Bài 20 trang 217 Sách bài tập (SBT) Toán Đại số 10

Chứng minh rằng

Chứng minh rằng

a) ({{sqrt {1 + cos alpha }  + sqrt {1 - cos alpha } } over {sqrt {1 + cos alpha }  - sqrt {1 - cos alpha } }} = cot ({alpha  over 2} + {pi  over 4})) ((pi  < alpha  < 2pi ))

b) ({{cos 4a an 2a - sin 4a} over {cos 4acot 2a + sin 4a}} =  - { an ^2}2a)

c) ({{{{sin }^2}2a + 4{{sin }^2}a - 4} over {1 - 8{{sin }^2}a - cos 4a}} = {1 over 2}{cot ^4}a)

d) (1 + 2cos 7a = {{sin 10,5a} over {sin 3,5a}})

e) ({{ an 3a} over { an a}} = {{3 - {{ an }^2}a} over {1 - 3{{ an }^2}a}})

Gợi ý làm bài

a) Vì (sqrt {1 + cos alpha }  =  - sqrt 2 cos {alpha  over 2}(do{pi  over 2} < {alpha  over 2} < pi ))

(sqrt {1 - cos alpha }  = sqrt 2 sin {alpha  over 2}) cho nên

({{sqrt {1 + cos alpha }  + sqrt {1 - cos alpha } } over {sqrt {1 + cos alpha }  - sqrt {1 - cos alpha } }} = {{ - sqrt 2 cos {alpha  over 2} + sqrt 2 cos {alpha  over 2}} over { - sqrt 2 cos {alpha  over 2} - sqrt 2 cos {alpha  over 2}}})

( = {{cos {alpha  over 2} - sin {alpha  over 2}} over {cos {alpha  over 2} + sin {alpha  over 2}}} = {{1 - an {alpha  over 2}} over {1 + an {alpha  over 2}}} = an ({pi  over 4} - {alpha  over 2}))

( = cot ({alpha  over 2} + {pi  over 4}))

b) 

(eqalign{
& = {{cos 4a an 2a - sin 4a} over {cos 4acot 2a + sin 4a}} cr
& = {{cos 4asin 2a - sin 4acos 2a} over {cos 4acos 2a + sin 4asin 2a}}. an 2a cr} )

( = {{ - sin 2a} over {cos 2a}} an 2a =  - { an ^2}2a$)

c) 

(eqalign{
& {{{{sin }^2}2a + 4{{sin }^2}4a} over {1 - {{sin }^2}a - cos 4a}} cr
& = {{4{{sin }^2}a{{cos }^2}a + 4({{sin }^2}a - 1)} over {1 - 8{{sin }^2}a - (1 - 2{{sin }^2}2a)}} cr} )

({{4{{cos }^2}a({{sin }^2}a - 1)} over {8{{sin }^2}a(co{s^2}a - 1)}} = {1 over 2}{cot ^4}a.)

d) 

(eqalign{
& {{sin 10,5a} over {sin 3,5a}} = {{sin (7 + 3,5a)} over {sin 3,5a}} cr
& = {{sin 7acos 3,5a + cos 7asin 3,5a} over {sin 3,5a}} cr} )

( = {{sin 3,5a(2{{cos }^2}3,5a + cos 7a)} over {sin 3,5a}})

( = (2{cos ^2}3,5a - 1) + 1 + cos7a)

( = 2cos7a + 1.)

e) 

(eqalign{
& {{ an (a + 2a)} over { an a}} = {{ an a + an 2a} over { an a(1 - {mathop{ m tanatan} olimits} 2a}} cr
& = {{ an a + {{2 an a} over {1 - {{ an }^2}a}}} over { an a(1 - {{2{{ an }^2}a} over {1 - {{ an }^2}a}})}} cr} )

( = {{3 - {{ an }^2}a} over {1 - 3{{ an }^2}a}}$)

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