Tính: \(H=cos\dfrac{\pi}{19}+cos\dfrac{3\pi}{19}+...+cos\dfrac{17\pi}{19}\)
Những câu hỏi liên quan
tính \(A=cos\dfrac{\pi}{11}.cos\dfrac{3\pi}{11}.cos\dfrac{5\pi}{11}.....cos\dfrac{9\pi}{11}\)
\(A=cos\dfrac{\pi}{11}.cos\dfrac{3\pi}{11}.cos\dfrac{5\pi}{11}.cos\left(\pi-\dfrac{4\pi}{11}\right)cos\left(\pi-\dfrac{2\pi}{11}\right)\)
\(=cos\dfrac{\pi}{11}.cos\dfrac{3\pi}{11}cos\dfrac{5\pi}{11}\left(-cos\dfrac{4\pi}{11}\right)\left(-cos\dfrac{2\pi}{11}\right)\)
\(=cos\dfrac{\pi}{11}cos\dfrac{2\pi}{11}cos\dfrac{3\pi}{11}cos\dfrac{4\pi}{11}cos\dfrac{5\pi}{11}\)
\(\Rightarrow2A.sin\dfrac{\pi}{11}=2sin\dfrac{\pi}{11}cos\dfrac{\pi}{11}cos\dfrac{2\pi}{11}cos\dfrac{4\pi}{11}cos\dfrac{3\pi}{11}cos\dfrac{5\pi}{11}\)
\(=sin\dfrac{2\pi}{11}cos\dfrac{2\pi}{11}cos\dfrac{4\pi}{11}cos\dfrac{3\pi}{11}cos\dfrac{5\pi}{11}\)
\(=\dfrac{1}{2}sin\dfrac{4\pi}{11}cos\dfrac{4\pi}{11}cos\dfrac{3\pi}{11}cos\dfrac{5\pi}{11}\)
\(=\dfrac{1}{4}sin\dfrac{8\pi}{11}.cos\dfrac{3\pi}{11}.cos\left(\pi-\dfrac{6\pi}{11}\right)\)
\(=-\dfrac{1}{4}sin\left(\pi-\dfrac{3\pi}{11}\right)cos\dfrac{3\pi}{11}cos\dfrac{6\pi}{11}=-\dfrac{1}{4}sin\dfrac{3\pi}{11}cos\dfrac{3\pi}{11}cos\dfrac{6\pi}{11}\)
\(=-\dfrac{1}{8}sin\dfrac{6\pi}{11}cos\dfrac{6\pi}{11}=-\dfrac{1}{16}sin\dfrac{12\pi}{11}=-\dfrac{1}{16}sin\left(\pi+\dfrac{\pi}{11}\right)\)
\(=\dfrac{1}{16}sin\dfrac{\pi}{11}\)
\(\Rightarrow A=\dfrac{1}{32}\)
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27 tháng 5 2021 lúc 8:58
Tính
A = \(\dfrac{1}{cos\dfrac{\pi}{7}}+\dfrac{1}{cos\dfrac{3\pi}{7}}+\dfrac{1}{cos\dfrac{5\pi}{7}}\)
A\(=\dfrac{cos\dfrac{5\pi}{7}.cos\dfrac{3\pi}{7}+cos\dfrac{5\pi}{7}.cos\dfrac{\pi}{7}+cos\dfrac{3\pi}{7}.cos\dfrac{\pi}{7}}{cos\dfrac{\pi}{7}.cos\dfrac{3\pi}{7}.cos\dfrac{5\pi}{7}}\)
Đặt tử là Y; mẫu là U
Có \(Y=\)\(cos\dfrac{5\pi}{7}.cos\dfrac{3\pi}{7}+\left(cos\dfrac{5\pi}{7}.cos\dfrac{\pi}{7}+cos\dfrac{3\pi}{7}.cos\dfrac{\pi}{7}\right)\)
\(=cos\left(\pi-\dfrac{2\pi}{7}\right).cos\left(\pi-\dfrac{4\pi}{7}\right)+cos\dfrac{\pi}{7}\left(cos\dfrac{5\pi}{7}+cos\dfrac{3\pi}{7}\right)\)
\(=cos\dfrac{2\pi}{7}.cos\dfrac{4\pi}{7}+cos\dfrac{\pi}{7}.2cos\dfrac{4\pi}{7}.cos\dfrac{\pi}{7}\)\(=cos\dfrac{2\pi}{7}.cos\dfrac{4\pi}{7}+2.cos^2\dfrac{\pi}{7}.cos\dfrac{4\pi}{7}\)
