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Chứng minh các đẳng thức sau
a) cos4α –sin4α = 2cos2α – 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:
cos4α –sin4α = (cos2α + sin2α)(cos2α – sin2α)
= cos2α – sin2α = cos2α – (1 – cos2α) = 2cos2α – 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:
\(\eqalign{
& {{1 + {{\sin }^2}\alpha } \over {1 – {{\sin }^2}\alpha }} = {{1 + {{\sin }^2}\alpha } \over {{{\cos }^2}\alpha }} ={1 \over {{{\cos }^2}\alpha }} + {\tan ^2}\alpha \cr
& = 1 + 2{\tan ^2}\alpha \cr} \)