Tính:
\(\dfrac{Cos\alpha+Sin\alpha}{Cos\alpha-Sin\alpha}\) với \(tan\alpha=\dfrac{1}{3}\)
Cho \(\tan\alpha=\dfrac{3}{5}\). Tính giá trị của các biểu thức sau:
M=\(\dfrac{\sin\alpha+\cos\alpha}{\sin\alpha-\cos\alpha}\)
N=\(\dfrac{\sin\alpha\times\cos\alpha}{\sin^2\alpha-\cos^2\alpha}\)
Lời giải:
\(M=\frac{\frac{\sin a}{\cos a}+1}{\frac{\sin a}{\cos a}-1}=\frac{\tan a+1}{\tan a-1}=\frac{\frac{3}{5}+1}{\frac{3}{5}-1}=-4\)
\(N = \frac{\frac{\sin a\cos a}{\cos ^2a}}{\frac{\sin ^2a-\cos ^2a}{\cos ^2a}}=\frac{\frac{\sin a}{\cos a}}{(\frac{\sin a}{\cos a})^2-1}=\frac{\tan a}{\tan ^2a-1}=\frac{\frac{3}{5}}{\frac{3^2}{5^2}-1}=\frac{-15}{16}\)
6. CM đẳng thức
a) \(\dfrac{sin^3\alpha+cos^3\alpha}{sin\alpha+cos\alpha}=1-sin\alpha.cos\alpha\)
c) sin4α + cos4α - sin6α - cos6α = sin2α . cos2α
b) \(\dfrac{sin^2\alpha-cos^2\alpha}{1+2sin\alpha.cos\alpha}=\dfrac{tan\alpha-1}{tan\alpha+1}\)
a: \(VT=\dfrac{\left(sina+cosa\right)^3-3\cdot sina\cdot cosa\left(sina+cosa\right)}{sina+cosa}\)
=(sina+cosa)^2-3*sina*cosa
=sin^2a+cos^2a-sina*cosa
=1-sina*cosa=VP
c: VT=(sin^2a+cos^2a)^2-2*sin^2a*cos^2a-(sin^2a+cos^2a)^3+3*sin^2a*cos^2a*(sin^2a+cos^2a)
=1-2sin^2a*cos^2a-1+3*sin^2a*cos^2a
=sin^2a*cos^2a=VP
1/ Cho \(cot\alpha=\sqrt{5}\) . Tính \(C=sin^2\alpha-sin\alpha cos\alpha+cos^2\alpha\)
2/ Cho \(tan\alpha=3\) . Tính \(B=\dfrac{sin\alpha-cos\alpha}{sin^3\alpha+3cos^3\alpha+2sin\alpha}\)
1) \(cot\alpha=\sqrt[]{5}\Rightarrow tan\alpha=\dfrac{1}{\sqrt[]{5}}\)
\(C=sin^2\alpha-sin\alpha.cos\alpha+cos^2\alpha\)
\(\Leftrightarrow C=\dfrac{1}{cos^2\alpha}\left(tan^2\alpha-tan\alpha+1\right)\)
\(\Leftrightarrow C=\left(1+tan^2\alpha\right)\left(tan^2\alpha-tan\alpha+1\right)\)
\(\Leftrightarrow C=\left(1+\dfrac{1}{5}\right)\left(\dfrac{1}{5}-\dfrac{1}{\sqrt[]{5}}+1\right)\)
\(\Leftrightarrow C=\dfrac{6}{5}\left(\dfrac{6}{5}-\dfrac{\sqrt[]{5}}{5}\right)=\dfrac{6}{25}\left(6-\sqrt[]{5}\right)\)
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
\(1+tan^2a=\dfrac{1}{cos^2a}\)
=>\(\dfrac{1}{cos^2a}=1+9=10\)
=>\(cos^2a=\dfrac{1}{10}\)
\(B=\dfrac{3\cdot cosa-cosa}{27\cdot cos^3a+3\cdot cos^3a+2\cdot3\cdot cosa}\)
\(=\dfrac{2\cdot cosa}{30cos^3a+6cosa}=\dfrac{2}{30cos^2a+6}\)
\(=\dfrac{2}{3+6}=\dfrac{2}{9}\)
