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) \(\dfrac{sin^2\alpha-cos^2\alpha}{1+2sin\alpha.cos\alpha}=\dfrac{tan\alpha-1}{tan\alpha+1}\)
Cho cot α = 3. Tính giá trị của các biểu thức sau
a) \(A=\dfrac{3sin\alpha-cos\alpha}{2sin\alpha+cos\alpha}\)
b)\(B=\dfrac{sin^2\alpha-3sin\alpha.cos\alpha+2}{2sin^2\alpha+sin\alpha.cos\alpha+cos^2\alpha}\)
Giúp em với ạ, em đang cần gấp!
\(A=\dfrac{\dfrac{3sina}{sina}-\dfrac{cosa}{sina}}{\dfrac{2sina}{sina}+\dfrac{cosa}{sina}}=\dfrac{3-cota}{2+cota}=\dfrac{3-3}{2+3}=0\)
\(B=\dfrac{\dfrac{sin^2a}{sin^2a}-\dfrac{3sina.cosa}{sin^2a}+\dfrac{2}{sin^2a}}{\dfrac{2sin^2a}{sin^2a}+\dfrac{sina.cosa}{sin^2a}+\dfrac{cos^2a}{sin^2a}}=\dfrac{1-3cota+2\left(1+cot^2a\right)}{2+cota+cot^2a}=\dfrac{1-3.3+2\left(1+3^2\right)}{2+3+3^2}=...\)
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 \(\dfrac{cos\alpha}{sin\alpha}=3,cos^2\alpha+sin^2\alpha=1\Rightarrow sin^2\alpha=\dfrac{1}{10}\)
\(B=\dfrac{1-3.3+\dfrac{2}{\dfrac{1}{10}}}{2+3+3^2}=\dfrac{6}{7}\)
CM đẳng thức
a) cos4α - sin4α = 2cos2α - 1
b) \(\dfrac{cos^2\alpha+tan^2\alpha-1}{sin^2\alpha}=tan^2\alpha\)
\(a,cos^4a-sin^4a=2cos^2a-1\\ VT=\left(cos^2a-sin^2a\right)\left(cos^2a+sin^2a\right)\\ =cos^2a-sin^2a\\ =cos2a=2cos^2a-1\)
\(b,VT=\dfrac{cos^2a+\dfrac{sin^2a}{cos^2a}-1}{sin^2a}\\ =\dfrac{\dfrac{cos^4a+sin^2a-cos^2a}{cos^2a}}{sin^2a}\\ =\dfrac{\dfrac{cos^4a+\left(1-cos^2a\right)-cos^2a}{cos^2a}}{sin^2a}\\ =\dfrac{\dfrac{cos^4a+1-2cos^2a}{cos^2a}}{sin^2a}\\ =\dfrac{\dfrac{\left(1-cos^2a\right)^2}{cos^2a}}{sin^2a}\\ =\dfrac{sin^4a}{cos^2a}:sin^2a\\ =\dfrac{sin^4a}{cos^2a}\times\dfrac{1}{sin^2a}\\ =\dfrac{sin^2a}{cos^2a}=tan^2a\)
Rút gọn:
A= \(\sin^6\alpha+cos^6\alpha+3sin^2\alpha.cos^2\alpha\)
B= \(\left(cos\alpha-sin\alpha\right)^2+\left(cos\alpha+sin\alpha\right)^2\)
C= \(\dfrac{\left(cos\alpha-sin\alpha\right)^2-\left(cos\alpha+sin\alpha\right)^2}{sin\alpha.cos\alpha}\)
Lời giải:
\(A=(\sin ^2a)^3+(\cos ^2a)^3+3\sin ^2a\cos ^2a(\sin ^2a+\cos ^2a)\)
\(=(\sin ^2a+\cos ^2a)^3=1^3=1\)
\(B=(\cos ^2a+\sin ^2a-2\sin a\cos a)+(\cos ^2a+\sin ^2a+2\sin a\cos a)\)
\(=(1-2\sin a\cos a)+(1+2\sin a\cos a)=2\)
\(C=\frac{(\cos ^2a+\sin ^2a-2\sin a\cos a)-(\cos ^2a+\sin ^2a+2\sin a\cos a)}{\sin a\cos a}=\frac{(1-2\sin a\cos a)-(1+2\sin a\cos a)}{\sin a\cos a}\)
$=\frac{-4\sin a\cos a}{\sin a\cos a}=-4$
Rút gọn biểu thức:
\(A=\dfrac{1+2sin\alpha.cos\alpha}{cos^2\alpha-sin\alpha}\)
Đề bài ko chính xác, biểu thức này không rút gọn được (có thể coi việc biến đổi khả dĩ duy nhất \(1+2sina.cosa=\left(sina+cosa\right)^2\) không phải là hành động rút gọn)