\(=cos\dfrac{2\pi}{7}.cos\dfrac{4\pi}{7}+\left(cos\dfrac{2\pi}{7}+1\right).cos\dfrac{4\pi}{7}\)\(=2.cos\dfrac{2\pi}{7}.cos\dfrac{4\pi}{7}+cos\dfrac{4\pi}{7}\)
\(=cos\dfrac{6\pi}{7}+cos\dfrac{2\pi}{7}+cos\dfrac{4\pi}{7}\)
\(\Rightarrow sin\dfrac{\pi}{7}.Y=sin\dfrac{\pi}{7}.cos\dfrac{2\pi}{7}+sin\dfrac{\pi}{7}.cos\dfrac{4\pi}{7}+sin\dfrac{\pi}{7}.cos\dfrac{6\pi}{7}\)
\(=\dfrac{1}{2}\left(-sin\dfrac{\pi}{7}+sin\dfrac{3\pi}{7}\right)+\dfrac{1}{2}\left(-sin\dfrac{3\pi}{7}+sin\dfrac{5\pi}{7}\right)+\dfrac{1}{2}\left(-sin\dfrac{5\pi}{7}+sin\pi\right)\)
\(=\dfrac{1}{2}\left(sin\pi-sin\dfrac{\pi}{7}\right)\)\(=-\dfrac{1}{2}sin\dfrac{\pi}{7}\)
\(\Rightarrow Y=-\dfrac{1}{2}\)
Có \(sin\dfrac{\pi}{7}.U=sin\dfrac{\pi}{7}.cos\dfrac{\pi}{7}.cos\dfrac{3\pi}{5}.cos\dfrac{5\pi}{7}\)
\(=\dfrac{1}{2}.sin\dfrac{2\pi}{7}.cos\left(\pi-\dfrac{2\pi}{7}\right).cos\dfrac{3\pi}{5}\)
\(=-\dfrac{1}{4}.sin\dfrac{4\pi}{7}.cos\left(\pi-\dfrac{4\pi}{5}\right)\)
\(=\dfrac{1}{8}.sin\dfrac{8\pi}{7}\)\(=\dfrac{1}{8}.sin\left(\pi+\dfrac{\pi}{7}\right)=-\dfrac{1}{8}.sin\dfrac{\pi}{7}\)
\(\Rightarrow U=-\dfrac{1}{8}\)
Vậy \(A=\dfrac{Y}{U}=4\)
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1. Cho 2cosleft(alpha+betaright)cosalphacosleft(pi+betaright)Tính Adfrac{1}{2sin^2alpha+3cos^2alpha}+dfrac{1}{2sin^2beta+3cos^2beta}2. Rút gọn: a) A4cosdfrac{2x}{3}cosdfrac{pi+2x}{3}cosdfrac{pi-2x}{3}b) Bdfrac{sinleft(a-bright).sinleft(a+bright)}{cos^2a.sin^2b}-tan^2a.cot^2b3. Chứng minh rằng: Nếu 2tan atanleft(a+bright) thì:a) sin bsin a.cosleft(a+bright)b) 3sin bsinleft(2a+bright)
Đọc tiếp
1. Cho \(2\cos\left(\alpha+\beta\right)=\cos\alpha\cos\left(\pi+\beta\right)\)
Tính \(A=\dfrac{1}{2\sin^2\alpha+3\cos^2\alpha}+\dfrac{1}{2\sin^2\beta+3\cos^2\beta}\)
2. Rút gọn: a) \(A=4\cos\dfrac{2x}{3}\cos\dfrac{\pi+2x}{3}\cos\dfrac{\pi-2x}{3}\)
b) \(B=\dfrac{\sin\left(a-b\right).\sin\left(a+b\right)}{\cos^2a.\sin^2b}-\tan^2a.\cot^2b\)
3. Chứng minh rằng: Nếu \(2\tan a=\tan\left(a+b\right)\) thì:
a) \(\sin b=\sin a.\cos\left(a+b\right)\)
b) \(3\sin b=\sin\left(2a+b\right)\)
1.
