Bài 26 trang 70 SBT Toán 11 - Kết nối tri thức: Chứng minh rằng: ${\rm{sin}}3x = 4{\rm{sin}}x{\rm{sin}}\left( {{{60}^ \circ } - x} \right){\rm{sin}}\left( {{{60}^ \circ } + x} \right)$;...

Chứng minh rằng:

a) ${\rm{sin}}3x = 4{\rm{sin}}x{\rm{sin}}\left( {{{60}^ \circ } - x} \right){\rm{sin}}\left( {{{60}^ \circ } + x} \right)$;

b) $\frac{{{\rm{sin}}x - {\rm{sin}}2x + {\rm{sin}}3x}}{{{\rm{cos}}x - {\rm{cos}}2x + {\rm{cos}}3x}} = {\rm{tan}}2x$.

Áp dụng các công thức lượng giác

$\sin a\sin b = \frac{1}{2}\left[ {\cos \left( {a - b} \right) - \cos \left( {a + b} \right)} \right]$

$\sin a\cos b = \frac{1}{2}\left[ {\sin \left( {a - b} \right) + \sin \left( {a + b} \right)} \right]$

$\cos u + \cos v = 2\cos \frac{{u + v}}{2}\cos \frac{{u - v}}{2}$

$\sin u + \sin v = 2\sin \frac{{u + v}}{2}\cos \frac{{u - v}}{2}$

a) Vế phải $= 4{\rm{sin}}x{\rm{sin}}\left( {{{60}^ \circ } - x} \right){\rm{sin}}\left( {{{60}^ \circ } + x} \right) = 2{\rm{sin}}x\left[ {{\rm{cos}}2x - {\rm{cos}}{{120}^ \circ }} \right]$

$= 2{\rm{sin}}x{\rm{cos}}2x + {\rm{sin}}x = {\rm{sin}}3x + {\rm{sin}}\left( { - x} \right) + {\rm{sin}}x = {\rm{sin}}3x{\rm{.}}$$= \frac{{\sin x - \sin 2x + \sin 3x}}{{\cos x - \cos 2x + \cos 3x}} = \frac{{2\sin 2x\cos x - \sin 2x}}{{2\cos 2x\cos x - \cos 2x}}{\rm{\;}}$

b) VT$= \frac{{{\rm{sin}}2x\left( {2{\rm{cos}}x - 1} \right)}}{{{\rm{cos}}2x\left( {2{\rm{cos}}x - 1} \right)}}$$= \frac{{{\rm{sin}}2x}}{{{\rm{cos}}2x}} = {\rm{tan}}2x =$vế phải