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\(cos\frac{\pi}{4}=2cos^2\frac{\pi}{8}-1\Rightarrow cos^2\frac{\pi}{8}=\frac{cos\frac{\pi}{4}+1}{2}\)
\(\Rightarrow cos^2\frac{\pi}{8}=\frac{2+\sqrt{2}}{4}\Rightarrow cos\frac{\pi}{8}=\frac{\sqrt{2+\sqrt{2}}}{2}\) (do \(0< \frac{\pi}{8}< \frac{\pi}{2}\) nên \(cos\frac{\pi}{8}>0\))
\(M=cos\frac{\pi}{7}-cos\frac{2\pi}{7}+cos\frac{3\pi}{7}\)
\(\Rightarrow2M.sin\frac{\pi}{7}=2sin\frac{\pi}{7}cos\frac{\pi}{7}-2sin\frac{\pi}{7}cos\frac{2\pi}{7}+2sin\frac{\pi}{7}cos\frac{3\pi}{7}\)
\(=sin\frac{2\pi}{7}-sin\frac{3\pi}{7}+sin\frac{\pi}{7}+sin\frac{4\pi}{7}-sin\frac{2\pi}{7}\)
\(=-sin\frac{3\pi}{7}+sin\frac{\pi}{7}+sin\left(\pi-\frac{3\pi}{7}\right)\)
\(=-sin\frac{3\pi}{7}+sin\frac{\pi}{7}+sin\frac{3\pi}{7}=sin\frac{\pi}{7}\)
\(\Rightarrow M=\frac{sin\frac{\pi}{7}}{2sin\frac{\pi}{7}}=\frac{1}{2}\)
\(sin\left(\frac{\pi}{7}\right)H=sin\left(\frac{\pi}{7}\right)cos\left(\frac{2\pi}{7}\right)+sin\left(\frac{\pi}{7}\right)cos\left(\frac{4\pi}{7}\right)+sin\left(\frac{\pi}{7}\right)cos\left(\frac{6\pi}{7}\right)\)
\(=\frac{1}{2}\left[sin\left(\frac{3\pi}{7}\right)-sin\left(\frac{\pi}{7}\right)+sin\left(\frac{5\pi}{7}\right)-sin\left(\frac{3\pi}{7}\right)+sin\pi-sin\left(\frac{5\pi}{7}\right)\right]\)
\(=-\frac{1}{2}sin\left(\frac{\pi}{7}\right)\)
\(\Rightarrow H=-\frac{1}{2}\)
\(sinA+sinB+sinC=2sin\left(\frac{A+B}{2}\right)cos\left(\frac{A-B}{2}\right)+2sin\left(\frac{C}{2}\right)cos\left(\frac{C}{2}\right)\)
\(=2cos\frac{C}{2}cos\left(\frac{A-B}{2}\right)+2cos\left(\frac{A+B}{2}\right)cos\frac{C}{2}\)
\(=2cos\frac{C}{2}\left[cos\left(\frac{A-B}{2}\right)+cos\left(\frac{A+B}{2}\right)\right]\)
\(=4cos\frac{C}{2}cos\frac{A}{2}cos\frac{B}{2}\)
\(cos\left(2\pi+\frac{\pi}{16}\right).sin\frac{5\pi}{16}.cos\frac{5\pi}{16}.cos\left(\frac{\pi}{2}-\frac{\pi}{16}\right)\)
\(=\frac{1}{4}.2cos\frac{\pi}{16}.sin\frac{\pi}{16}.2sin\frac{5\pi}{16}.cos\frac{5\pi}{16}\)
\(=\frac{1}{4}sin\frac{2\pi}{16}.sin\frac{10\pi}{16}=\frac{1}{4}sin\frac{\pi}{8}.sin\frac{5\pi}{8}\)
\(=\frac{1}{4}sin\frac{\pi}{8}.sin\left(\frac{\pi}{2}+\frac{\pi}{8}\right)\)
\(=\frac{1}{4}sin\frac{\pi}{8}.cos\frac{\pi}{8}=\frac{1}{8}sin\frac{2\pi}{8}\)
\(=\frac{1}{8}sin\frac{\pi}{4}=\frac{\sqrt{2}}{16}\)
Đề sai hoặc bạn gõ thiếu số 1 ở dưới mẫu
- Xét \(sin\frac{x}{5}=0\Rightarrow C=...\)
- Với \(sin\frac{x}{5}\ne0\)
\(C.sin\frac{x}{5}=sin\frac{x}{5}.cos\frac{x}{5}.cos\frac{2x}{5}cos\frac{4x}{5}cos\frac{8x}{5}\)
\(=\frac{1}{2}sin\frac{2x}{5}cos\frac{2x}{5}cos\frac{4x}{5}cos\frac{8x}{5}\)
\(=\frac{1}{4}sin\frac{4x}{5}cos\frac{4x}{5}cos\frac{8x}{5}=\frac{1}{8}sin\frac{8x}{5}cos\frac{8x}{5}\)
