a.
[tex]sin\frac{2\pi}{5}=sin(\pi-\frac{3\pi}{5})\\\rightarrow sin2\frac{\pi}{5}=sin3\frac{\pi}{5}\\\rightarrow 2sin\frac{\pi}{5}cos\frac{\pi}{5}=3sin\frac{\pi}{5}-4sin^3\frac{\pi}{5}\\\rightarrow 2cos\frac{\pi}{5}=3-4sin^2\frac{\pi}{5}\\\rightarrow 2cos\frac{\pi}{5}=-1+4cos^2\frac{\pi}{5}\\\rightarrow 4cos^2\frac{\pi}{5}-2cos\frac{\pi}{5}-1=0[/tex]
Giải PT tìm [tex]cos\frac{\pi}{5}[/tex] thế vào:
[tex]cos\frac{\pi}{5}-cos2\frac{\pi}{5}=cos\frac{\pi}{5}-2cos^2\frac{\pi}{5}+1[/tex]
(Lưu ý [tex]0<\frac{\pi}{5}<\frac{\pi}{2}[/tex])
là ra
b. Quy đồng
[tex]\frac{cos15+sin15}{cos15-sin15}=\frac{cos^215-sin^215}{(cos15-sin15)^2}=\frac{cos30}{1-2sin15cos15}=\frac{cos30}{1-sin30}=...[/tex]