Rút gọn biểu thức sau:
\(cos\dfrac{2\pi}{9}+cos\dfrac{4\pi}{9}+cos\dfrac{8\pi}{9}\)
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=\(cos\left(x-\dfrac{\pi}{3}\right).cos\left(x+\dfrac{\pi}{4}\right)+cos\left(x+\dfrac{\pi}{6}\right).cos\left(x+\dfrac{3\pi}{4}\right)\)
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:
\(A=cos\dfrac{\pi}{7}cos\dfrac{2\pi}{7}cos\dfrac{4\pi}{7}\)
\(A.sin\dfrac{\pi}{7}=sin\left(\dfrac{\pi}{7}\right)cos\left(\dfrac{\pi}{7}\right)cos\left(\dfrac{2\pi}{7}\right)cos\left(\dfrac{4\pi}{7}\right)\)
\(=\dfrac{1}{2}sin\left(\dfrac{2\pi}{7}\right)cos\left(\dfrac{2\pi}{7}\right)cos\left(\dfrac{4\pi}{7}\right)\)
\(=\dfrac{1}{4}sin\left(\dfrac{4\pi}{7}\right)cos\left(\dfrac{4\pi}{7}\right)\)
\(=\dfrac{1}{8}sin\left(\dfrac{8\pi}{7}\right)\)
\(=\dfrac{1}{8}sin\left(\pi+\dfrac{\pi}{7}\right)=\dfrac{1}{8}sin\left(-\dfrac{\pi}{7}\right)\)
\(=-\dfrac{1}{8}sin\left(\dfrac{\pi}{7}\right)\)
\(\Rightarrow A=-\dfrac{1}{8}\)
tính \(A=cos\dfrac{\pi}{11}.cos\dfrac{3\pi}{11}.cos\dfrac{5\pi}{11}.....cos\dfrac{9\pi}{11}\)
\(A=cos\dfrac{\pi}{11}.cos\dfrac{3\pi}{11}.cos\dfrac{5\pi}{11}.cos\left(\pi-\dfrac{4\pi}{11}\right)cos\left(\pi-\dfrac{2\pi}{11}\right)\)
\(=cos\dfrac{\pi}{11}.cos\dfrac{3\pi}{11}cos\dfrac{5\pi}{11}\left(-cos\dfrac{4\pi}{11}\right)\left(-cos\dfrac{2\pi}{11}\right)\)
\(=cos\dfrac{\pi}{11}cos\dfrac{2\pi}{11}cos\dfrac{3\pi}{11}cos\dfrac{4\pi}{11}cos\dfrac{5\pi}{11}\)
\(\Rightarrow2A.sin\dfrac{\pi}{11}=2sin\dfrac{\pi}{11}cos\dfrac{\pi}{11}cos\dfrac{2\pi}{11}cos\dfrac{4\pi}{11}cos\dfrac{3\pi}{11}cos\dfrac{5\pi}{11}\)
\(=sin\dfrac{2\pi}{11}cos\dfrac{2\pi}{11}cos\dfrac{4\pi}{11}cos\dfrac{3\pi}{11}cos\dfrac{5\pi}{11}\)
\(=\dfrac{1}{2}sin\dfrac{4\pi}{11}cos\dfrac{4\pi}{11}cos\dfrac{3\pi}{11}cos\dfrac{5\pi}{11}\)
\(=\dfrac{1}{4}sin\dfrac{8\pi}{11}.cos\dfrac{3\pi}{11}.cos\left(\pi-\dfrac{6\pi}{11}\right)\)
\(=-\dfrac{1}{4}sin\left(\pi-\dfrac{3\pi}{11}\right)cos\dfrac{3\pi}{11}cos\dfrac{6\pi}{11}=-\dfrac{1}{4}sin\dfrac{3\pi}{11}cos\dfrac{3\pi}{11}cos\dfrac{6\pi}{11}\)
\(=-\dfrac{1}{8}sin\dfrac{6\pi}{11}cos\dfrac{6\pi}{11}=-\dfrac{1}{16}sin\dfrac{12\pi}{11}=-\dfrac{1}{16}sin\left(\pi+\dfrac{\pi}{11}\right)\)
\(=\dfrac{1}{16}sin\dfrac{\pi}{11}\)
\(\Rightarrow A=\dfrac{1}{32}\)
Rút gọn cac biểu thức sau:
\(A=sin\left(\dfrac{5\pi}{2}-\alpha\right)+cos\left(13\pi+\alpha\right)-3sin\left(\alpha-5\pi\right)\)
\(B=sin\left(x+\dfrac{85\pi}{2}\right)+cos\left(2017\pi+x\right)+sin^2\left(33\pi+x\right)+sin^2\left(x-\dfrac{5\pi}{2}\right)+cos\left(x+\dfrac{3\pi}{2}\right)\)\(C=sin\left(x+\dfrac{2017\pi}{2}\right)+2sin^2\left(x-\pi\right)+cos\left(x+2019\pi\right)+cos2x+sin\left(x+\dfrac{9\pi}{2}\right)\)
\(A=sin\left(\dfrac{\pi}{2}-\alpha+2\pi\right)+cos\left(\pi+\alpha+12\pi\right)-3sin\left(\alpha-\pi-4\pi\right)\)