\(=\frac{1}{16}sin\frac{16x}{5}\Rightarrow C=\frac{sin\frac{16x}{5}}{16.sin\frac{x}{5}}\)
\(D=sin\frac{x}{7}+sin\frac{5x}{7}+2sin\frac{3x}{7}\)
\(=2sin\frac{3x}{7}cos\frac{2x}{7}+2sin\frac{3x}{7}\)
\(=2sin\frac{3x}{7}\left(cos\frac{2x}{7}+1\right)=4cos^2\frac{x}{7}.sin\frac{3x}{7}\)
\(A=cos\frac{\pi}{7}cos\frac{3\pi}{7}cos\frac{5\pi}{7}=cos\frac{\pi}{7}cos\frac{4\pi}{7}cos\frac{2\pi}{7}\)
\(\Rightarrow A.sin\frac{\pi}{7}=sin\frac{\pi}{7}.cos\frac{\pi}{7}.cos\frac{2\pi}{7}cos\frac{4\pi}{7}\)
\(=\frac{1}{2}sin\frac{2\pi}{7}cos\frac{2\pi}{7}cos\frac{4\pi}{7}=\frac{1}{4}sin\frac{4\pi}{7}cos\frac{4\pi}{7}\)
\(=\frac{1}{8}sin\frac{8\pi}{7}=\frac{1}{8}sin\left(\pi+\frac{\pi}{7}\right)=-\frac{1}{8}sin\frac{\pi}{7}\)
\(\Rightarrow A=-\frac{1}{8}\)
\(B=sin6.cos48.cos24.cos12\)
\(B.cos6=sin6.cos6.cos12.cos24.cos48\)
\(=\frac{1}{2}sin12.cos12.cos24.cos48=\frac{1}{4}sin24.cos24.cos48\)
\(=\frac{1}{8}sin48.cos48=\frac{1}{16}sin96\)
\(=\frac{1}{16}sin\left(90+6\right)=\frac{1}{16}cos6\Rightarrow B=\frac{1}{16}\)
\(sina.sin\left(\frac{\pi}{3}-a\right)sin\left(\frac{\pi}{3}+a\right)\)
\(=-\frac{1}{2}sina\left[cos\frac{2\pi}{3}-cos2a\right]=-\frac{1}{2}sina\left(-\frac{1}{2}-cos2a\right)\)
\(=\frac{1}{4}sina+\frac{1}{2}sina.cos2a=\frac{1}{4}sina+\frac{1}{4}sin3a-\frac{1}{4}sina\)
\(=\frac{1}{4}sin3a\)
\(sin\frac{\pi}{9}sin\frac{2\pi}{9}sin\frac{4\pi}{9}=sin\frac{\pi}{9}sin\left(\frac{\pi}{3}-\frac{\pi}{9}\right)sin\left(\frac{\pi}{3}+\frac{\pi}{9}\right)=\frac{1}{4}sin\frac{\pi}{3}=\frac{\sqrt{3}}{8}\)
\(cosa.cos\left(\frac{\pi}{3}-a\right)cos\left(\frac{\pi}{3}+a\right)=\frac{1}{2}cosa\left(cos\frac{2\pi}{3}+cos2a\right)\)
\(=\frac{1}{2}cosa\left(cos2a-\frac{1}{2}\right)=\frac{1}{2}cosa.cos2a-\frac{1}{4}cosa\)
\(=\frac{1}{4}cos3a+\frac{1}{4}cosa-\frac{1}{4}cosa=\frac{1}{4}cos3a\)
\(cos\frac{\pi}{18}cos\frac{5\pi}{18}cos\frac{7\pi}{18}=cos\frac{\pi}{18}.cos\left(\frac{\pi}{3}-\frac{\pi}{18}\right).cos\left(\frac{\pi}{3}+\frac{\pi}{18}\right)=\frac{1}{4}cos\frac{\pi}{6}=\frac{\sqrt{3}}{8}\)
\(A=cos\left(\pi+\frac{\pi}{2}-a\right)-sin\left(\pi+\frac{\pi}{2}-a\right)+cos\left(a+\frac{\pi}{2}-4\pi\right)-sin\left(a+\frac{\pi}{2}-4\pi\right)\)
\(=-cos\left(\frac{\pi}{2}-a\right)+sin\left(\frac{\pi}{2}-a\right)+cos\left(a+\frac{\pi}{2}\right)-sin\left(a+\frac{\pi}{2}\right)\)
\(=-sina+cosa-sina-cosa=-2sina\)
\(A=cos\left(32^0+28^0\right)=cos60^0=\frac{1}{2}\)
\(B=cos\left(220^0+170^0\right)=cos390^0=cos\left(30^0+360^0\right)=cos30^0=\frac{\sqrt{3}}{2}\)
\(C=sin\left(\frac{7\pi}{18}-\frac{5\pi}{9}\right)=sin\left(-\frac{\pi}{6}\right)=-sin\left(\frac{\pi}{6}\right)=-\frac{1}{2}\)