Hãy đơn giản các biểu thức

Giải SBT Toán 9 Bài 6: Ôn tập chương 1

Bài 81 trang 119 SBT Toán lớp 9 Tập 1: Hãy đơn giản các biểu thức

a) $1-{\sin }^2\alpha$

b) $(1-cos\alpha )(1+cos\alpha )$

c) $1+{\sin }^2\alpha +co{s}^2\alpha$

d) $\sin \alpha -\sin \alpha co{s}^2\alpha$

e) ${\sin }^4\alpha +co{s}^4\alpha +2{\sin }^2\alpha co{s}^2\alpha$

g) ${\tan }^2\alpha -{\sin }^2\alpha .{\tan }^2\alpha$

h) $co{s}^2\alpha +{\tan }^2\alpha .co{s}^2\alpha$

i) ${\tan }^2\alpha .(2co{s}^2\alpha +{\sin }^2\alpha -1)$

Lời giải:

a) $1-{\sin }^2\alpha =({\sin }^2\alpha +co{s}^2\alpha )-{\sin }^2\alpha$

$={\sin }^2\alpha +co{s}^2\alpha -{\sin }^2\alpha =co{s}^2\alpha$

b) $(1-cos\alpha )(1+cos\alpha )=1-co{s}^2\alpha$

$=({\sin }^2\alpha +co{s}^2\alpha )-co{s}^2\alpha ={\sin }^2\alpha +co{s}^2\alpha -co{s}^2\alpha ={\sin }^2\alpha$

c) $1+{\sin }^2\alpha +co{s}^2\alpha$

$=1+\left({\sin }^2\alpha +co{s}^2\alpha \right)=1+1=2$

d) $\sin \alpha -\sin \alpha co{s}^2\alpha =\sin \alpha (1-co{s}^2\alpha )$

$=\sin \alpha ({\sin }^2\alpha +co{s}^2\alpha -co{s}^2\alpha )={\sin }^3\alpha$

e) ${\sin }^4\alpha +co{s}^4\alpha +2{\sin }^2\alpha co{s}^2\alpha$

$={\left({\sin }^2\alpha \right)}^2+2{\sin }^2\alpha co{s}^2\alpha +{\left(co{s}^2\alpha \right)}^2={\left({\sin }^2\alpha +co{s}^2\alpha \right)}^2=1^2=1$

g) ${\tan }^2\alpha -{\sin }^2\alpha .{\tan }^2\alpha ={\tan }^2\alpha (1-{\sin }^2\alpha )$

$={\tan }^2\alpha ({\sin }^2\alpha +co{\mathrm{s}}^2\alpha -{\sin }^2\alpha )={\tan }^2\alpha .co{s}^2\alpha =\frac{{\sin }^2\alpha }{co{s}^2\alpha }.co{s}^2\alpha ={\sin }^2\alpha$

h) $co{s}^2\alpha +{\tan }^2\alpha .co{s}^2\alpha$

$=co{s}^2\alpha +\frac{{\sin }^2\alpha }{co{s}^2\alpha }.co{s}^2\alpha =co{s}^2\alpha +{\sin }^2\alpha =1$

i) ${\tan }^2\alpha .(2co{s}^2\alpha +{\sin }^2\alpha -1)$

$={\tan }^2\alpha .(co{s}^2\alpha +co{s}^2\alpha +{\sin }^2\alpha -1)={\tan }^2\alpha .(co{s}^2\alpha +1-1)={\tan }^2\alpha .co{s}^2\alpha =\frac{{\sin }^2\alpha }{co{s}^2\alpha }.co{s}^2\alpha ={\sin }^2\alpha$