\(A=cos\frac{\pi}{7}cos\frac{3\pi}{7}cos\frac{5\pi}{7}=cos\frac{\pi}{7}cos\frac{4\pi}{7}cos\frac{2\pi}{7}\)

\(\Rightarrow A.sin\frac{\pi}{7}=sin\frac{\pi}{7}.cos\frac{\pi}{7}.cos\frac{2\pi}{7}cos\frac{4\pi}{7}\)

\(=\frac{1}{2}sin\frac{2\pi}{7}cos\frac{2\pi}{7}cos\frac{4\pi}{7}=\frac{1}{4}sin\frac{4\pi}{7}cos\frac{4\pi}{7}\)

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

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

\(B=sin6.cos48.cos24.cos12\)

\(B.cos6=sin6.cos6.cos12.cos24.cos48\)

\(=\frac{1}{2}sin12.cos12.cos24.cos48=\frac{1}{4}sin24.cos24.cos48\)

\(=\frac{1}{8}sin48.cos48=\frac{1}{16}sin96\)

\(=\frac{1}{16}sin\left(90+6\right)=\frac{1}{16}cos6\Rightarrow B=\frac{1}{16}\)

- Xét \(sin\frac{x}{5}=0\Rightarrow C=...\)

- Với \(sin\frac{x}{5}\ne0\)

\(C.sin\frac{x}{5}=sin\frac{x}{5}.cos\frac{x}{5}.cos\frac{2x}{5}cos\frac{4x}{5}cos\frac{8x}{5}\)

\(=\frac{1}{2}sin\frac{2x}{5}cos\frac{2x}{5}cos\frac{4x}{5}cos\frac{8x}{5}\)

\(=\frac{1}{4}sin\frac{4x}{5}cos\frac{4x}{5}cos\frac{8x}{5}=\frac{1}{8}sin\frac{8x}{5}cos\frac{8x}{5}\)

\(=\frac{1}{16}sin\frac{16x}{5}\Rightarrow C=\frac{sin\frac{16x}{5}}{16.sin\frac{x}{5}}\)

\(D=sin\frac{x}{7}+sin\frac{5x}{7}+2sin\frac{3x}{7}\)

\(=2sin\frac{3x}{7}cos\frac{2x}{7}+2sin\frac{3x}{7}\)

\(=2sin\frac{3x}{7}\left(cos\frac{2x}{7}+1\right)=4cos^2\frac{x}{7}.sin\frac{3x}{7}\)

ADCT: sin2a=2sina.cosa

cos2a=2cos²a-1 (a ở đây có thể là: x, 2x,3x, pi/2-x,......)

a)

pt<=>4sin2x.cos2x=cos2.(\(\dfrac{\Pi}{4}\)-4x)

<=>2sin4x=2cos²(\(\dfrac{\Pi}{4}\)-4x)-1

<=>2sin4x=2.(\(\dfrac{\sqrt{2}}{2}\))².(cos4x+sin4x)²-1

<=>2sin4x=(cos²4x+sin²4x)+2sin4x.cos4x-1

<=>2sin4x=1+2sin4x.cos4x-1

<=>2sin4x(1-cos4x)=0

Tới đây đơn giản rồi bạn tự giải đi!

b)

Pt<=>(sinx.cos\(\dfrac{\Pi}{2}\)+cosx.sin\(\dfrac{\Pi}{2}\))⁴-sin⁴x=sin4x

<=>cos⁴x-sin⁴x=sin4x

<=>(cos²x-sin²x)(cos²x+sin²x)-sin4x=0

cos²x+sin²x=1, cos²x-sin²x=cos2x

<=>cos2x-2sin2x.cos2x=0

<=>cos2x(1-2sin2x)=0

Tự giải dc rồi chứ????

a) \(A = \frac{{\sin \frac{\pi }{{15}}\cos \frac{\pi }{{10}} + \sin \frac{\pi }{{10}}\cos \frac{\pi }{{15}}}}{{\cos \frac{{2\pi }}{{15}}\cos \frac{\pi }{5} - \sin \frac{{2\pi }}{{15}}\sin \frac{\pi }{5}}} = \frac{{\sin \left( {\frac{\pi }{{15}} + \frac{\pi }{{10}}} \right)}}{{\cos \left( {\frac{{2\pi }}{{15}} + \frac{\pi }{5}} \right)}} = \frac{{\sin \frac{\pi }{6}}}{{\cos \frac{\pi }{3}}} = 1\)

b) \(B = \sin \frac{\pi }{{32}}\cos \frac{\pi }{{32}}\cos \frac{\pi }{{16}}\cos \frac{\pi }{8} = \frac{1}{2}\sin \frac{\pi }{{16}}.\cos \frac{\pi }{{16}}.\cos \frac{\pi }{8} = \frac{1}{4}\sin \frac{\pi }{8}.\cos \frac{\pi }{8} = \frac{1}{8}\sin \frac{\pi }{4} = \frac{1}{8}.\frac{{\sqrt 2 }}{2} = \frac{{\sqrt 2 }}{{16}}\;.\)

\(\Leftrightarrow4cosx\left(cos2x+cos\frac{4\pi}{3}\right)\left(cos6x+cos\frac{4\pi}{3}\right)=sin9x\)

\(\Leftrightarrow cosx\left(2cos2x-1\right)\left(2cos6x-1\right)=sin9x\)

\(\Leftrightarrow\left(2cos2x.cosx-cosx\right)\left(2cos6x-1\right)=sin9x\)

\(\Leftrightarrow\left(cos3x+cosx-cosx\right)\left(2cos6x-1\right)=sin9x\)

\(\Leftrightarrow cos3x\left(2cos6x-1\right)=sin9x\)

\(\Leftrightarrow2cos6x.cos3x-cos3x=sin9x\)

\(\Leftrightarrow cos9x+cos3x-cos3x=sin9x\)

\(\Leftrightarrow cos9x=sin9x\)

\(C=cos\frac{\pi}{9}+cos\frac{8\pi}{9}+cos\frac{2\pi}{9}+cos\frac{7\pi}{9}+...+cos\frac{4\pi}{9}+cos\frac{5\pi}{9}+cos\pi\)

\(C=cos\frac{\pi}{9}+cos\left(\pi-\frac{\pi}{9}\right)+cos\frac{2\pi}{9}+cos\left(\pi-\frac{2\pi}{9}\right)+...+cos\frac{4\pi}{9}+cos\left(\pi-\frac{4\pi}{9}\right)-1\)

\(C=cos\frac{\pi}{9}-cos\frac{\pi}{9}+cos\frac{2\pi}{9}-cos\frac{2\pi}{9}+...+cos\frac{4\pi}{9}-cos\frac{4\pi}{9}-1\)

\(C=-1\)