Bài 46 trang 215 SGK Đại số 10 Nâng cao

Chứng minh rằng:

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

b)

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

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

Đáp án

a) Ta có:

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

\( = {\rm{ }}2{\rm{ }}sin\alpha {\rm{ }}co{s^2}\alpha {\rm{ }} + {\rm{ }}sin\alpha {\rm{ }}(1{\rm{ }}-{\rm{ }}2si{n^2}\alpha )\)

\(= {\rm{ }}2sin\alpha {\rm{ }}(1{\rm{ }}-{\rm{ }}si{n^2}\alpha ){\rm{ }} + {\rm{ }}sin(1{\rm{ }}-{\rm{ }}si{n^2}\alpha ){\rm{ }}\)

\(= {\rm{ }}3sin\alpha {\rm{ }}-{\rm{ }}4si{n^3}\alpha \)

\(cos3α = cos (2α  + α) = cos 2α  cosα  - sin2α sinα\)

\(= {\rm{ }}(2co{s^2}\alpha {\rm{ }}-{\rm{ }}1)cos\alpha {\rm{ }}-{\rm{ }}2si{n^2}\alpha {\rm{ }}cos\alpha \)

\( = {\rm{ }}2co{s^3}\alpha {\rm{ }}-{\rm{ }}cos\alpha {\rm{ }}-{\rm{ }}2cos\alpha {\rm{ }}(1{\rm{ }}-{\rm{ }}co{s^2}\alpha ){\rm{ }} \)

\(= {\rm{ }}4co{s^3}\alpha {\rm{ }}-{\rm{ }}3cos\alpha \)

b) Ta có:

\(\eqalign{
& \sin \alpha \sin ({\pi \over 3} - \alpha )\sin ({\pi \over 3} + \alpha ) \cr&= sin\alpha .{1 \over 2}(cos2\alpha - \cos {{2\pi } \over 3}) \cr
& = {1 \over 2}\sin \alpha (1 - 2{\sin ^2}\alpha + {1 \over 2}) = {1 \over 4}\sin \alpha (3 - 4{\sin ^2}\alpha ) \cr
& = {1 \over 4}\sin 3\alpha \cr
& \cos \alpha \cos ({\pi \over 3} - \alpha )cos({\pi \over 3} + \alpha ) \cr&= \cos \alpha .{1 \over 2}(cos\alpha + \cos {{2\pi } \over 3}) \cr
& = {1 \over 2}\cos \alpha (2{\cos ^2}\alpha - 1 - {1 \over 2}) \cr&= {1 \over 4}\cos \alpha (4{\cos ^2}\alpha - 3) = {1 \over 4}\cos 3\alpha \cr} \)

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 : \(\tan {20^0}\tan {40^0}\tan {80^0} = \sqrt 3 \)