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a) \(\dfrac{\sin2\text{a}+\cos a}{1+\cos2\text{a}+\cos a}=2\tan a\)
a) \(\dfrac{sin2\alpha+sin\alpha}{1+cos2\alpha+cos\alpha}=\dfrac{2sin\alpha cos\alpha+sin\alpha}{2cos^2\alpha+cos\alpha}\)\(=\dfrac{sin\alpha\left(2cos\alpha+1\right)}{cos\alpha\left(2cos\alpha+1\right)}=\dfrac{sin\alpha}{cos\alpha}=tan\alpha\).
\(\frac{1}{cos^2a}=1+tan^2a\Rightarrow cos^2a=\frac{1}{1+tan^2a}=\frac{1}{10}\)
a/ \(\frac{sina-cosa}{sina+cosa}=\frac{\frac{sina}{cosa}-\frac{cosa}{cosa}}{\frac{sina}{cosa}+\frac{cosa}{cosa}}=\frac{tana-1}{tana+1}=\frac{3-1}{3+1}\)
b/ \(\frac{2sina+3cosa}{3sina-5cosa}=\frac{3tana+3}{3tana-5}=\frac{3.3+3}{3.3-5}\)
c/ \(\frac{1+2cos^2a}{1-cos^2a-cos^2a}=\frac{1+2cos^2a}{1-2cos^2a}=\frac{1+2.\frac{1}{10}}{1-2.\frac{1}{10}}\)
d/ \(\frac{\left(1-cos^2a\right)^2+\left(cos^2a\right)^2}{1+1-cos^2a}=\frac{\left(1-\frac{1}{10}\right)^2+\left(\frac{1}{10}\right)^2}{2-\frac{1}{10}}\)
a) Ta có: \(sin^2x+sin^2\left(90-x\right)=sin^2x+cos^2x=1.\)
áp dụng: A = 2
b)Ta có: \(cos\left(x\right)=-cos\left(180-x\right)\)
áp dụng: B = 0
c) Ta có: \(tan\left(x\right)\cdot tan\left(90-x\right)=\frac{sinx}{cosx}\cdot\frac{sin\left(90-x\right)}{cos\left(90-x\right)}=\frac{sinx}{cosx}\cdot\frac{cosx}{sinx}=1\)
áp dụng: C = 1
a) \(M = \sin {45^o}.\cos {45^o} + \sin {30^o}\)
Ta có: \(\left\{ \begin{array}{l}\sin {45^o} = \cos {45^o} = \frac{{\sqrt 2 }}{2};\;\\\sin {30^o} = \frac{1}{2}\end{array} \right.\)
Thay vào M, ta được: \(M = \frac{{\sqrt 2 }}{2}.\frac{{\sqrt 2 }}{2} + \frac{1}{2} = \frac{2}{4} + \frac{1}{2} = 1\)
b) \(N = \sin {60^o}.\cos {30^o} + \frac{1}{2}.\sin {45^o}.\cos {45^o}\)
Ta có: \(\sin {60^o} = \frac{{\sqrt 3 }}{2};\;\;\cos {30^o} = \frac{{\sqrt 3 }}{2};\;\sin {45^o} = \frac{{\sqrt 2 }}{2};\, \cos {45^o}= \frac{{\sqrt 2 }}{2}\)
Thay vào N, ta được: \(N = \frac{{\sqrt 3 }}{2}.\frac{{\sqrt 3 }}{2} + \frac{1}{2}.\frac{{\sqrt 2 }}{2}.\frac{{\sqrt 2 }}{2} = \frac{3}{4} + \frac{1}{4} = 1\)
c) \(P = 1 + {\tan ^2}{60^o}\)
Ta có: \(\tan {60^o} = \sqrt 3 \)
Thay vào P, ta được: \(Q = 1 + {\left( {\sqrt 3 } \right)^2} = 4.\)
d) \(Q = \frac{1}{{{{\sin }^2}{{120}^o}}} - {\cot ^2}{120^o}.\)
Ta có: \(\sin {120^o} = \frac{{\sqrt 3 }}{2};\;\;\cot {120^o} = \frac{{ - 1}}{{\sqrt 3 }}\)
Thay vào P, ta được: \(Q = \frac{1}{{{{\left( {\frac{{\sqrt 3 }}{2}} \right)}^2}}} - \;{\left( {\frac{{ - 1}}{{\sqrt 3 }}} \right)^2} = \frac{1}{{\frac{3}{4}}} - \;\frac{1}{3} = \;\frac{4}{3} - \;\frac{1}{3} = 1.\)
