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\(A=\frac{\sin\left(-234\right)-cos216}{\sin144-cos126}\cdot\tan36\)
\(=\frac{-\sin234-cos216}{\sin\left(360-216\right)-cos\left(360-234\right)}\cdot\tan36\)
\(=\frac{-\sin\left(180+54\right)-cos\left(180^0+36^0\right)}{\sin\left(180^0-36^0\right)-cos\left(90^0+36^0\right)}\cdot\tan36^0=\frac{\sin54^0+cos36^0}{\sin36^0+\sin36^0}\cdot\tan36^0\)
\(=\frac{2\cdot cos36^0}{2\cdot\sin36^0}\cdot\frac{\sin36^0}{cos36^0}=1\)
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:
\(cos10.cos30.cos50.cos70=cos10.cos30.\frac{1}{2}\left(cos120+cos20\right)\)
\(=\frac{1}{2}cos30\left(cos10.cos120+cos10.cos20\right)\)
\(=\frac{1}{2}cos30\left(cos10.cos120+\frac{1}{2}\left(cos30+cos10\right)\right)\)
\(=\frac{1}{2}cos30\left(cos10.cos120+\frac{1}{2}cos30+\frac{1}{2}cos10\right)\)
\(=\frac{1}{2}.\frac{\sqrt{3}}{2}\left(-\frac{1}{2}cos10+\frac{1}{2}\frac{\sqrt{3}}{2}+\frac{1}{2}cos10\right)\)
\(=\frac{3}{16}\)
\(B=\frac{sin126^0-cos144^0}{sin144^0-cos126^0}.tan36^0=\frac{cos36^0+sin54^0}{cos54^0+sin36^0}.tan36^0\)
\(=\frac{cos36^0+cos36^0}{sin36^0+sin36^0}.tan36^0=cot36^0.tan36^0=1\)
\(B=\frac{-\sin\left(\frac{\pi}{2}+144^0\right)-\cos126^0}{\sin144^0-\cos126^0}.\tan\left(\pi-144^0\right)\)
\(B=\frac{-\cos144^0-\cos126^0}{\sin144^0-\cos126^0}.\left(-\tan144^0\right)\)
\(B=\frac{\sin144^0.\cos144^0+\sin144^0.\cos126^0}{\sin144^0.\cos144^0-\cos144^0.\cos126^0}\)
\(B=\frac{\sin\left(\pi+\frac{\pi}{2}-126^0\right)[\cos\left(\pi+\frac{\pi}{2}-126^0\right)+\cos126^0]}{\cos\left(\pi+\frac{\pi}{2}-126^0\right)[\sin\left(\pi+\frac{\pi}{2}-126^0\right)-\cos126^0]}\)
\(\sin\left(\pi+\frac{\pi}{2}-126^0\right)=-\sin\left(\frac{\pi}{2}-126^0\right)=-\cos126^0\)
\(\cos\left(\pi+\frac{\pi}{2}-126^0\right)=-\cos\left(\frac{\pi}{2}-126^0\right)=-\sin126^0\)
\(\Rightarrow B=\frac{-\cos126^0\left(-\sin126^0+\cos126^0\right)}{-\sin126^0\left(-\cos126^0-\cos126^0\right)}\)
\(=\cot126^0.\frac{\sin126^0-\cos126^0}{2\cos126^0}\)
\(=\cot126^0\left(\frac{1}{2}.\tan126^0-\frac{1}{2}\right)\)
\(=\frac{1}{\tan126^0}.\frac{1}{2}.\tan126^0-\frac{1}{2}.\cot126^0=\frac{1}{2}\left(1-\cot126^0\right)\)
Thế này là gọn nhất rồi đấy :<
b) \(S=\frac{1}{2}\sqrt{AB^2.AC^2-\left(\overrightarrow{AB}.\overrightarrow{AC}\right)^2}\)
\(=\frac{1}{2}\sqrt{AB^2.AC^2-AB^2.AC^2.cos^2A}\)
\(=\frac{1}{2}\sqrt{AB^2AC^2.sin^2A}\)
\(=\frac{1}{2}.AB.AC.\sin A\) (đpcm)
a) P = cos(\(\frac{\Pi}{2}\) + x) + cos(2π - x) + cos(3π + x) = -sinx + cosx - cosx = -sinx
\(A=\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}+\frac{1}{2}cosa}}}\)
\(=\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}+\frac{1}{2}\left(2cos^2\frac{a}{2}-1\right)}}}\)
\(=\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}+cos^2\frac{a}{2}-\frac{1}{2}}}}\)
\(=\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}+\frac{1}{2}cos\frac{a}{2}}}\)
\(=\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}+\frac{1}{2}\left(2cos^2\frac{a}{4}-1\right)}}\)
\(=\sqrt{\frac{1}{2}+\frac{1}{2}cos\frac{a}{4}}=\sqrt{\frac{1}{2}+\frac{1}{2}\left(cos^2\frac{a}{8}-1\right)}\)
\(=cos\frac{a}{8}\Rightarrow n=8\)
Bài 1. Ta có: \(a\left(a+2\right)\left(a-1\right)^2\ge0\therefore\frac{1}{4a^2-2a+1}\ge\frac{1}{a^4+a^2+1}\)
Thiết lập tương tự 2 BĐT còn lại và cộng theo vế rồi dùng Vasc (https://olm.vn/hoi-dap/detail/255345443802.html)
Bài 5: Bất đẳng thức này đúng với mọi a, b, c là các số thực. Chứng minh:
Quy đồng và chú ý các mẫu thức đều không âm, ta cần chứng minh:
\(\frac{1}{2}\left(a^2+b^2+c^2-ab-bc-ca\right)\Sigma\left[\left(a^2+b^2\right)+2c^2\right]\left(a-b\right)^2\ge0\)
Đây là điều hiển nhiên.
\(=\frac{cos36-sin\left(180+54\right)}{sin\left(180-36\right)-cos\left(90+36\right)}.cos54=\frac{cos36+sin54}{sin36+sin36}.cos54\)
\(=\frac{cos36+cos36}{2sin36}.sin36=cos36\)