\(2cos\left(a+b\right)=cosa.cos\left(\pi+b\right)\)
\(\Leftrightarrow2cosa.cosb-2sina.sinb=-cosa.cosb\)
\(\Leftrightarrow2sina.sinb=3cosa.cosb\Rightarrow4sin^2a.sin^2b=9cos^2a.cos^2b\)
\(\Rightarrow4\left(1-cos^2a\right)\left(1-cos^2b\right)=9cos^2a.cos^2b\)
\(\Leftrightarrow4-4\left(cos^2a+cos^2b\right)=5cos^2a.cos^2b\)
\(A=\dfrac{1}{cos^2a+2\left(sin^2a+cos^2a\right)}+\dfrac{1}{cos^2b+2\left(sin^2b+cos^2b\right)}\)
\(=\dfrac{1}{2+cos^2a}+\dfrac{1}{2+cos^2b}=\dfrac{4+cos^2a+cos^2b}{4+2\left(cos^2a+cos^2b\right)+cos^2a.cos^2b}\)
\(=\dfrac{4+cos^2a+cos^2b}{4+2\left(cos^2a+cos^2b\right)+\dfrac{4}{5}-\dfrac{4}{5}\left(cos^2a+cos^2b\right)}=\dfrac{4+cos^2a+cos^2b}{\dfrac{24}{5}+\dfrac{6}{5}\left(cos^2a+cos^2b\right)}=\dfrac{5}{6}\)
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2.
\(A=2cos\dfrac{2x}{3}\left(cos\dfrac{2\pi}{3}+cos\dfrac{4x}{3}\right)=2cos\dfrac{2x}{3}\left(cos\dfrac{4x}{3}-\dfrac{1}{2}\right)\)
\(=2cos\dfrac{2x}{3}.cos\dfrac{4x}{3}-cos\dfrac{2x}{3}\)
\(=cos3x+cos\dfrac{2x}{3}-cos\dfrac{2x}{3}\)
\(=cos3x\)
\(B=\dfrac{cos2b-cos2a}{cos^2a.sin^2b}-tan^2a.cot^2b=\dfrac{1-2sin^2b-\left(1-2sin^2a\right)}{cos^2a.sin^2b}-tan^2a.cot^2b\)
\(=\dfrac{2sin^2a-2sin^2b}{cos^2a.sin^2b}-tan^2a.cot^2b=2tan^2a\left(1+cot^2b\right)-2\left(1+tan^2a\right)-tan^2a.cot^2b\)
\(=2tan^2a+2tan^2a.cot^2b-2-2tan^2a-tan^2a.cot^2b\)
\(=tan^2a.cot^2b-2\)
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3.
\(\dfrac{2sina}{cosa}=\dfrac{sin\left(a+b\right)}{cos\left(a+b\right)}\Leftrightarrow2sina.cos\left(a+b\right)=cosa.sin\left(a+b\right)\)
\(\Leftrightarrow sina.cos\left(a+b\right)=sin\left(a+b\right).cosa-cos\left(a+b\right)sina\)
\(\Leftrightarrow sina.cos\left(a+b\right)=sin\left(a+b-a\right)\)
\(\Leftrightarrow sina.cos\left(a+b\right)=sinb\)
b.
