Đề bài - bài 81 trang 119 sbt toán 9 tập 1

\(\eqalign{& h)\,{\cos ^2}\alpha + ta{n^2}\alpha .c{\rm{o}}{{\rm{s}}^2}\alpha \cr& = c{\rm{o}}{{\rm{s}}^2}\alpha + {{{{\sin }^2}\alpha } \over {c{\rm{o}}{{\rm{s}}^2}\alpha }}.c{\rm{o}}{{\rm{s}}^2}\alpha \cr& = c{\rm{o}}{{\rm{s}}^2}\alpha + {\sin ^2}\alpha = 1 \cr} \)

Đề bài

Hãy đơn giản các biểu thức:

a) \(1 - {\sin ^2}\alpha \);

b) \((1 - \cos \alpha )(1 + \cos \alpha )\);

c) \(1 + {\sin ^2}\alpha + {\cos ^2}\alpha \);

d) \(\sin \alpha - \sin \alpha .{\cos ^2}\alpha \);

e) \({\sin ^4}\alpha + {\cos ^4}\alpha + 2.{\sin ^2}\alpha .{\cos ^2}\alpha \);

g) \(ta{n^2}\alpha - {\sin ^2}\alpha .ta{n^2}\alpha \);

h) \({\cos ^2}\alpha + ta{n^2}\alpha .c{\rm{o}}{{\rm{s}}^2}\alpha \);

i) \(ta{n^2}\alpha (2.{\cos ^2}\alpha + {\sin ^2}\alpha - 1).\)

Phương pháp giải - Xem chi tiết

Áp dụng các kiến thức:

1) \({\sin ^2}\alpha + {\cos ^2}\alpha =1\)

2) \(ta{n^2}\alpha =\displaystyle {{{{\sin }^2}\alpha } \over {{{\cos }^2}\alpha }}\)

Lời giải chi tiết

a) \(1 - {\sin ^2}\alpha = ({\sin ^2}\alpha + {\cos ^2}\alpha ) - {\sin ^2}\alpha \)

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

b)

\(\eqalign{
&(1 - \cos \alpha )(1 + \cos \alpha ) = 1 - {\cos ^2}\alpha \cr
& = ({\sin ^2}\alpha + {\cos ^2}\alpha ) - {\cos ^2}\alpha \cr} \)

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

c)

\(\eqalign{
& 1 + {\sin ^2}\alpha + {\cos ^2}\alpha \cr
& = 1 + ({\sin ^2}\alpha + {\cos ^2}\alpha ) = 1 + 1 = 2 \cr} \)

d) \(\sin \alpha - \sin \alpha .{\cos ^2}\alpha\)\(= \sin \alpha (1 - {\cos ^2}\alpha )\)

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

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

\( = \sin \alpha .{\sin ^2}\alpha = {\sin ^3}\alpha \)

\(\eqalign{
& e)\,{\sin ^4}\alpha + {\cos ^4}\alpha + 2.{\sin ^2}\alpha .{\cos ^2}\alpha \cr
& = {({\sin ^2}\alpha + {\cos ^2}\alpha )^2} = {1^2} = 1 \cr} \)

g) \(ta{n^2}\alpha - {\sin ^2}\alpha .ta{n^2}\alpha \)\(= ta{n^2}\alpha (1 - {\sin ^2}\alpha )\)

\( = ta{n^2}\alpha.\left[ {\left( {{{\sin }^2}\alpha + {{\cos }^2}\alpha } \right) - {{\sin }^2}\alpha } \right]\)

\( = ta{n^2}\alpha .{\cos ^2}\alpha = \displaystyle {{{{\sin }^2}\alpha } \over {{{\cos }^2}\alpha }}.{\cos ^2}\alpha\)\( = {\sin ^2}\alpha \)

\(\eqalign{
& h)\,{\cos ^2}\alpha + ta{n^2}\alpha .c{\rm{o}}{{\rm{s}}^2}\alpha \cr
& = c{\rm{o}}{{\rm{s}}^2}\alpha + {{{{\sin }^2}\alpha } \over {c{\rm{o}}{{\rm{s}}^2}\alpha }}.c{\rm{o}}{{\rm{s}}^2}\alpha \cr
& = c{\rm{o}}{{\rm{s}}^2}\alpha + {\sin ^2}\alpha = 1 \cr} \)

i)

\(ta{n^2}\alpha (2.{\cos ^2}\alpha + {\sin ^2}\alpha - 1) \)
\( = ta{n^2}\alpha .\)\(\left[ {{{\cos }^2}\alpha + \left( {{{\cos }^2}\alpha + {{\sin }^2}\alpha } \right) - 1} \right] \)

\( = ta{n^2}\alpha .({\cos ^2}\alpha + 1 - 1)\)\( = ta{n^2}\alpha .{\cos ^2}\alpha \)

\( = \displaystyle {{{{\sin }^2}\alpha } \over {{{\cos }^2}\alpha }}.{\cos ^2}\alpha = {\sin ^2}\alpha \)