Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
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
1234567890-01234567890-=qưertyuiop[]\';;lkjhfgdsazxcvbnm,./\'l;[]7894561230.+-
a) Ta có A=\dfrac{\tan \alpha+3 \dfrac{1}{\tan \alpha}}{\tan \alpha+\dfrac{1}{\tan \alpha}}=\dfrac{\tan ^{2} \alpha+3}{\tan ^{2} \alpha+1}=\dfrac{\dfrac{1}{\cos ^{2} \alpha}+2}{\dfrac{1}{\cos ^{2} \alpha}}=1+2 \cos ^{2} \alphaA=tanα+tanα1tanα+3tanα1=tan2α+1tan2α+3=cos2α1cos2α1+2=1+2cos2α Suy ra A=1+2 \cdot \dfrac{9}{16}=\dfrac{17}{8}A=1+2⋅169=817.
b) B=\dfrac{\dfrac{\sin \alpha}{\cos ^{3} \alpha}-\dfrac{\cos \alpha}{\cos ^{3} \alpha}}{\dfrac{\sin ^{3} \alpha}{\cos ^{3} \alpha}+\dfrac{3 \cos ^{3} \alpha}{\cos ^{3} \alpha}+\dfrac{2 \sin \alpha}{\cos ^{3} \alpha}}=\dfrac{\tan \alpha\left(\tan ^{2} \alpha+1\right)-\left(\tan ^{2} \alpha+1\right)}{\tan ^{3} \alpha+3+2 \tan \alpha\left(\tan ^{2} \alpha+1\right)}B=cos3αsin3α+cos3α3cos3α+cos3α2sinαcos3αsinα−cos3αcosα=tan3α+3+2tanα(tan2α+1)tanα(tan2α+1)−(tan2α+1).
Suy ra B=\dfrac{\sqrt{2}(2+1)-(2+1)}{2 \sqrt{2}+3+2 \sqrt{2}(2+1)}=\dfrac{3(\sqrt{2}-1)}{3+8 \sqrt{2}}B=22+3+22(2+1)2(2+1)−(2+1)=3+823(2−1).
a) Vì 90^{\circ}<\alpha<180^{\circ}90∘<α<180∘ nên \cos \alpha<0cosα<0 mặt khác \sin ^{2} \alpha+\cos ^{2} \alpha=1sin2α+cos2α=1 suy ra \cos \alpha=-\sqrt{1-\sin ^{2} \alpha}=-\sqrt{1-\dfrac{1}{9}}=-\dfrac{2 \sqrt{2}}{3}cosα=−1−sin2α=−1−91=−322.
Do đó \tan \alpha=\dfrac{\sin \alpha}{\cos \alpha}=\dfrac{\dfrac{1}{3}}{-\dfrac{2 \sqrt{2}}{3}}=-\dfrac{1}{2 \sqrt{2}}tanα=cosαsinα=−32231=−221.
b) Vì \sin ^{2} \alpha+\cos ^{2} \alpha=1sin2α+cos2α=1 nên \sin \alpha=\sqrt{1-\cos ^{2} \alpha}=\sqrt{1-\dfrac{4}{9}}=\dfrac{\sqrt{5}}{3}sinα=1−cos2α=1−94=35 và \cot \alpha=\dfrac{\cos \alpha}{\sin \alpha}=\dfrac{-\dfrac{2}{3}}{\dfrac{\sqrt{5}}{3}}=-\dfrac{2}{\sqrt{5}}cotα=sinαcosα=35−32=−52.
c) Vì \tan \gamma=-2 \sqrt{2}<0 \Rightarrow \cos \alpha<0tanγ=−22<0⇒cosα<0 mặt khác \tan ^{2} \alpha+1=\dfrac{1}{\cos ^{2} \alpha}tan2α+1=cos2α1 nên \cos \alpha=-\sqrt{\dfrac{1}{\tan ^{2}+1}}=-\sqrt{\dfrac{1}{8+1}}=-\dfrac{1}{3}cosα=−tan2+11=−8+11=−31.
Ta có \tan \alpha=\dfrac{\sin \alpha}{\cos \alpha} \Rightarrow \sin \alpha=\tan \alpha \cdot \cos \alpha=-2 \sqrt{2} \cdot\left(-\dfrac{1}{3}\right)=\dfrac{2 \sqrt{2}}{3}tanα=cosαsinα⇒sinα=tanα⋅cosα=−22⋅(−31)=322 \Rightarrow \cot \alpha=\dfrac{\cos \alpha}{\sin \alpha}=\dfrac{-\dfrac{1}{3}}{\dfrac{2 \sqrt{2}}{3}}=-\dfrac{1}{2 \sqrt{2}}⇒cotα=sinαcosα=322−31=−221.
a: 
b: \(B=3-sin^290^0+2\cdot cos^260^0-3\cdot tan^245^0\)
\(=3-1+2\cdot\left(\dfrac{1}{2}\right)^2-3\cdot1^2\)
\(=2-3+2\cdot\dfrac{1}{4}=-1+\dfrac{1}{2}=-\dfrac{1}{2}\)
c: \(C=sin^245^0-2\cdot sin^250^0+3\cdot cos^245^0-2\cdot sin^240^0+4\cdot tan55\cdot tan35\)
\(=\left(\dfrac{\sqrt{2}}{2}\right)^2+3\cdot\left(\dfrac{\sqrt{2}}{2}\right)^2-2\cdot\left(sin^250^0+sin^240^0\right)+4\)
\(=\dfrac{1}{2}+3\cdot\dfrac{1}{2}-2+4\)
\(=2-2+4=4\)
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: \(Q = \frac{1}{{{{\left( {\frac{{\sqrt 3 }}{2}} \right)}^2}}} - \;{\left( {\frac{{ - 1}}{{\sqrt 3 }}} \right)^2} = \frac{1}{{\frac{3}{4}}} - \;\frac{1}{3} = \;\frac{4}{3} - \;\frac{1}{3} = 1.\)
Do a, b, c là độ dài 3 cạnh của tam giác ABC nên \(a+b-c\ne0\). Như vậy, \(\dfrac{a^3+b^3-c^3}{a+b-c}=c^2\)
\(\Leftrightarrow a^3+b^3-c^3=c^2a+c^2b-c^3\)
\(\Leftrightarrow a^3+b^3-c^2a-c^2b=0\)
\(\Leftrightarrow\left(a+b\right)\left(a^2-ab+b^2\right)-c^2\left(a+b\right)=0\)
\(\Leftrightarrow\left(a+b\right)\left(a^2-ab+b^2-c^2\right)=0\)
\(\Leftrightarrow a^2-ab+b^2-c^2=0\) (do \(a+b\ne0\))
\(\Leftrightarrow c^2=a^2+b^2-ab\) (1)
Mặt khác, theo định lý cosin, ta có \(c^2=a^2+b^2-2ab.\cos C\) (2)
Từ (1) và (2), ta thu được \(2\cos C=1\Leftrightarrow\cos C=\dfrac{1}{2}\Leftrightarrow\widehat{C}=60^o\)
Vậy \(\widehat{C}=60^o\)
A=a2sin90∘+b2cos90∘+c2cos180∘A=a2sin90∘+b2cos90∘+c2cos180∘
=a2*1+b2* 0 +c2* (-1
=a2 - c2
B=3−sin290∘+2cos260∘−3tan245∘B=3−sin290∘+2cos260∘−3tan245∘.
= 3 - 1 + 1/2 - 3 = -1/2
What did you see at the zoo?
I saw crocodiles.