\(=sin\left(\dfrac{\pi}{2}-\alpha\right)+cos\left(\pi+\alpha\right)-3sin\left(\alpha-\pi\right)\)
\(=cos\alpha-cos\alpha+3sin\left(\pi-\alpha\right)\)\(=3sin\alpha\)
\(B=sin\left(x+\dfrac{\pi}{2}+42\pi\right)+cos\left(x+\pi+2016\pi\right)+sin^2\left(x+\pi+32\pi\right)+sin^2\left(x-\dfrac{\pi}{2}-2\pi\right)+cos\left(x-\dfrac{\pi}{2}+2\pi\right)\)
\(=sin\left(x+\dfrac{\pi}{2}\right)+cos\left(x+\pi\right)+sin^2\left(x+\pi\right)+sin^2\left(x-\dfrac{\pi}{2}\right)+cos\left(x-\dfrac{\pi}{2}\right)\)
\(=cosx-cosx+sin^2x+cos^2x+sinx\)
\(=1+sinx\)
\(C=sin\left(x+\dfrac{\pi}{2}+1008\pi\right)+2sin^2\left(\pi-x\right)+cos\left(x+\pi+2018\pi\right)+cos2x+sin\left(x+\dfrac{\pi}{2}+4\pi\right)\)
\(=sin\left(x+\dfrac{\pi}{2}\right)+2sin^2\left(\pi-x\right)+cos\left(x+\pi\right)+cos2x+sin\left(x+\dfrac{\pi}{2}\right)\)
\(=cosx+2sin^2x-cosx+1-2sin^2x+cosx\)
\(=1+cosx\)
1; tính B \(=4sin^4\dfrac{\pi}{16}+2cos\dfrac{\pi}{8}\)
2;tính C= \(\dfrac{\sin\dfrac{\pi}{5}-\sin\dfrac{2\pi}{15}}{\cos\dfrac{\pi}{5}-\cos\dfrac{2\pi}{15}}\)
3; tính D=\(\sin\dfrac{\pi}{9}-sin\dfrac{5\pi}{9}+sin\dfrac{7\pi}{9}\)
a) Biến đổi \(\sin\alpha-1\)thành tích
b) Rút gọn biểu thức \(P=\dfrac{\cos\alpha+2\cos3\alpha+\cos5a}{\sin\alpha+2\sin3\alpha+\sin5a}\)
c) Tính giá trị biểu thức \(P=\sin30.\cos60+\sin60.\cos30\)
d) Giá đúng của \(cos\dfrac{2\pi}{7}+\cos\dfrac{4\pi}{7}+\cos\dfrac{6\pi}{7}\)
e) Giá trị đúng của \(\tan\dfrac{\pi}{24}+\tan\dfrac{7\pi}{24}\)
a/\(sina-1=2sin\dfrac{a}{2}.cos\dfrac{a}{2}-sin^2\dfrac{a}{2}-cos^2\dfrac{a}{2}=-\left(sin\dfrac{a}{2}-cos\dfrac{a}{2}\right)^2\)
b/\(P=\dfrac{cosa+cos5a+2cos3a}{sina+sin5a+2sin3a}=\dfrac{2cos3a.cos2a+2cos3a}{2sin3a.cos2a+2sin3a}=\dfrac{2cos3a\left(cos2a+1\right)}{2sin3a\left(cos2a+1\right)}=cot3a\)
c/\(P=sin\left(30+60\right)=sin90=1\)
d/
\(A=cos\dfrac{2\pi}{7}+cos\dfrac{6\pi}{7}+cos\dfrac{4\pi}{7}\Rightarrow A.sin\dfrac{\pi}{7}=sin\dfrac{\pi}{7}.cos\dfrac{2\pi}{7}+sin\dfrac{\pi}{7}cos\dfrac{4\pi}{7}+sin\dfrac{\pi}{7}.cos\dfrac{6\pi}{7}\)
\(=\dfrac{1}{2}sin\dfrac{3\pi}{7}-\dfrac{1}{2}sin\dfrac{\pi}{7}+\dfrac{1}{2}sin\dfrac{5\pi}{7}-\dfrac{1}{2}sin\dfrac{3\pi}{7}+\dfrac{1}{2}sin\dfrac{7\pi}{7}-\dfrac{1}{2}sin\dfrac{5\pi}{7}\)
\(=-\dfrac{1}{2}sin\dfrac{\pi}{7}\Rightarrow A=-\dfrac{1}{2}\)
e/
\(tan\dfrac{\pi}{24}+tan\dfrac{7\pi}{24}=\dfrac{sin\dfrac{\pi}{24}}{cos\dfrac{\pi}{24}}+\dfrac{sin\dfrac{7\pi}{24}}{cos\dfrac{7\pi}{24}}=\dfrac{sin\dfrac{\pi}{24}cos\dfrac{7\pi}{24}+sin\dfrac{7\pi}{24}cos\dfrac{\pi}{24}}{cos\dfrac{\pi}{24}.cos\dfrac{7\pi}{24}}\)
\(=\dfrac{sin\left(\dfrac{\pi}{24}+\dfrac{7\pi}{24}\right)}{\dfrac{1}{2}cos\dfrac{\pi}{4}+\dfrac{1}{2}cos\dfrac{\pi}{3}}=\dfrac{2sin\dfrac{\pi}{3}}{cos\dfrac{\pi}{4}+cos\dfrac{\pi}{3}}=\dfrac{\sqrt{3}}{\dfrac{\sqrt{2}}{2}+\dfrac{1}{2}}=\dfrac{2\sqrt{3}}{\sqrt{2}+1}\)
sina - 1 = sina - sin\(\dfrac{\pi}{2}\)
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.