Chứng minh các đẳng thức sau:
a, \(\sin^4\alpha-\cos^4\alpha+1=2\sin^2\alpha\)
b,\(\dfrac{\sin^2\alpha+2\cos^2\alpha-1}{\cot^2\alpha}=\sin^2\alpha\)
c, \(\dfrac{1-\sin^2\alpha.\cos^2\alpha}{\cos^2\alpha}-\cos^2\alpha=\tan^2\alpha\)
d, \(\dfrac{\sin^2\alpha-\tan^2\alpha}{\cos^2\alpha-\cot^2\alpha}=\tan^6\alpha\)
e, \(\left(1+\cot\alpha\right)\sin^3\alpha+\left(1+\tan\alpha\right)\cos^3\alpha=\sin\alpha.\cos\alpha\)
f,\(\dfrac{\left(\sin\alpha+\cos\alpha\right)^2-1}{\cot\alpha-\sin\alpha.\cos\alpha}=2\tan^2\alpha\)
a)
\(\sin ^4a-\cos ^4a+1=(\sin ^2a-\cos ^2a)(\sin ^2a+\cos^2a)+1\)
\(=(\sin ^2a-\cos ^2a).1+1=\sin ^2a-\cos ^2a+\sin ^2a+\cos ^2a\)
\(=2\sin ^2a\)
b) \(\sin ^2a+2\cos ^2a-1=(\sin ^2a+\cos^2a)+\cos ^2a-1\)
\(=1+\cos ^2a-1=\cos ^2a\)
\(\Rightarrow \frac{\sin ^2a+2\cos ^2a-1}{\cot ^2a}=\frac{\cos ^2a}{\cot ^2a}=\frac{\cos ^2a}{\frac{\cos ^2a}{\sin ^2a}}=\sin ^2a\)
c)
\(\frac{1-\sin ^2a\cos ^2a}{\cos ^2a}-\cos ^2a=\frac{1}{\cos ^2a}-\sin ^2a-\cos ^2a\)
\(=\frac{1}{\cos ^2a}-(\sin ^2a+\cos ^2a)=\frac{1}{\cos ^2a}-1\)
\(=\frac{1-\cos ^2a}{\cos ^2a}=\frac{\sin ^2a}{\cos ^2a}=\tan ^2a\)
d)
\(\frac{\sin ^2a-\tan ^2a}{\cos ^2a-\cot ^2a}=\frac{\sin ^2a-\frac{\sin ^2a}{\cos ^2a}}{\cos ^2a-\frac{\cos ^2a}{\sin ^2a}}\) \(=\frac{\sin ^2a(1-\frac{1}{\cos ^2a})}{\cos ^2a(1-\frac{1}{\sin ^2a})}\)
\(=\frac{\sin ^2a.\frac{\cos ^2a-1}{\cos ^2a}}{\cos ^2a.\frac{\sin ^2a-1}{\sin ^2a}}\) \(=\frac{\sin ^2a.\frac{-\sin ^2a}{\cos ^2a}}{\cos ^2a.\frac{-\cos ^2a}{\sin ^2a}}=\frac{\sin ^6a}{\cos ^6a}=\tan ^6a\)
f)
\(\frac{(\sin a+\cos a)^2-1}{\cot a-\sin a\cos a}=\frac{\sin ^2a+\cos ^2a+2\sin a\cos a-1}{\frac{\cos a}{\sin a}-\sin a\cos a}\)
\(=\sin a.\frac{1+2\sin a\cos a-1}{\cos a-\cos a\sin ^2a}\)
\(=\sin a. \frac{2\sin a\cos a}{\cos a(1-\sin ^2a)}=\sin a. \frac{2\sin a\cos a}{\cos a. \cos^2 a}=\frac{2\sin ^2a}{\cos ^2a}=2\tan ^2a\)
e)
\((1+\cot a)\sin ^3a+(1+\tan a)\cos ^3a\)
\(=(\sin ^3a+\cos ^3a)+\cot a.\sin ^3a+\tan a.\cos^3a\)
\(=(\sin a+\cos a)(\sin ^2a-\sin a\cos a+\cos ^2a)+\frac{\cos a}{\sin a}.\sin ^3a+\frac{\sin a}{\cos a}.\cos ^3a\)
\(=(\sin a+\cos a)(1-\sin a\cos a)+\cos a\sin ^2a+\sin a\cos ^2a\)
\(=\sin a+\cos a-\sin a\cos a(\sin a+\cos a)+\cos a\sin a(\sin a+\cos a)\)
\(=\sin a+\cos a\)
