Rút gọn \(\frac{\cos36-\sin234}{\sin144-\cos126}\cdot\cos54\)
Rút gọn bt A= \(\dfrac{\sin\left(-234\right)-\cos216}{\sin144-\cos126}\) . \(\tan36\), ta có A bằng ?
1. Biểu thức A = \(\frac{1}{2\sin10}-2\sin70\) có gái trị bằng bao nhiêu ?
2. Tích số cos10.cos30.cos50.cos70 = ?
3. Tích số \(cos\frac{\pi}{7}.cos\frac{4\pi}{7}.cos\frac{5\pi}{7}\) = ?
4. Tính A = \(\frac{tan30+tan40+tan50+tan60}{cos20}\)=?
5.Rút gọn biểu thức : cos54.cos4 - cos36.cos86
=> P/S : (Làm theo công thức lượng giác lớp 10 ở tất cả các câu)
Câu 3:
\(A=cos\frac{\pi}{7}.cos\frac{5\pi}{7}.cos\frac{4\pi}{7}=cos\frac{\pi}{7}.cos\left(\pi-\frac{2\pi}{7}\right).cos\frac{4\pi}{7}\)
\(A=-cos\frac{\pi}{7}.cos\frac{2\pi}{7}.cos\frac{4\pi}{7}\)
\(\Rightarrow sin\frac{\pi}{7}.A=-\frac{1}{2}.2sin\frac{\pi}{7}.cos\frac{\pi}{7}.cos\frac{2\pi}{7}.cos\frac{4\pi}{7}\)
\(\Rightarrow sin\frac{\pi}{7}.A=-\frac{1}{2}.sin\frac{2\pi}{7}.cos\frac{2\pi}{7}.cos\frac{4\pi}{7}\)
\(\Rightarrow sin\frac{\pi}{7}.A=-\frac{1}{4}sin\frac{4\pi}{7}.cos\frac{4\pi}{7}\)
\(\Rightarrow sin\frac{\pi}{7}.A=-\frac{1}{8}sin\frac{8\pi}{7}=-\frac{1}{8}sin\left(\pi+\frac{\pi}{7}\right)=\frac{1}{8}sin\frac{\pi}{7}\)
\(\Rightarrow A=\frac{1}{8}\)
Câu 4:
Đầu tiên ta chứng minh công thức:
\(tana+tanb=\frac{sina}{cosa}+\frac{sinb}{cosb}=\frac{sina.cosb+cosa.sinb}{cosa.cosb}=\frac{sin\left(a+b\right)}{cosa.cosb}\)
Áp dụng để biến đổi tử số:
\(tan30+tan60+tan40+tan50=\frac{sin90}{cos30.cos60}+\frac{sin90}{cos40.cos50}=\frac{1}{cos30.cos60}+\frac{1}{cos40.cos50}\)
\(=\frac{2}{cos90+cos30}+\frac{2}{cos90+cos10}=\frac{2}{cos30}+\frac{2}{cos10}=2\left(\frac{cos30+cos10}{cos30.cos10}\right)\)
\(=2\left(\frac{2cos20.cos10}{cos30.cos10}\right)=\frac{4.cos20}{cos30}=\frac{8\sqrt{3}}{3}.cos20\)
\(\Rightarrow A=\frac{\frac{8\sqrt{3}}{3}cos20}{cos20}=\frac{8\sqrt{3}}{3}\)
Câu 5:
\(cos54.cos4-cos36.cos86=cos54.cos4-cos\left(90-54\right).cos\left(90-4\right)\)
\(=cos54.cos4-sin54.sin4=cos\left(54+4\right)=cos58\)
Câu 1:
\(A=\frac{1}{2sin10}-2sin70=\frac{1-4sin10.sin70}{2sin10}=\frac{1+2\left(cos80-cos60\right)}{2sin10}\)
\(=\frac{1+2cos80-1}{2sin10}=\frac{2cos80}{2sin10}=\frac{sin10}{sin10}=1\)
Câu 2:
\(cos10.cos30.cos50.cos70=cos10.cos30.\frac{1}{2}\left(cos120+cos20\right)\)
\(=\frac{1}{2}cos30\left(cos10.cos120+cos10.cos20\right)\)
\(=\frac{1}{2}cos30\left(cos10.cos120+\frac{1}{2}\left(cos30+cos10\right)\right)\)
\(=\frac{1}{2}cos30\left(cos10.cos120+\frac{1}{2}cos30+\frac{1}{2}cos10\right)\)
\(=\frac{1}{2}.\frac{\sqrt{3}}{2}\left(-\frac{1}{2}cos10+\frac{1}{2}\frac{\sqrt{3}}{2}+\frac{1}{2}cos10\right)\)
\(=\frac{3}{16}\)
Rút gọn biểu thức A=\(\frac{3}{4}\cdot\frac{8}{9}\cdot\frac{15}{16}\cdot\frac{24}{25}\cdot....\cdot\frac{899}{900}\)ta được A=......