chỉnh lại đề 1 chút: \(A=\dfrac{1+2sin\alpha.cos\alpha}{cos^2\alpha-sin^2\alpha}=\dfrac{cos^2\alpha+sin^2\alpha+2sin\alpha.cos\alpha}{\left(cos\alpha-sin\alpha\right)\left(cos\alpha+sin\alpha\right)}\)
\(=\dfrac{\left(cos\alpha+sin\alpha\right)^2}{\left(cos\alpha-sin\alpha\right)\left(cos\alpha+sin\alpha\right)}=\dfrac{cos\alpha+sin\alpha}{cos\alpha-sin\alpha}\)
Rút gọn biểu thức:
a) A= \(\frac{1+2sin\alpha.cos\alpha}{cos^2\alpha-sin^2\alpha}\)
b) B= ( 1 + tan2α)( 1 + sin2α) - ( 1 + cot2α)( 1 - cos2α)
c) C= sin6α + cos6α + 3 sin2α. cos2α
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)
\(C=\left(sin^2a+cos^2a\right)^3-3.sin^2a.cos^2a\left(sin^2a+cos^2a\right)+3sin^2a.cos^2a\)
\(=1-3sin^2a.cos^2a\left(1-1\right)=1\)
Tính
A= \(\frac{2sin\alpha+cos\alpha}{3sin\alpha-4cos\alpha}\)
B= \(sin^2\alpha+2sin\alpha.cos\alpha-3cos^3\alpha\)
C= \(\frac{sin^2\alpha-sin\alpha.cos\alpha-cos^2\alpha}{2sin\alpha.cos\alpha}\)
Giúp mik với, ai làm được mik sẽ tick cho. Cảm ơn trước nhé
Những biểu thức này đều không tính toán ra được giá trị cụ thể nên không phù hợp với yêu cầu "tính". Mình nghĩ bạn nên xem xét lại yêu cầu đề.
Lời giải:
Biểu thức $A$ dạng như vậy là gọn rồi bạn ạ. Biến đổi thêm cũng không có ý nghĩa.
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\(B=\sin ^2a+\sin 2a-3\cos ^3a\)
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\(C=\frac{\sin ^2a-\sin a\cos a-\cos ^2a}{2\sin a\cos a}=\frac{\sin a}{2\cos a}-\frac{1}{2}-\frac{\cos a}{2\sin a}\)
\(=\frac{\tan a-1-\cot a}{2}\)
Biết tanB=2 tính
\(A=\frac{2sin\alpha+cos\alpha}{3sin\alpha-4cos\alpha}\)
\(B=sin^2\alpha+2sin\alpha.cos\alpha-3cos^2\alpha\)
\(C=\frac{sin^2\alpha-sin\alpha.cos\alpha-cos^2\alpha}{2sin\alpha.cos\alpha}\)
Giúp mik với, ai làm xong mik sẽ tick cho cảm ơn nhiều
hỏi tí chớ \(TanB=2\) hay \(Tan\alpha=2\) vậy .
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
\(1+tan^2a=\dfrac{1}{cos^2a}\)
=>\(\dfrac{1}{cos^2a}=1+9=10\)
=>\(cos^2a=\dfrac{1}{10}\)
\(B=\dfrac{3\cdot cosa-cosa}{27\cdot cos^3a+3\cdot cos^3a+2\cdot3\cdot cosa}\)
\(=\dfrac{2\cdot cosa}{30cos^3a+6cosa}=\dfrac{2}{30cos^2a+6}\)
\(=\dfrac{2}{3+6}=\dfrac{2}{9}\)
\(\dfrac{sin^3\alpha.cos\alpha+sin\alpha.cos^3\alpha}{sin^4\alpha+cos^4\alpha}\) Cho biết tan alpha = 2, tính giá trị của biểu thức trên
tan a=2 nên sina/cosa=2
=>sina=2cosa
\(A=\dfrac{sinacosa\left(sin^2a+cos^2a\right)}{\left(sin^2a+cos^2a\right)^2-2\cdot sin^2a\cdot cos^2a}=\dfrac{sina\cdot cosa}{1-2\cdot\left(sina\cdot cosa\right)^2}\)
\(=\dfrac{2cosa\cdot cosa}{1-2\cdot\left(2cosa\cdot cosa\right)^2}=\dfrac{2cos^2a}{1-8cos^2a}\)