Cho $\tan \alpha = 3$. Tính
a) \(\dfrac{2\sin\alpha+3\cos\alpha}{3\sin\alpha-4\cos\alpha}.\)
b) \(\dfrac{\sin\alpha\cos\alpha}{\sin^2\alpha-\sin\alpha\cos\alpha+\cos^2\alpha}.\)
a) \(\dfrac{2sina+3cosa}{3sina-4cosa}=\dfrac{9}{5}\)
b) \(\dfrac{sina.cosa}{sin^2a-sina.cosa+cos^2a}=0\)
\(a.\dfrac{2\sin\alpha+3\cos\alpha}{3\sin\alpha-4\cos\alpha}=\dfrac{2\left(3cos\alpha\right)+3cos\alpha}{3\left(3cos\alpha\right)-4cos\alpha}=\dfrac{9cos\alpha}{5cos\alpha}=\dfrac{9}{5}\)
\(b.\dfrac{sin\alpha cos\alpha}{sin^2\alpha-sin\alpha cos\alpha+cos^2\alpha}=\dfrac{3cos^2\alpha}{9cos^2\alpha-3cos^2\alpha+cos^2\alpha}=\dfrac{3cos^2\alpha}{7cos^2\alpha}=\dfrac{3}{7}\)
a) Cho $\cos \alpha=\dfrac{3}{4}$ với $0^{\circ}<\alpha<90^{\circ}$. Tính $A=\dfrac{\tan \alpha+3 \cot \alpha}{\tan \alpha+\cot \alpha}$.
b) Cho $\tan \alpha=\sqrt{2}$. Tính $B=\dfrac{\sin \alpha-\cos \alpha}{\sin ^{3} \alpha+3 \cos ^{3} \alpha+2 \sin \alpha}$.
Cho tan\(\alpha\)= 2 . Tính B =\(\dfrac{\sin\alpha-\cos\alpha}{\sin^3\alpha+3\cos^3\alpha+2\sin\alpha}\)
Biết \(tan\alpha=2.\) Tính \(\dfrac{sin\alpha+cos\alpha}{sin\alpha-cos\alpha}\)
\(\dfrac{sina+cosa}{sina-cosa}=\dfrac{\dfrac{sina+cosa}{cosa}}{\dfrac{sina-cosa}{cosa}}=\dfrac{tana+1}{tana-1}=\dfrac{3}{1}=3\)
Có \(\dfrac{sin\alpha}{cos\alpha}=tan\alpha=2\)\(\Rightarrow sin\alpha=2cos\alpha\)
\(\dfrac{sin\alpha+cos\alpha}{sin\alpha-cos\alpha}=\dfrac{2cos\alpha+cos\alpha}{2cos\alpha-cos\alpha}=\dfrac{3cos\alpha}{cos\alpha}=3\)
Chứng minh các công thức sau :
\(Tan\alpha=\dfrac{sin\alpha}{cos\alpha}\)
\(Cot\alpha=\dfrac{cos\alpha}{sin\alpha}\)
\(sin^2\alpha+cos^2\alpha=1\)
\(1+tan^2\alpha=\dfrac{1}{cos^2\alpha}\)
\(1+cos^2\alpha=\dfrac{1}{sin^2\alpha}\)
\(cos^4\alpha-sin^4\alpha=2cos^2\alpha-1\)
Ta có:
\(sin=\dfrac{doi}{huyen}\); \(cos=\dfrac{ke}{chuyen}\);\(tan=\dfrac{doi}{ke}\); \(cot=\dfrac{ke}{doi}\)
Dùng cái này làm được hết mấy câu đó.
nếu bn thấy dùng cách của hùng có hới dài thì bn chỉ cần sử dụng cách đó cho 3 ý trên thôi . còn 3 ý dưới bn có thể sử dụng công thức \(sin^2x+cos^2x=1\) vừa chứng minh xong để giải quyết .
CMR:\(1,\tan\alpha\cdot\cot\alpha=1\)
\(2,\sin^2\alpha+\cos^2\alpha=1\)
\(3,\tan\alpha=\dfrac{\sin\alpha}{\cos\alpha};\cot\alpha=\dfrac{\cos\alpha}{\tan\alpha}\)