Bài 22 trang 201 SGK Đại số 10 Nâng cao
Câu 22. Chứng minh các đẳng thức sau
a) cos 4 α –sin 4 α = 2cos 2 α - 1
b) \(1 - {\cot ^4}\alpha = {2 \over {{{\sin }^2}\alpha }} - {1 \over {{{\sin }^4}\alpha }}\,\,\,(\sin \alpha \ne 0)\)
c) \({{1 + {{\sin }^2}\alpha } \over {1 - {{\sin }^2}\alpha }} = 1 + 2{\tan ^2}\alpha \,\,\,(\sin \alpha \ne \pm 1)\)
Đáp án
a) Ta có:
cos 4 α –sin 4 α = (cos 2 α + sin 2 α)(cos 2 α – sin 2 α)
= cos 2 α – sin 2 α = cos 2 α – (1 – cos 2 α) = 2cos 2 α – 1
b) Ta có:
\(\eqalign{
& 1 - {\cot ^4}\alpha \cr
& = {1 \over {{{\sin }^2}\alpha }}(1 - {{{{\cos }^2}\alpha } \over {{{\sin }^2}\alpha }}) \cr&= {1 \over {{{\sin }^2}\alpha }}{\rm{[}}{{{{\sin }^2}\alpha - (1 - {{\sin }^2}\alpha )} \over {{{\sin }^2}\alpha }}{\rm{]}} \cr
& = {{2{{\sin }^2}\alpha - 1} \over {{{\sin }^4}\alpha }} = {2 \over {{{\sin }^2}\alpha }} - {1 \over {{{\sin }^4}\alpha }} \cr} \)
c) Ta có:
\(\eqalign{
& {{1 + {{\sin }^2}\alpha } \over {1 - {{\sin }^2}\alpha }} = {{1 + {{\sin }^2}\alpha } \over {{{\cos }^2}\alpha }} ={1 \over {{{\cos }^2}\alpha }} + {\tan ^2}\alpha \cr
& = 1 + 2{\tan ^2}\alpha \cr} \)