqrt[]{5}}\)
\(C=sin^2\alpha-sin\alpha.cos\alpha+cos^2\alpha\)
\(\Leftrightarrow C=\dfrac{1}{cos^2\alpha}\left(tan^2\alpha-tan\alpha+1\right)\)
\(\Leftrightarrow C=\left(1+tan^2\alpha\right)\left(tan^2\alpha-tan\alpha+1\right)\)
\(\Leftrightarrow C=\left(1+\dfrac{1}{5}\right)\left(\dfrac{1}{5}-\dfrac{1}{\sqrt[]{5}}+1\right)\)
\(\Leftrightarrow C=\dfrac{6}{5}\left(\dfrac{6}{5}-\dfrac{\sqrt[]{5}}{5}\right)=\dfrac{6}{25}\left(6-\sqrt[]{5}\right)\)
12 tháng 9 2023
1: \(cota=\sqrt{5}\)
=>\(cosa=\sqrt{5}\cdot sina\)
\(1+cot^2a=\dfrac{1}{sin^2a}\)
=>\(\dfrac{1}{sin^2a}=1+5=6\)
=>\(sin^2a=\dfrac{1}{6}\)
\(C=sin^2a-sina\cdot\sqrt{5}\cdot sina+\left(\sqrt{5}\cdot sina\right)^2\)
\(=sin^2a\left(1-\sqrt{5}+5\right)=\dfrac{1}{6}\cdot\left(6-\sqrt{5}\right)\)
2: tan a=3
=>sin a=3*cosa
\(1+tan^2a=\dfrac{1}{cos^2a}\)
=>\(\dfrac{1}{cos^2a}=1+9=10\)
=>\(cos^2a=\dfrac{1}{10}\)
\(B=\dfrac{3\cdot cosa-cosa}{27\cdot cos^3a+3\cdot cos^3a+2\cdot3\cdot cosa}\)
\(=\dfrac{2\cdot cosa}{30cos^3a+6cosa}=\dfrac{2}{30cos^2a+6}\)
\(=\dfrac{2}{3+6}=\dfrac{2}{9}\)
NV
Nguyễn Việt Lâm
Giáo viên
8 tháng 2 2022
\(A=\dfrac{\dfrac{3sina}{sina}-\dfrac{cosa}{sina}}{\dfrac{2sina}{sina}+\dfrac{cosa}{sina}}=\dfrac{3-cota}{2+cota}=\dfrac{3-3}{2+3}=0\)
\(B=\dfrac{\dfrac{sin^2a}{sin^2a}-\dfrac{3sina.cosa}{sin^2a}+\dfrac{2}{sin^2a}}{\dfrac{2sin^2a}{sin^2a}+\dfrac{sina.cosa}{sin^2a}+\dfrac{cos^2a}{sin^2a}}=\dfrac{1-3cota+2\left(1+cot^2a\right)}{2+cota+cot^2a}=\dfrac{1-3.3+2\left(1+3^2\right)}{2+3+3^2}=...\)
AM
8 tháng 2 2022
a. \(A=\dfrac{3sin\alpha-cos\alpha}{2sin\alpha+cos\alpha}=\dfrac{3\dfrac{sin\alpha}{cos\alpha}-1}{2\dfrac{sin\alpha}{cos\alpha}+1}=\dfrac{3.\dfrac{1}{3}-1}{2.\dfrac{1}{3}+1}=0\)
b.\(B=\dfrac{sin^2\alpha-3sin\alpha.cos\alpha+2}{2sin^2\alpha+sin\alpha.cos\alpha+cos^2\alpha}\)\(=\dfrac{1-\dfrac{3cos\alpha}{sin\alpha}+\dfrac{2}{sin^2\alpha}}{2+\dfrac{cos\alpha}{sin\alpha}+\dfrac{cos^2\alpha}{sin^2\alpha}}=\dfrac{1-3.3+\dfrac{2}{sin^2\alpha}}{2+3+3^2}\)
Mà \(\dfrac{cos\alpha}{sin\alpha}=3,cos^2\alpha+sin^2\alpha=1\Rightarrow sin^2\alpha=\dfrac{1}{10}\)
\(B=\dfrac{1-3.3+\dfrac{2}{\dfrac{1}{10}}}{2+3+3^2}=\dfrac{6}{7}\)
AH
Akai Haruma
Giáo viên
29 tháng 3 2019
Lời giải:
a)
\(\frac{1-\cos x}{\sin x}=\frac{(1-\cos x)(1+\cos x)}{\sin x(1+\cos x)}=\frac{1-\cos ^2x}{\sin x(1+\cos x)}=\frac{\sin ^2x}{\sin x(1+\cos x)}=\frac{\sin x}{1+\cos x}\)
b)
\((\sin x+\cos x-1)(\sin x+\cos x+1)=(\sin x+\cos x)^2-1^2\)
\(=\sin ^2x+\cos ^2x+2\sin x\cos x-1=1+2\sin x\cos x-1=2\sin x\cos x\)
c)
\(\frac{\sin ^2x+2\cos x-1}{2+\cos x-\cos ^2x}=\frac{1-\cos ^2x+2\cos x-1}{2+\cos x-\cos ^2x}=\frac{-\cos ^2x+2\cos x}{2+\cos x-\cos ^2x}\)
\(=\frac{\cos x(2-\cos x)}{(2-\cos x)(\cos x+1)}=\frac{\cos x}{\cos x+1}\)
d)
\(\frac{\cos ^2x-\sin ^2x}{\cot ^2x-\tan ^2x}=\frac{\cos ^2x-\sin ^2x}{\frac{\cos ^2x}{\sin ^2x}-\frac{\sin ^2x}{\cos ^2x}}=\frac{\sin ^2x\cos ^2x(\cos ^2x-\sin ^2x)}{\cos ^4x-\sin ^4x}\)
\(=\frac{\sin ^2x\cos ^2x(\cos ^2x-\sin ^2x)}{(\cos ^2x-\sin ^2x)(\cos ^2x+\sin ^2x)}=\frac{\sin ^2x\cos ^2x}{\sin ^2x+\cos ^2x}=\sin ^2x\cos ^2x\)
e)
\(1-\cot ^4x=1-\frac{\cos ^4x}{\sin ^4x}=\frac{\sin ^4x-\cos ^4x}{\sin ^4x}=\frac{(\sin ^2x-\cos ^2x)(\sin ^2x+\cos ^2x)}{\sin ^4x}\)
\(=\frac{\sin ^2x-\cos ^2x}{\sin ^4x}=\frac{\sin ^2x-(1-\sin ^2x)}{\sin ^4x}=\frac{2\sin ^2x-1}{\sin ^4x}=\frac{2}{\sin ^2x}-\frac{1}{\sin ^4x}\)
Ta có ddpcm.