a) \(A = \cos {0^o} + \cos {40^o} + \cos {120^o} + \cos {140^o}\)
Tra bảng giá trị lượng giác của một số góc đặc biệt, ta có:
\(\cos {0^o} = 1;\;\cos {120^o} = - \frac{1}{2}\)
Lại có: \(\cos {140^o} = - \cos \left( {{{180}^o} - {{40}^o}} \right) = - \cos {40^o}\)
\(\begin{array}{l} \Rightarrow A = 1 + \cos {40^o} + \left( { - \frac{1}{2}} \right) - \cos {40^o}\\ \Leftrightarrow A = \frac{1}{2}.\end{array}\)
b) \(B = \sin {5^o} + \sin {150^o} - \sin {175^o} + \sin {180^o}\)
Tra bảng giá trị lượng giác của một số góc đặc biệt, ta có:
\(\sin {150^o} = \frac{1}{2};\;\sin {180^o} = 0\)
Lại có: \(\sin {175^o} = \sin \left( {{{180}^o} - {{175}^o}} \right) = \sin {5^o}\)
\(\begin{array}{l} \Rightarrow B = \sin {5^o} + \frac{1}{2} - \sin {5^o} + 0\\ \Leftrightarrow B = \frac{1}{2}.\end{array}\)
c) \(C = \cos {15^o} + \cos {35^o} - \sin {75^o} - \sin {55^o}\)
Ta có: \(\sin {75^o} = \cos\left( {{{90}^o} - {{75}^o}} \right) = \cos {15^o}\); \(\sin {55^o} = \cos\left( {{{90}^o} - {{55}^o}} \right) = \cos {35^o}\)
\(\begin{array}{l} \Rightarrow C = \cos {15^o} + \cos {35^o} - \cos {15^o} - \cos {35^o}\\ \Leftrightarrow C = 0.\end{array}\)
d) \(D = \tan {25^o}.\tan {45^o}.\tan {115^o}\)
Ta có: \(\tan {115^o} = - \tan \left( {{{180}^o} - {{115}^o}} \right) = - \tan {65^o}\)
Mà: \(\tan {65^o} = \cot \left( {{{90}^o} - {{65}^o}} \right) = \cot {25^o}\)
\(\begin{array}{l} \Rightarrow D = \tan {25^o}.\tan {45^o}.(-\cot {25^o})\\ \Leftrightarrow D =- \tan {45^o} = -1\end{array}\)
e) \(E = \cot {10^o}.\cot {30^o}.\cot {100^o}\)
Ta có: \(\cot {100^o} = - \cot \left( {{{180}^o} - {{100}^o}} \right) = - \cot {80^o}\)
Mà: \(\cot {80^o} = \tan \left( {{{90}^o} - {{80}^o}} \right) = \tan {10^o}\Rightarrow \cot {100^o} =- \tan {10^o}\)
\(\begin{array}{l} \Rightarrow E = \cot {10^o}.\cot {30^o}.(-\tan {10^o})\\ \Leftrightarrow E = -\cot {30^o} =- \sqrt 3 .\end{array}\)
a/
\(\frac{1}{sinx}+\frac{cosx}{sinx}=\frac{1+cosx}{sinx}=\frac{1+2cos^2\frac{x}{2}-1}{2sin\frac{x}{2}cos\frac{x}{2}}=\frac{2cos^2\frac{x}{2}}{2sin\frac{x}{2}cos\frac{x}{2}}=\frac{cos\frac{x}{2}}{sin\frac{x}{2}}=cot\frac{x}{2}\)
b/
\(\frac{1-cosx}{sinx}=\frac{1-\left(1-2sin^2\frac{x}{2}\right)}{2sin\frac{x}{2}cos\frac{x}{2}}=\frac{2sin^2\frac{x}{2}}{2sin\frac{x}{2}cos\frac{x}{2}}=\frac{sin\frac{x}{2}}{cos\frac{x}{2}}=tan\frac{x}{2}\)
c/
\(tan\frac{x}{2}\left(\frac{1}{cosx}+1\right)=\left(\frac{1-cosx}{sinx}\right)\left(\frac{1}{cosx}+1\right)=\frac{\left(1-cosx\right)\left(1+cosx\right)}{sinx.cosx}=\frac{1-cos^2x}{sinx.cosx}\)
