Sửa đề

\(A=cos^212+cos^223+cos^234+cos^245+cos^256+cos^267+\)

\(=\left(cos^212+cos^278\right)+\left(cos^223+cos^267\right)+\left(cos^234+cos^256\right)+cos^245\)

\(=\left(cos^212+sin^212\right)+\left(cos^223+sin^223\right)+\left(cos^234+sin^234\right)+cos^245\)

\(=1+1+1+\frac{\sqrt{2}}{2}=\frac{6+\sqrt{2}}{2}\)

\(P.sin\left(\dfrac{\pi}{7}\right)=sin\dfrac{\pi}{7}.cos\dfrac{\pi}{7}.cos\dfrac{2\pi}{7}.cos\dfrac{4\pi}{7}\)

\(\Leftrightarrow P.sin\dfrac{\pi}{7}=\dfrac{1}{2}sin\dfrac{2\pi}{7}cos\dfrac{2\pi}{7}cos\dfrac{4\pi}{7}\)

\(\Leftrightarrow P.sin\dfrac{\pi}{7}=\dfrac{1}{4}sin\dfrac{4\pi}{7}cos\dfrac{4\pi}{7}\)

\(\Leftrightarrow P.sin\dfrac{\pi}{7}=\dfrac{1}{8}sin\dfrac{8\pi}{7}=\dfrac{1}{8}sin\left(\pi+\dfrac{\pi}{7}\right)\)

\(\Leftrightarrow P.sin\dfrac{\pi}{7}=-\dfrac{1}{8}sin\dfrac{\pi}{7}\)

\(\Rightarrow P=-\dfrac{1}{8}\)

Chọn B.

Ta có: B = ( cos0⁰ + cos180⁰) + (cos20⁰ + cos160⁰)  +...+ cos80⁰ + cos100⁰)

= (cos0⁰ - cos0⁰)  + (cos20⁰ - cos 20⁰) + ... + (cos80⁰ - cos80⁰) = 0

\(A.sin\dfrac{\pi}{7}=sin\left(\dfrac{\pi}{7}\right)cos\left(\dfrac{\pi}{7}\right)cos\left(\dfrac{2\pi}{7}\right)cos\left(\dfrac{4\pi}{7}\right)\)

\(=\dfrac{1}{2}sin\left(\dfrac{2\pi}{7}\right)cos\left(\dfrac{2\pi}{7}\right)cos\left(\dfrac{4\pi}{7}\right)\)

\(=\dfrac{1}{4}sin\left(\dfrac{4\pi}{7}\right)cos\left(\dfrac{4\pi}{7}\right)\)

\(=\dfrac{1}{8}sin\left(\dfrac{8\pi}{7}\right)\)

\(=\dfrac{1}{8}sin\left(\pi+\dfrac{\pi}{7}\right)=\dfrac{1}{8}sin\left(-\dfrac{\pi}{7}\right)\)

\(=-\dfrac{1}{8}sin\left(\dfrac{\pi}{7}\right)\)

\(\Rightarrow A=-\dfrac{1}{8}\)

\(B = \left( {\cos \frac{\pi }{9} + \cos \frac{{5\pi }}{9}} \right) + \cos \frac{{11\pi }}{9} = \left( {2\cos \frac{{\frac{\pi }{9} + \frac{{5\pi }}{9}}}{2}\cos \frac{{\frac{\pi }{9} - \frac{{5\pi }}{9}}}{2}} \right) + \cos \frac{{11\pi }}{9} = 2\cos \frac{\pi }{3}\cos \frac{{2\pi }}{9} + \cos \frac{{11\pi }}{9}\)

\( = \cos \frac{{2\pi }}{9} + \cos \frac{{11\pi }}{9} = 2\cos \frac{{\frac{{2\pi }}{9} + \frac{{11\pi }}{9}}}{2}\cos \frac{{\frac{{2\pi }}{9} - \frac{{11\pi }}{9}}}{2} = 2\cos \frac{{13\pi }}{{18}}\cos \frac{\pi }{2} = 0\)

\(E=\)\(cos^273+1-sin^247+cos73\left(cos120.cos73+sin120.sin73\right)\)

\(=cos^273+1-\left(sin120.cos73-cos120.sin73\right)^2-\dfrac{1}{2}.cos^273+\dfrac{\sqrt{3}}{2}cos73.sin73\)

\(=cos^273+1-\left(\dfrac{\sqrt{3}}{2}.cos73+\dfrac{1}{2}.sin73\right)^2-\dfrac{1}{2}.cos73^2+\dfrac{\sqrt{3}}{2}cos73.sin73\)

\(=\dfrac{1}{2}cos^273+1-\left(\dfrac{3}{4}cos^273+\dfrac{\sqrt{3}}{2}.cos73.sin73+\dfrac{1}{4}sin^273\right)+\dfrac{\sqrt{3}}{2}.cos73.sin73\)

\(=1-\dfrac{1}{4}.cos^273-\dfrac{1}{4}.sin^273\)

\(=1-\dfrac{1}{4}=\dfrac{3}{4}\)

Đề có sai không bạn nhỉ?

Chọn D.

Ta có : sin2a = 2.sina. cosa và sin²a = 1 - cos²a.

Do đó;

Botay.com.vn

\(\cos^21^o+\cos^289^o=\cos^21^o+\cos^2\left(90^o-1^o\right)=\cos^21^o+\sin^21^o=1\)

\(\cos^22^o+\cos^288^o=\cos^22^o+\cos^2\left(90^o-2^o\right)=\cos^22^o+\sin^22^o=1\)

.......

\(\cos^244^o+\cos^246^o=\cos^244^o+\cos^2\left(90^o-44^o\right)=\cos^244^o+\sin^244^o=1\)

\(\cos^245^o=\left(\frac{\sqrt{2}}{2}\right)^2=\frac{1}{2}\)

=> \(A=1.44+\frac{1}{2}-\frac{1}{2}=44\)

Ta có : \(cos^215^o=sin^275^o;cos^225^o=sin^265^o;cos^235^o=sin^255^o;\frac{cos^245^o}{2}=\frac{sin^245^o}{2}\)

Khi đó \(N=sin^275^o+cos^275^o-\left(sin^265^o+cos^265^o\right)+sin^255^o+cos^255^o-\left(\frac{sin^245^0+cos^245^o}{2}\right)\)

Áp dụng công thức \(sin^2a+cos^2a=1\)ta được 

\(N=1-1+1-\frac{1}{2}=\frac{1}{2}\)

Vậy N = 1/2 

câu b chờ chút mình làm cho nhé <33

Ta có : \(cos^21^o=sin^289^o;cos^22^o=sin^288^o;...;cos^244^o=sin^246^o;\frac{cos^245^o}{2}=\frac{sin^245^o}{2}\)

Khi đó \(A=\frac{sin^245^o+cos^245^o}{2}+\left(sin^246^0+cos^246^o\right)+...+\left(sin^289^o+cos^289^o\right)\)

Áp dụng ct \(sin^2a+cos^2a=1\)ta được \(A=\frac{1}{2}+1+1+...+1=...\)

P/S : bạn tự đếm xem bao nhiêu cặp nhé ;) tìm ssh á