\(\dfrac{2sina}{cosa}=\dfrac{sin\left(a+b\right)}{cos\left(a+b\right)}\Leftrightarrow2sina.cos\left(a+b\right)=cosa.sin\left(a+b\right)\)
\(\Leftrightarrow sin\left(2a+b\right)+sin\left(-b\right)=\dfrac{1}{2}sin\left(2a+b\right)+\dfrac{1}{2}sinb\)
\(\Leftrightarrow\dfrac{1}{2}sin\left(2a+b\right)=\dfrac{3}{2}sinb\)
\(\Leftrightarrow sin\left(2a+b\right)=3sinb\)
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8 tháng 2 2021 lúc 22:34
tính tổng vô hạn \(1+cos\dfrac{\pi}{6}+cos^2\dfrac{\pi}{6}+...+cos^n\dfrac{\pi}{6}+...\)
\(cos^2\dfrac{\pi}{7}+cos^2\dfrac{2\pi}{7}+cos^2\dfrac{3\pi}{7}\)
\(cos^2\left(\dfrac{pi}{7}\right)+cos^2\left(\dfrac{2pi}{7}\right)+cos^2\left(\dfrac{3pi}{7}\right)\)
\(=1-2\cdot cos\left(\dfrac{pi}{7}\right)\cdot cos\left(\dfrac{2pi}{7}\right)\cdot cos\left(\dfrac{3pi}{7}\right)\)
\(=1-2\cdot\dfrac{1}{2}\left[cos\left(\dfrac{pi}{7}+\dfrac{3pi}{7}\right)+cos\left(\dfrac{3pi}{7}-\dfrac{pi}{7}\right)\right]\cdot cos\left(\dfrac{2pi}{7}\right)\)
\(=1-cos\left(\dfrac{2pi}{7}\right)\cdot cos\left(\dfrac{4pi}{7}\right)-cos\left(\dfrac{2pi}{7}\right)\cdot cos\left(\dfrac{2pi}{7}\right)\)
\(=1-cos^2\left(\dfrac{2pi}{7}\right)-cos\left(\dfrac{2pi}{7}\right)\cdot cos\left(\dfrac{4pi}{7}\right)\)
\(=sin^2\left(\dfrac{2pi}{7}\right)-cos\left(\dfrac{2pi}{7}\right)\cdot\left[2\cdot cos^2\left(\dfrac{2pi}{7}\right)-1\right]\)
\(=sin^2\left(\dfrac{2pi}{7}\right)-2\cdot cos^3\left(\dfrac{2pi}{7}\right)+cos\left(\dfrac{2pi}{7}\right)\)
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Tính giá trị của biểu thức:
A= \(\dfrac{cos\dfrac{9\pi}{5}-cos\dfrac{6\pi}{5}+cos\dfrac{11\pi}{5}}{cos\dfrac{3\pi}{10}-sin\dfrac{6\pi}{5}}.tan\dfrac{16\pi}{5}\).
Tính giá trị của biểu thức sau:
\(A=cos\dfrac{\pi}{7}cos\dfrac{2\pi}{7}cos\dfrac{4\pi}{7}\)
\(A.sin\dfrac{\pi}{7}=sin\left(\dfrac{\pi}{7}\right)cos\left(\dfrac{\pi}{7}\right)cos\left(\dfrac{2\pi}{7}\right)cos\left(\dfrac{4\pi}{7}\right)\)
\(=\dfrac{1}{2}sin\left(\dfrac{2\pi}{7}\right)cos\left(\dfrac{2\pi}{7}\right)cos\left(\dfrac{4\pi}{7}\right)\)
\(=\dfrac{1}{4}sin\left(\dfrac{4\pi}{7}\right)cos\left(\dfrac{4\pi}{7}\right)\)
\(=\dfrac{1}{8}sin\left(\dfrac{8\pi}{7}\right)\)
\(=\dfrac{1}{8}sin\left(\pi+\dfrac{\pi}{7}\right)=\dfrac{1}{8}sin\left(-\dfrac{\pi}{7}\right)\)
\(=-\dfrac{1}{8}sin\left(\dfrac{\pi}{7}\right)\)
\(\Rightarrow A=-\dfrac{1}{8}\)
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Rút gọn:
C= \(sin^2\dfrac{\pi}{3}+sin^2\dfrac{5\pi}{6}+sin^2\dfrac{\pi}{9}+sin^2\dfrac{11\pi}{18}+sin^2\dfrac{13\pi}{18}+sin^2\dfrac{2\pi}{9}\)
D=\(cos\left(x-\dfrac{\pi}{3}\right).cos\left(x+\dfrac{\pi}{4}\right)+cos\left(x+\dfrac{\pi}{6}\right).cos\left(x+\dfrac{3\pi}{4}\right)\)
Với \(\alpha\in R\). Biểu thức
\(A=cos\alpha+cos\left(\alpha+\dfrac{\pi}{6}\right)+cos\left(\alpha+\dfrac{2\pi}{6}\right)+cos\left(\alpha+\dfrac{3\pi}{6}\right)+...+cos\left(\alpha+\dfrac{11\pi}{6}\right)\)
A. A = 10
B. A = -12
C. A = 0
D. A = 12