\(=\frac{sin^2x}{sinx.cosx}=\frac{sinx}{cosx}=tanx\)
d/
\(\frac{sin2a}{2cosa\left(1+cosa\right)}=\frac{2sina.cosa}{2cosa\left(1+2cos^2\frac{a}{2}-1\right)}=\frac{sina}{2cos^2\frac{a}{2}}=\frac{2sin\frac{a}{2}cos\frac{a}{2}}{2cos^2\frac{a}{2}}=tan\frac{a}{2}\)
e/
\(cotx+tan\frac{x}{2}=\frac{cosx}{sin}+\frac{1-cosx}{sinx}=\frac{cosx+1-cosx}{sinx}=\frac{1}{sinx}\)
Các câu c, e đều sử dụng kết quả từ câu b
f/
\(3-4cos2x+cos4x=3-4cos2x+2cos^22x-1\)
\(=2cos^22x-4cos2x+2=2\left(cos^22x-2cos2x+1\right)\)
\(=2\left(cos2x-1\right)^2=2\left(1-2sin^2x-1\right)^2\)
\(=2.\left(-2sin^2x\right)^2=8sin^4x\)
g/
\(\frac{1-cosx}{sinx}=\frac{sinx\left(1-cosx\right)}{sin^2x}=\frac{sinx\left(1-cosx\right)}{1-cos^2x}=\frac{sinx\left(1-cosx\right)}{\left(1-cosx\right)\left(1+cosx\right)}=\frac{sinx}{1+cosx}\)
h/
\(sinx+cosx=\sqrt{2}\left(sinx.\frac{\sqrt{2}}{2}+cosx.\frac{\sqrt{2}}{2}\right)\)
\(=\sqrt{2}\left(sinx.cos\frac{\pi}{4}+cosx.sin\frac{\pi}{4}\right)=\sqrt{2}sin\left(x+\frac{\pi}{4}\right)\)
i/
\(sinx-cosx=\sqrt{2}\left(sinx.\frac{\sqrt{2}}{2}-cosx.\frac{\sqrt{2}}{2}\right)\)
\(=\sqrt{2}\left(sinx.cos\frac{\pi}{4}-cosx.sin\frac{\pi}{4}\right)=\sqrt{2}sin\left(x-\frac{\pi}{4}\right)\)
j/
\(cosx-sinx=\sqrt{2}\left(cosx.\frac{\sqrt{2}}{2}-sinx\frac{\sqrt{2}}{2}\right)\)
\(=\sqrt{2}\left(cosx.cos\frac{\pi}{4}-sinx.sin\frac{\pi}{4}\right)=\sqrt{2}cos\left(x+\frac{\pi}{4}\right)\)
\(A=\frac{2sinx.cosx+sinx}{1+2cos^2x-1+cosx}=\frac{sinx\left(2cosx+1\right)}{cosx\left(2cosx+1\right)}=\frac{sinx}{cosx}=tanx\)
\(B=\frac{cosa}{sina}\left(\frac{1+sin^2a}{cosa}-cosa\right)=\frac{cosa}{sina}\left(\frac{1+sin^2a-cos^2a}{cosa}\right)=\frac{cosa}{sina}.\frac{2sin^2a}{cosa}=2sina\)
\(C=\frac{1+cos2x+cosx+cos3x}{2cos^2x-1+cosx}=\frac{1+2cos^2x-1+2cos2x.cosx}{cos2x+cosx}=\frac{2cosx\left(cosx+cos2x\right)}{cos2x+cosx}=2cosx\)
\(D=\frac{2sinx.cosx.\left(-tanx\right)}{-tanx.sinx}-2cosx=2cosx-2cosx=0\)
\(E=cos^2x.cot^2x-cot^2x+cos^2x+2cos^2x+2sin^2x\)
\(E=cot^2x\left(cos^2x-1\right)+cos^2x+2=\frac{cos^2x}{sin^2x}\left(-sin^2x\right)+cos^2x+2=2\)
\(F=\frac{sin^2x\left(1+tan^2x\right)}{cos^2x\left(1+tan^2x\right)}=\frac{sin^2x}{cos^2x}=tan^2x\)
Câu G mẫu số có gì đó sai sai, sao lại là \(2sina-sina?\)
\(H=sin^4\left(\frac{\pi}{2}+a\right)-cos^4\left(\frac{3\pi}{2}-a\right)+1=cos^4a-sin^4a+1\)
\(=\left(cos^2a-sin^2a\right)\left(cos^2a+sin^2a\right)+1=cos^2a-\left(1-cos^2a\right)+1=2cos^2a\)
Chọn A.
Ta có: