Trang chủ Lớp 10 Toán lớp 10 Nâng cao Bài 51 trang 216 Đại số 10 Nâng cao: Chứng minh rằng...

Bài 51 trang 216 Đại số 10 Nâng cao: Chứng minh rằng nếu ∝ + β + γ = π thì...

Chứng minh rằng nếu ∝ + β + γ = π thì. Bài 51 trang 216 SGK Đại số 10 Nâng cao – Bài 4: Một số công thức lượng giác

Chứng minh rằng nếu \(∝ + β + γ = π\) thì

a) \(\sin \alpha  + \sin \beta  + \sin \gamma  = 4\cos {\alpha  \over 2}\cos {\beta  \over 2}\cos {\gamma  \over 2}\)

b) \(\cos \alpha  + \cos \beta  + \cos \gamma  = 1 + 4\sin {\alpha  \over 2}\sin {\beta  \over 2}\sin {\gamma  \over 2}\)

c) \(sin2∝ + sin2β + sin2γ = 4sin∝ sinβ sin γ\)

d) \(co{s^2} \propto + {\rm{ }}co{s^2}\beta + co{s^2}\gamma {\rm{ }}= 1 – 2cos∝ cosβ cosγ\)

 Đáp án

a) Ta có:

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\(\eqalign{
& \sin \alpha + \sin \beta + \sin \gamma\cr& = \sin \alpha + 2\sin {{\beta + \gamma } \over 2}\cos {{\beta – \gamma } \over 2} \cr
& = \sin \alpha + 2\sin {{\pi – \alpha } \over 2}\cos {{\beta – \gamma } \over 2} \cr&= 2\sin {\alpha \over 2}\cos {\alpha \over 2} + 2\cos {\alpha \over 2}  \cos {{\beta – \gamma } \over 2} \cr
& = 2\cos {\alpha \over 2}(\sin {\alpha \over 2} + \cos {{\beta – \gamma } \over 2})\cr& = 2\cos {\alpha \over 2}{\rm{[sin}}{{\pi – (\beta + \gamma )} \over 2} + \cos{{\beta – \gamma } \over 2}{\rm{]}} \cr
& = 2\cos {\alpha \over 2}(cos{{\beta + \gamma } \over 2} + \cos {{\beta – \gamma } \over 2}) \cr
& =4\cos {\alpha \over 2}\cos {\beta \over 2}\cos {\gamma \over 2} \cr} \)

b) Ta có:

\(\eqalign{
& \cos \alpha + \cos \beta + \cos \gamma \cr&= 2\cos {{\alpha + \beta } \over 2}\cos {{\alpha – \beta } \over 2} + 1 – 2\sin {{2\gamma } \over 2} \cr
& = 2\cos ({\pi \over 2} – {\gamma \over 2})cos{{\alpha – \beta } \over 2} + 1 – 2{\sin ^2}{\gamma \over 2} \cr&= 1 + 2\sin {\gamma \over 2}(cos{{\alpha – \beta } \over 2} – \sin {\gamma \over 2}) \cr
& = 1 + 2\sin {\gamma \over 2}(cos{{\alpha – \beta } \over 2} – cos{{\alpha + \beta } \over 2}) \cr
& = 1 + 4\sin {\alpha \over 2}\sin {\beta \over 2}\sin {\gamma \over 2} \cr} \)

c) \(sin2∝ + sin2β + sin2γ\)

\(= 2sin (∝ + β)cos(∝ – β ) + 2sinγcosγ\)

\(= 2sinγ (cos(∝ – β ) – cos(∝ + β)) \)

\(= 4sin∝ sinβ sin γ\)

d) Ta có:

\(\eqalign{
& co{s^2} \propto + {\rm{ }}co{s^2}\beta + co{s^2}\gamma {\rm{ }} \cr
& {\rm{ = }}{{1 + \cos 2\alpha } \over 2} + {{1\cos 2\beta } \over 2} + {\cos ^2}\gamma \cr
& = 1 + {1 \over 2}(cos2\alpha + \cos 2\beta ) + {\cos ^2}\gamma \cr
& = 1 + \cos (\alpha + \beta )cos(\alpha – \beta ) + {\cos ^2}\gamma \cr
& = 1 + \cos \gamma (\cos \gamma – \cos (\alpha – \beta )) \cr&= 1 – \cos \gamma {\rm{[cos(}}\alpha {\rm{ + }}\beta {\rm{) + cos(}}\alpha {\rm{ – }}\beta ){\rm{]}} \cr
& = {\rm{ }}1{\rm{ }}-{\rm{ }}2cos \propto {\rm{ }}cos\beta {\rm{ }}cos\gamma \cr} \)

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