(Nhập kết quả dạng phân số tối giản)
~ So sad :( !! ~
\(A=\frac{31}{60}\)
I thinks so ! Sad
Rút Gọn :\frac{4^5\cdot 9^4-2\cdot 6^9}{2^{10}\cdot 3^8+6^8\cdot 20}
Rút gọn: \(\frac{x^2}{\left(x+y\right)\cdot\left(1-y\right)}-\frac{y^2}{\left(x+y\right)\cdot\left(1+x\right)}-\frac{x^2\cdot y^2}{\left(x+1\right)\cdot\left(1-y\right)}\)
MTC: (x+y)(x+1)(1-y)
\(=\frac{x^2\left(1+x\right)-y^2\left(1-y\right)-x^2y^2\left(x+y\right)}{\left(x+y\right)\left(1+x\right)\left(1-y\right)}=\frac{\left(x+y\right)\left(1+x\right)\left(1-y\right)\left(x-y+xy\right)}{\left(x+y\right)\left(1+x\right)\left(1-y\right)}\)
\(=x-y+xy\)
Với \(x\ne-1;x\ne-y;y\ne1\)thì giá trị biểu thức được xác định
Rút gọn biểu thức A= \(\frac{\left(\frac{2}{3}\right)^3\cdot\left(-\frac{3}{4}\right)^2\cdot\left(-1\right)^{2017}}{\left(\frac{2}{5}\right)^2\cdot\left(-\frac{5}{12}\right)^3}-\frac{71}{5}\)
\(\left(\frac{4\cdot\sqrt{x}}{2+\sqrt{x}}\right)+\left(\frac{8\cdot x}{4-x}\right)\) rút gọn giúp mình với
ĐK x khác 4 và x không âm
\(=\frac{4\sqrt{x}\left(2-\sqrt{x}\right)+8x}{4-x}\\ =\frac{8\sqrt{x}+4x}{4-x}\\ =\frac{4\sqrt{x}\left(2+\sqrt{x}\right)}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}\\ =\frac{4\sqrt{x}}{2-\sqrt{x}}\)
Rút gọn phân số
\(\frac{48}{168}\cdot\frac{132}{32}\)
\(\frac{48}{168}\cdot\frac{132}{32}=\frac{2^4\cdot3}{2^3\cdot3\cdot7}\cdot\frac{2^2\cdot3\cdot11}{2^5}=\frac{2\cdot1}{1\cdot1\cdot7}\cdot\frac{1\cdot3\cdot11}{2^3}=\frac{33}{28}\)
Rút gọn
B= \(\frac{\sin\left(-234^o\right)-\cos216^o}{\sin\left(144^o\right)-\cos126^o}.\tan36^o\)
\(B=\frac{-\sin\left(\frac{\pi}{2}+144^0\right)-\cos126^0}{\sin144^0-\cos126^0}.\tan\left(\pi-144^0\right)\)
\(B=\frac{-\cos144^0-\cos126^0}{\sin144^0-\cos126^0}.\left(-\tan144^0\right)\)
\(B=\frac{\sin144^0.\cos144^0+\sin144^0.\cos126^0}{\sin144^0.\cos144^0-\cos144^0.\cos126^0}\)
\(B=\frac{\sin\left(\pi+\frac{\pi}{2}-126^0\right)[\cos\left(\pi+\frac{\pi}{2}-126^0\right)+\cos126^0]}{\cos\left(\pi+\frac{\pi}{2}-126^0\right)[\sin\left(\pi+\frac{\pi}{2}-126^0\right)-\cos126^0]}\)
\(\sin\left(\pi+\frac{\pi}{2}-126^0\right)=-\sin\left(\frac{\pi}{2}-126^0\right)=-\cos126^0\)
\(\cos\left(\pi+\frac{\pi}{2}-126^0\right)=-\cos\left(\frac{\pi}{2}-126^0\right)=-\sin126^0\)
\(\Rightarrow B=\frac{-\cos126^0\left(-\sin126^0+\cos126^0\right)}{-\sin126^0\left(-\cos126^0-\cos126^0\right)}\)
\(=\cot126^0.\frac{\sin126^0-\cos126^0}{2\cos126^0}\)
\(=\cot126^0\left(\frac{1}{2}.\tan126^0-\frac{1}{2}\right)\)
\(=\frac{1}{\tan126^0}.\frac{1}{2}.\tan126^0-\frac{1}{2}.\cot126^0=\frac{1}{2}\left(1-\cot126^0\right)\)
Thế này là gọn nhất rồi đấy :<
\(B=\frac{sin126^0-cos144^0}{sin144^0-cos126^0}.tan36^0=\frac{cos36^0+sin54^0}{cos54^0+sin36^0}.tan36^0\)
\(=\frac{cos36^0+cos36^0}{sin36^0+sin36^0}.tan36^0=cot36^0.tan36^0=1\)