Xét tính đúng, sai của các mệnh đề sau:

Đáp án: a) Đúng. b) Sai. c) Sai. d) Đúng.

a) Đúng. sin2x = 2sinx∙cosx.

b) Sai. \(\cos a - \cos b = - 2\sin \frac{{a + b}}{2} \cdot \sin \frac{{a - b}}{2}\).

c) Sai. sina + sin2a + sin3a + sin4a + sin5a + sin6a

= (sina + sin6a) + (sin2a + sin5a) + (sin3a + sin4a)

\( = 2\sin \frac{{7a}}{2} \cdot \cos \frac{{5a}}{2} + 2\sin \frac{{7a}}{2} \cdot \cos \frac{{3a}}{2} + 2\sin \frac{{7a}}{2} \cdot \cos \frac{a}{2}\)

\( = 2\left( {\cos \frac{{5a}}{2} + \cos \frac{{3a}}{2} + \cos \frac{a}{2}} \right) \cdot \sin \frac{{7a}}{2} = 2\left( {2\cos \frac{{3a}}{2} \cdot \cos a + \cos \frac{{3a}}{2}} \right) \cdot \sin \frac{{7a}}{2}\)

\( = 2\left( {2\cos a + 1} \right) \cdot \cos \frac{{3a}}{2} \cdot \sin \frac{{7a}}{2} = 4\left( {\cos a + \frac{1}{2}} \right) \cdot \cos \frac{{3a}}{2} \cdot \sin \frac{{7a}}{2}\)

\( = 4\left( {\cos a + \cos \frac{\pi }{3}} \right) \cdot \cos \frac{{3a}}{2} \cdot \sin \frac{{7a}}{2}\)

\( = 8\cos \left( {\frac{a}{2} + \frac{\pi }{6}} \right) \cdot \cos \left( {\frac{a}{2} - \frac{\pi }{6}} \right) \cdot \cos \frac{{3a}}{2} \cdot \sin \frac{{7a}}{2}\).

d) Đúng.

Vế trái:

\(\frac{{\sin A + \sin B}}{{\cos A + \cos B}} = \frac{{2\sin \frac{{A + B}}{2} \cdot \cos \frac{{A - B}}{2}}}{{2\cos \frac{{A + B}}{2} \cdot \cos \frac{{A - B}}{2}}} = \frac{{2\sin \frac{{A + B}}{2}}}{{2\cos \frac{{A + B}}{2}}} = \tan \frac{{A + B}}{2} = \frac{{\sin \frac{{A + B}}{2}}}{{\cos \frac{{A + B}}{2}}}\).

Vế phải:

\(\frac{1}{2}\left( {\tan A + \tan B} \right) = \frac{1}{2} \cdot \left( {\frac{{\sin A}}{{\cos A}} + \frac{{\sin B}}{{\cos B}}} \right) = \frac{1}{2}\left( {\frac{{\sin A\cos B + \cos A\sin B}}{{\cos A\cos B}}} \right)\)

\( = \frac{1}{2}\frac{{\sin \left( {A + B} \right)}}{{\cos A\cos B}} = \frac{{\sin \frac{{A + B}}{2}\cos \frac{{A + B}}{2}}}{{\cos A\cos B}}\).

Khi đó \(\frac{{\sin \frac{{A + B}}{2}}}{{\cos \frac{{A + B}}{2}}} = \frac{{\sin \frac{{A + B}}{2}\cos \frac{{A + B}}{2}}}{{\cos A\cos B}}\)\( \Leftrightarrow \cos A\cos B = {\cos ^2}\frac{{A + B}}{2}\)

\( \Leftrightarrow \frac{1}{2}\left[ {\cos \left( {A - B} \right) + \cos \left( {A + B} \right)} \right] = \frac{{1 + \cos \left( {A + B} \right)}}{2}\)\( \Leftrightarrow \cos \left( {A - B} \right) = 1 \Leftrightarrow A - B = 0 \Leftrightarrow A = B\) hay \(\widehat A = \widehat B\) ⇔ tam giác ABC cân tại C.