Bài 46 trang 215 Đại số 10 Nâng cao: Chứng minh rằng:...

Chứng minh rằng:

a) (sin3α = 3sinα – 4si{n^3}alpha ) ; ( cos3α =4co{s^3}alpha – 3cosα)

b)

(eqalign{
& sin alpha sin ({pi over 3} – alpha )sin ({pi over 3} + alpha ) = {1 over 4}sin 3alpha cr
& cos alpha cos ({pi over 3} – alpha )cos({pi over 3} + alpha ) = {1 over 4}cos 3alpha cr} )

Ứng dụng: Tính: sin 20⁰ sin 40⁰ sin 80⁰ và tan 20⁰ tan 40⁰ tan 80⁰

Đáp án

a) Ta có:

(sin3α = sin (2α  + α) = sin 2α  cosα + sinα cos 2α)

( = { m{ }}2{ m{ }}sinalpha { m{ }}co{s^2}alpha { m{ }} + { m{ }}sinalpha { m{ }}(1{ m{ }}-{ m{ }}2si{n^2}alpha ))

(= { m{ }}2sinalpha { m{ }}(1{ m{ }}-{ m{ }}si{n^2}alpha ){ m{ }} + { m{ }}sin(1{ m{ }}-{ m{ }}si{n^2}alpha ){ m{ }})

(= { m{ }}3sinalpha { m{ }}-{ m{ }}4si{n^3}alpha )

(cos3α = cos (2α  + α) = cos 2α  cosα  – sin2α sinα)

(= { m{ }}(2co{s^2}alpha { m{ }}-{ m{ }}1)cosalpha { m{ }}-{ m{ }}2si{n^2}alpha { m{ }}cosalpha )

( = { m{ }}2co{s^3}alpha { m{ }}-{ m{ }}cosalpha { m{ }}-{ m{ }}2cosalpha { m{ }}(1{ m{ }}-{ m{ }}co{s^2}alpha ){ m{ }} )

(= { m{ }}4co{s^3}alpha { m{ }}-{ m{ }}3cosalpha )

b) Ta có:

(eqalign{
& sin alpha sin ({pi over 3} – alpha )sin ({pi over 3} + alpha ) cr&= sinalpha .{1 over 2}(cos2alpha – cos {{2pi } over 3}) cr
& = {1 over 2}sin alpha (1 – 2{sin ^2}alpha + {1 over 2}) = {1 over 4}sin alpha (3 – 4{sin ^2}alpha ) cr
& = {1 over 4}sin 3alpha cr
& cos alpha cos ({pi over 3} – alpha )cos({pi over 3} + alpha ) cr&= cos alpha .{1 over 2}(cosalpha + cos {{2pi } over 3}) cr
& = {1 over 2}cos alpha (2{cos ^2}alpha – 1 – {1 over 2}) cr&= {1 over 4}cos alpha (4{cos ^2}alpha – 3) = {1 over 4}cos 3alpha cr} )

Ứng dụng:

(eqalign{
& sin {20^0}sin {40^0}sin {80^0} cr&= sin {20^0}sin ({60^0} – {20^0})sin ({60^0} + {20^0}) cr
& = {1 over 4}sin ({3.20^0}) = {1 over 4}sin {60^0} = {{sqrt 3 } over 8} cr
& cos {20^0}cos {40^0}cos {80^0} = {1 over 4}cos ({3.20^0}) = {1 over 8} cr} ) 

Vậy : ( an {20^0} an {40^0} an {80^0} = sqrt 3 )

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