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已知 α,β,γ\alpha, \beta, \gamma 成公比为 2 的等比数列，且 α∈(0,2π)\alpha \in(0,2 \pi). 若 $\cos \alpha,...
Mar 8, 2024
已知 α,β,γ\alpha, \beta, \gamma 成公比为 2 的等比数列，且 α∈(0,2π)\alpha \in(0,2 \pi). 若 cos⁡α,3cos⁡β,sin⁡γ)n\cos \alpha, \sqrt{3} \cos \beta, \sin \gamma)^{n} 等比数列，则所有满足条件的 α\alpha 的和为
Solution by Steps
step 1
Given that α,β,γ \alpha, \beta, \gamma form a geometric sequence with a common ratio of 2, we can express β \beta and γ \gamma in terms of α \alpha as β=2α \beta = 2\alpha and γ=4α \gamma = 4\alpha
step 2
Since cos⁡α,3cos⁡β,sin⁡γ \cos \alpha, \sqrt{3} \cos \beta, \sin \gamma also form a geometric sequence, we can set up the relationship 3cos⁡(2α)=cos⁡2(α) \sqrt{3} \cos(2\alpha) = \cos^2(\alpha) and sin⁡(4α)=3cos⁡2(α) \sin(4\alpha) = 3\cos^2(\alpha)
step 3
Solve the equation 3cos⁡(2α)=cos⁡2(α) \sqrt{3} \cos(2\alpha) = \cos^2(\alpha) for α \alpha in the interval (0,2π) (0, 2\pi)
step 4
Solve the equation sin⁡(4α)=3cos⁡2(α) \sin(4\alpha) = 3\cos^2(\alpha) for α \alpha in the interval (0,2π) (0, 2\pi)
step 5
Find the values of α \alpha that satisfy both equations from steps 3 and 4
step 6
Sum all the valid values of α \alpha to find the total sum
Answer
[Insert final answer here]
Key Concept
Geometric Sequences and Trigonometric Identities
Explanation
To find the sum of all α \alpha that satisfy the given conditions, we need to use the properties of geometric sequences and apply trigonometric identities to solve the equations.
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