Câu hỏi 1.55 - Mục Bài tập trang 28

1. Nội dung câu hỏi

Rút gọn các biểu thức sau

a) \(\frac{{\sin ({{45}^0} + \alpha ) - \cos ({{45}^0} + \alpha )}}{{\sin ({{45}^0} + \alpha ) + \cos ({{45}^0} + \alpha )}}\);                    

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

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

d) \(\frac{{\sin \alpha  + \sin 3\alpha  + \sin 5\alpha }}{{\cos \alpha  + \cos 2\alpha  + \cos 5\alpha }}\).

3. Lời giải chi tiết 

a) Ta có

\(\begin{array}{l}\frac{{\sin ({{45}^0} + \alpha ) - \cos ({{45}^0} + \alpha )}}{{\sin ({{45}^0} + \alpha ) + \cos ({{45}^0} + \alpha )}}\\ = \frac{{\sin {{45}^0}\cos \alpha  + \cos {{45}^0}\sin \alpha  - (\cos {{45}^0}\cos \alpha  - \sin {{45}^0}\sin \alpha )}}{{\sin {{45}^0}\cos \alpha  + \cos {{45}^0}\sin \alpha  + (\cos {{45}^0}\cos \alpha  - \sin {{45}^0}\sin \alpha )}}\\ = \frac{{\frac{{\sqrt 2 }}{2}\cos \alpha  + \frac{{\sqrt 2 }}{2}\sin \alpha  - \left( {\frac{{\sqrt 2 }}{2}\cos \alpha  - \frac{{\sqrt 2 }}{2}\sin \alpha } \right)}}{{\frac{{\sqrt 2 }}{2}\cos \alpha  + \frac{{\sqrt 2 }}{2}\sin \alpha  + \left( {\frac{{\sqrt 2 }}{2}\cos \alpha  - \frac{{\sqrt 2 }}{2}\sin \alpha } \right)}}\\ = \frac{{\sqrt 2 \sin \alpha }}{{\sqrt 2 \cos \alpha }} = \tan \alpha .\end{array}\)

b) Ta có

\(\begin{array}{l}\frac{{\sin 2\alpha  + \sin \alpha }}{{1 + \cos 2\alpha  + \cos \alpha }} = \frac{{2\sin \alpha .\cos \alpha  + \sin \alpha }}{{1 + 2{{\cos }^2}\alpha  - 1 + \cos \alpha }}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{\sin \alpha (2\cos \alpha  + 1)}}{{2{{\cos }^2}\alpha  + \cos \alpha }}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{\sin \alpha .(2\cos \alpha  + 1)}}{{\cos \alpha .(2\cos \alpha  + 1)}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{\sin \alpha }}{{\cos \alpha }} = \tan \alpha .\end{array}\)

c) Ta có

\(\begin{array}{l}\frac{{1 + \cos \alpha  - \sin \alpha }}{{1 - \cos \alpha  - \sin \alpha }}\\ = \frac{{1 + 2{{\cos }^2}\frac{\alpha }{2} - 1 - 2\sin \frac{\alpha }{2}\cos \frac{\alpha }{2}}}{{1 - \left( {1 - 2{{\sin }^2}\frac{\alpha }{2}} \right) - 2\sin \frac{\alpha }{2}\cos \frac{\alpha }{2}}}\\ = \frac{{2{{\cos }^2}\frac{\alpha }{2} - 2\sin \frac{\alpha }{2}\cos \frac{\alpha }{2}}}{{2{{\sin }^2}\frac{\alpha }{2} - 2\sin \frac{\alpha }{2}\cos \frac{\alpha }{2}}}\\ = \frac{{2\cos \frac{\alpha }{2}.\left( {\cos \frac{\alpha }{2} - \sin \frac{\alpha }{2}} \right)}}{{2\sin \frac{\alpha }{2}.\left( {\sin \frac{\alpha }{2} - \cos \frac{\alpha }{2}} \right)}}\\ =  - \frac{{\cos \frac{\alpha }{2}}}{{\sin \frac{\alpha }{2}}} =  - \cot \frac{\alpha }{2}.\end{array}\)

d) Ta có:

\(\begin{array}{l}\frac{{\sin \alpha  + \sin 3\alpha  + \sin 5\alpha }}{{\cos \alpha  + \cos 3\alpha  + \cos 5\alpha }}\\ = \frac{{\left( {\sin \alpha  + \sin 5\alpha } \right) + \sin 3\alpha }}{{\left( {\cos \alpha  + \cos 5\alpha } \right) + \cos 3\alpha }}\\ = \frac{{2\sin \frac{{\alpha  + 5\alpha }}{2}\cos \frac{{\alpha  - 5\alpha }}{2} + \sin 3\alpha }}{{2\cos \frac{{\alpha  + 5\alpha }}{2}\cos \frac{{\alpha  - 5\alpha  + \cos 3\alpha }}{2}}}\\ = \frac{{2\sin 3\alpha .\cos ( - 2\alpha ) + \sin 3\alpha }}{{2\cos 3\alpha \cos ( - 2\alpha ) + \cos 3\alpha }}\\ = \frac{{\sin 3\alpha (2\cos ( - 2\alpha ) + 1)}}{{\cos 3\alpha (2\cos ( - 2\alpha ) + 1)}} = \frac{{\sin 3\alpha }}{{\cos 3\alpha }} = \tan 3\alpha .\end{array}\)