Rút gọn các biểu thức sau a) sin ( 45^0  + alpha ) - cos ( 45^0  + alpha)/sin ( 45^0  + alpha + cos ( 5^0  + alpha ); b) sin 2alpha  + sin alpha/1 + cos 2alpha  + cos alpha; c) 1 + cos al

Lời giải

a) \(\frac{{\sin \left( {45^\circ + \alpha } \right) - \cos \left( {45^\circ + \alpha } \right)}}{{\sin \left( {45^\circ + \alpha } \right) + \cos \left( {45^\circ + \alpha } \right)}}\)

\[ = \frac{{\left( {\sin 45^\circ \cos \alpha + \cos 45^\circ \sin \alpha } \right) - \left( {\cos 45^\circ \cos \alpha - \sin 45^\circ \sin \alpha } \right)}}{{\left( {\sin 45^\circ \cos \alpha + \cos 45^\circ \sin \alpha } \right) + \left( {\cos 45^\circ \cos \alpha - \sin 45^\circ \sin \alpha } \right)}}\]

\[ = \frac{{\left( {\frac{{\sqrt 2 }}{2}\cos \alpha + \frac{{\sqrt 2 }}{2}\sin \alpha } \right) - \left( {\frac{{\sqrt 2 }}{2}\cos \alpha - \frac{{\sqrt 2 }}{2}\sin \alpha } \right)}}{{\left( {\frac{{\sqrt 2 }}{2}\cos \alpha + \frac{{\sqrt 2 }}{2}\sin \alpha } \right) + \left( {\frac{{\sqrt 2 }}{2}\cos \alpha - \frac{{\sqrt 2 }}{2}\sin \alpha } \right)}}\]

\( = \frac{{\sqrt 2 \sin \alpha }}{{\sqrt 2 \cos \alpha }} = \tan \alpha \).

b) \(\frac{{\sin 2\alpha + \sin \alpha }}{{1 + \cos 2\alpha + \cos \alpha }}\)

\( = \frac{{2\sin \alpha \cos \alpha + \sin \alpha }}{{1 + \left( {2{{\cos }^2}\alpha - 1} \right) + \cos \alpha }}\)

\( = \frac{{2\sin \alpha \left( {\cos \alpha + 1} \right)}}{{2\cos \alpha \left( {\cos \alpha + 1} \right)}}\)

\( = \frac{{\sin \alpha }}{{\cos \alpha }} = \tan \alpha \).

c) \(\frac{{1 + \cos \alpha - \sin \alpha }}{{1 - \cos \alpha - \sin \alpha }}\)

\( = \frac{{1 + \left( {2{{\cos }^2}\frac{\alpha }{2} - 1} \right) - 2\sin \frac{\alpha }{2}\cos \frac{\alpha }{2}}}{{1 - \left( {1 - 2{{\sin }^2}\frac{\alpha }{2}} \right) - 2\sin \frac{\alpha }{2}\cos \frac{\alpha }{2}}}\)

\( = \frac{{2{{\cos }^2}\frac{\alpha }{2} - 2\sin \frac{\alpha }{2}\cos \frac{\alpha }{2}}}{{2{{\sin }^2}\frac{\alpha }{2} - 2\sin \frac{\alpha }{2}\cos \frac{\alpha }{2}}}\)

\( = \frac{{2\cos \frac{\alpha }{2}\left( {\cos \frac{\alpha }{2} - \sin \frac{\alpha }{2}} \right)}}{{2\sin \frac{\alpha }{2}\left( {\sin \frac{\alpha }{2} - \cos \frac{\alpha }{2}} \right)}}\)

\( = \frac{{\cos \frac{\alpha }{2}\left( {\cos \frac{\alpha }{2} - \sin \frac{\alpha }{2}} \right)}}{{\sin \frac{\alpha }{2}\left[ { - \left( {\cos \frac{\alpha }{2} - \sin \frac{\alpha }{2}} \right)} \right]}}\)

\( = - \frac{{\cos \frac{\alpha }{2}}}{{\sin \frac{\alpha }{2}}} = - \cot \frac{\alpha }{2}\).

d) \(\frac{{\sin \alpha + \sin 3\alpha + \sin 5\alpha }}{{\cos \alpha + \cos 3\alpha + \cos 5\alpha }}\)

\[ = \frac{{\left( {\sin 5\alpha + \sin \alpha } \right) + \sin 3\alpha }}{{\left( {\cos 5\alpha + \cos \alpha } \right) + \cos 3\alpha }}\]

\( = \frac{{2\sin \frac{{5\alpha + \alpha }}{2}\cos \frac{{5\alpha - \alpha }}{2} + \sin 3\alpha }}{{2\cos \frac{{5\alpha + \alpha }}{2}\cos \frac{{5\alpha - \alpha }}{2} + \cos 3\alpha }}\)

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

\( = \frac{{\sin 3\alpha \left( {2\cos 2\alpha + 1} \right)}}{{\cos 3\alpha \left( {2\cos 2\alpha + 1} \right)}}\)

\( = \frac{{\sin 3\alpha }}{{\cos 3\alpha }} = \tan 3\alpha \).