Bài 24 trang 218 SBT Toán Đại số 10: Rút gọn...

Rút gọn

a) ({{1 + cos a} over {1 – cos a}}{ an ^2}{a over 2} – {cos ^2}a)

b) (4{cos ^4}a – 2cos 2a – {1 over 2}cos 4a)

c) ({sin ^2}aleft( {1 + {1 over {sin a}} + cot a} ight)left( {1 – {1 over {sin a}} + cot a} ight))

d) ({{cos 2a} over {{{cos }^4}a – {{sin }^4}a}} – {{{{cos }^4}a + {{sin }^4}a} over {1 – {1 over 2}{{sin }^2}2a}})

Gợi ý làm bài

a) 

(eqalign{
& {{1 + cos a} over {1 – cos a}}{ an ^2}{a over 2} – {cos ^2}a cr
& = {{2{{cos }^2}{a over 2}} over {2{{sin }^2}{a over 2}}}{ an ^2}{a over 2} – {cos ^2}a = {sin ^2}a cr} )

b) (4{cos ^4}a – 2cos 2a – {1 over 2}cos 4a)

( = 4{cos ^4}a – 2(2{cos ^2}a – 1) – {1 over 2}(2{cos ^2}2a – 1))

( = 4{cos ^4}a – 4{cos ^2}a + 2 – {(2{cos ^2}a – 1)^2} + {1 over 2})

( = 4{cos ^4}a – 4{cos ^2}a + {5 over 2} – 4{cos ^4}a + 4{cos ^2}a – 1 = {3 over 2})

c) ({sin ^2}a(1 + {1 over {sin a}} + cot a)(1 – {1 over {sin a}} + cot a))

(eqalign{
& = {sin ^2}aleft[ {{{(1 + cota)}^2} – {1 over {{{sin }^2}a}}} ight] cr
& = {sin ^2}a(1 + {cot ^2}a + 2cot a) – 1 cr} )

(eqalign{
& = {sin ^2}a + {cos ^2}a + 2{sin ^2}a{{cos a} over {sin a}} – 1 cr
& = sin 2a cr} )

d) ({{cos 2a} over {{{cos }^4}a – {{sin }^4}a}} – {{{{cos }^4}a + {{sin }^4}a} over {1 – {1 over 2}{{sin }^2}2a}})

(= {{{{cos }^2}a – {{sin }^2}a} over {({{cos }^2}a + {{sin }^2}a)({{cos }^2}a – {{sin }^2}a)}} – {{{{cos }^4}a + {{sin }^4}a} over {1 – {1 over 2}{{(2sin acos a)}^2}}})

( = 1 – {{{{cos }^4}a + {{sin }^4}a} over {{{sin }^2}a – si{n^2}aco{s^2}a + {{cos }^2}a – {{sin }^2}a{{cos }^2}a}})

( = 1 – {{{{cos }^4}a + {{sin }^4}a} over {{{sin }^2}a(1 – co{s^2}a) + {{cos }^2}a(1 – {{sin }^2}a)}} = 0)