\(2cos\left(a+b\right)=cosa.cos\left(\pi+b\right)\)
\(\Leftrightarrow2cosa.cosb-2sina.sinb=-cosa.cosb\)
\(\Leftrightarrow2sina.sinb=3cosa.cosb\Rightarrow4sin^2a.sin^2b=9cos^2a.cos^2b\)
\(\Rightarrow4\left(1-cos^2a\right)\left(1-cos^2b\right)=9cos^2a.cos^2b\)
\(\Leftrightarrow4-4\left(cos^2a+cos^2b\right)=5cos^2a.cos^2b\)
\(A=\dfrac{1}{cos^2a+2\left(sin^2a+cos^2a\right)}+\dfrac{1}{cos^2b+2\left(sin^2b+cos^2b\right)}\)
\(=\dfrac{1}{2+cos^2a}+\dfrac{1}{2+cos^2b}=\dfrac{4+cos^2a+cos^2b}{4+2\left(cos^2a+cos^2b\right)+cos^2a.cos^2b}\)
\(=\dfrac{4+cos^2a+cos^2b}{4+2\left(cos^2a+cos^2b\right)+\dfrac{4}{5}-\dfrac{4}{5}\left(cos^2a+cos^2b\right)}=\dfrac{4+cos^2a+cos^2b}{\dfrac{24}{5}+\dfrac{6}{5}\left(cos^2a+cos^2b\right)}=\dfrac{5}{6}\)
2.
\(A=2cos\dfrac{2x}{3}\left(cos\dfrac{2\pi}{3}+cos\dfrac{4x}{3}\right)=2cos\dfrac{2x}{3}\left(cos\dfrac{4x}{3}-\dfrac{1}{2}\right)\)
\(=2cos\dfrac{2x}{3}.cos\dfrac{4x}{3}-cos\dfrac{2x}{3}\)
\(=cos3x+cos\dfrac{2x}{3}-cos\dfrac{2x}{3}\)
\(=cos3x\)
\(B=\dfrac{cos2b-cos2a}{cos^2a.sin^2b}-tan^2a.cot^2b=\dfrac{1-2sin^2b-\left(1-2sin^2a\right)}{cos^2a.sin^2b}-tan^2a.cot^2b\)
\(=\dfrac{2sin^2a-2sin^2b}{cos^2a.sin^2b}-tan^2a.cot^2b=2tan^2a\left(1+cot^2b\right)-2\left(1+tan^2a\right)-tan^2a.cot^2b\)
\(=2tan^2a+2tan^2a.cot^2b-2-2tan^2a-tan^2a.cot^2b\)
\(=tan^2a.cot^2b-2\)
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.
\(\dfrac{2sina}{cosa}=\dfrac{sin\left(a+b\right)}{cos\left(a+b\right)}\Leftrightarrow2sina.cos\left(a+b\right)=cosa.sin\left(a+b\right)\)
\(\Leftrightarrow sin\left(2a+b\right)+sin\left(-b\right)=\dfrac{1}{2}sin\left(2a+b\right)+\dfrac{1}{2}sinb\)
\(\Leftrightarrow\dfrac{1}{2}sin\left(2a+b\right)=\dfrac{3}{2}sinb\)
\(\Leftrightarrow sin\left(2a+b\right)=3sinb\)
a) Rút gọn biểu thức
\(A=\dfrac{\sin4x+2\sin2x}{\sin4x-2\sin2x}.\cot\left(\dfrac{3\pi}{2}-x\right)\) (khi biểu thức có nghĩa)
b) Cho \(\cot\alpha=\dfrac{4}{3},3\pi< \alpha< \dfrac{7\pi}{2}\). Tính \(\cos\left(\dfrac{2\pi}{3}-\alpha\right)\)