9 tháng 8 2023
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
AH
Akai Haruma
Giáo viên
26 tháng 10 2018
a)
\((\sin x+\cos x)^2=\sin ^2x+2\sin x\cos x+\cos ^2x\)
\(=(\sin ^2x+\cos ^2x)+2\sin x\cos x=1+2\sin x\cos x\)
b)
\(\sin ^4x+\cos ^4x=\sin ^4x+2\sin ^2x\cos ^2x+\cos ^4x-2\sin ^2\cos ^2x\)
\(=(\sin ^2x+\cos ^2x)^2-2\sin ^2x\cos ^2x\)
\(=1-2\sin ^2x\cos ^2x\)
c)
\(\tan ^2x-\sin ^2x=(\frac{\sin x}{\cos x})^2-\sin ^2x\)
\(=\sin ^2x\left(\frac{1}{\cos ^2x}-1\right)=\sin ^2x. \frac{1-\cos ^2x}{\cos ^2x}=\sin ^2x.\frac{\sin ^2x}{\cos ^2x}\)
\(=\sin ^2x\left(\frac{\sin x}{\cos x}\right)^2=\sin ^2x\tan ^2x\)
AH
Akai Haruma
Giáo viên
26 tháng 10 2018
d)
\(\sin ^6x+\cos ^6x=(\sin ^2x)^3+(\cos ^2x)^3\)
\(=(\sin ^2x+\cos ^2x)(\sin ^4x-\sin ^2x\cos ^2x+\cos ^4x)\)
\(=\sin ^4x-\sin ^2x\cos ^2x+\cos ^4x\)
\(=(\sin ^4x+\cos ^4x)-\sin ^2x\cos ^2x=1-2\sin ^2x\cos ^2x-\sin ^2x\cos ^2x\)
\(=1-3\sin ^2x\cos ^2x\) (theo kq phần b)
e)
\(\sin x\cos x(1+\tan x)(1+\cot x)=\sin x\cos x(1+\frac{\sin x}{\cos x})(1+\frac{\cos x}{\sin x})\)
\(=\sin x\cos x.\frac{\cos x+\sin x}{\cos x}.\frac{\sin x+\cos x}{\sin x}\)
\(=(\sin x+\cos x)^2=\sin ^2x+\cos ^2x+2\sin x\cos x\)
\(=1+2\sin x\cos x\)
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P/s: Nói chung cứ bám vào công thức \(\sin ^2x+\cos ^2x=1\)
B
24 tháng 7 2020
a) \(\frac{1+2sina.cosa}{cos^2a-sin^2a}=\frac{1+sin2a}{cos2a}\)
b) \(B=\left(1+tan^2a\right)\left(1-sin^2a\right)-\left(1+cot^2a\right)\left(1-cos^2a\right)\)
\(=\left(1+\frac{sin^2a}{cos^2a}\right)\left(sin^2a+cos^2a-sin^2a\right)-\left(1+\frac{cos^2a}{sin^2a}\right)\left(cos^2a+sin^2a-cos^2a\right)\)
\(=\left(\frac{cos^2a+sin^2a}{cos^2a}\right).cos^2a-\left(\frac{sin^2a+cos^2a}{sin^2a}\right).sin^2a\)
\(=\frac{1}{cos^2a}.cos^2a-\frac{1}{sin^2a}.sin^2a=1-1=0\)
c)
\(C=\left(sin^2a+cos^2a\right)^3-3.sin^2a.cos^2a\left(sin^2a+cos^2a\right)+3sin^2a.cos^2a\)
\(=1-3sin^2a.cos^2a\left(1-1\right)=1\)
4 tháng 8 2023
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
\(sina=\sqrt{1-cos^2a}=\dfrac{2}{\sqrt{5}}\)
+) ta có : \(A=\dfrac{sinx-cosx}{2sinx+cosx}=\dfrac{\dfrac{sinx}{sinx}-\dfrac{cosx}{sinx}}{\dfrac{2sinx}{sinx}+\dfrac{cosx}{sinx}}\) \(=\dfrac{1-cotx}{2+cotx}\)
\(=\dfrac{1-3\sqrt{8}}{2+3\sqrt{8}}=\dfrac{-37+9\sqrt{2}}{34}\)
+) ta có : \(A=\dfrac{1+sin^2x}{2+sinx.cosx}=\dfrac{sin^2x+cos^2x+sin^2x}{2sin^2x+2cos^2x+sinx.cosx}\) \(=\dfrac{2sin^2x+cos^2x}{2sin^2x+2cos^2x+sinx.cosx}=\dfrac{\dfrac{2sin^2x}{sin^2x}+\dfrac{cos^2x}{sin^2x}}{\dfrac{2sin^2x}{sin^2x}+\dfrac{2cos^2x}{sin^2x}+\dfrac{sinx.cosx}{sin^2x}}\) \(=\dfrac{2+cot^2x}{2+2cot^2x+cotx}=\dfrac{2+\left(3\sqrt{8}\right)^2}{2+2\left(3\sqrt{8}\right)^2+3\sqrt{8}}=\dfrac{74}{146+3\sqrt{8}}\)