Giải bài 1.55 trang 28 sách bài tập toán 11 - Kết nối tri thức với cuộc sống

Đề bài

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 2alpha + sin alpha }}{{1 + cos 2alpha + cos alpha }});

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

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

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