Bài 81 trang 119 SBT toán 9 tập 1

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

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

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

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

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

i)

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

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

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