Express As Function of x?
2020-07-21 10:11 pm
If sin(theta) = 2x and 0 < theta < pi/2, express theta + cos (2•theta) as a function of x.
回答 (3)
2020-07-21 11:00 pm
✔ 最佳答案
sin(θ) = 2xThen, θ = sin⁻¹(2x)
cos(2θ)
= 1 - 2 sin²(θ)
= 1 - 2 (2x)²
= 1 - 8x²
Hence, θ + cos(2θ) = sin⁻¹(2x) + 1 - 8x²
2020-07-21 11:15 pm
sin²θ + cos²θ = 1
so, cos²θ = 1 - sin²θ => 1 - (2x)²
i.e. cos²θ = 1 - 4x²
Now, cos2θ = 2cos²θ - 1 => 2(1 - 4x² ) - 1
Hence, cos2θ = 1 - 8x²
so, θ + cos2θ = sin⁻¹(2x) + 1 - 8x²
:)>
so, cos²θ = 1 - sin²θ => 1 - (2x)²
i.e. cos²θ = 1 - 4x²
Now, cos2θ = 2cos²θ - 1 => 2(1 - 4x² ) - 1
Hence, cos2θ = 1 - 8x²
so, θ + cos2θ = sin⁻¹(2x) + 1 - 8x²
:)>
2020-07-21 11:03 pm
sin(theta) = 2x >>> theta = invsin(2x)
2•theta = 2invsin(2x)cos (2•theta) = 1 - 2 sin^2(theta)theta + cos (2•theta) = invsin(2x) + cos (2invsin(2x)) <<< answer
2•theta = 2invsin(2x)cos (2•theta) = 1 - 2 sin^2(theta)theta + cos (2•theta) = invsin(2x) + cos (2invsin(2x)) <<< answer
收錄日期: 2021-04-18 18:33:53
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20200721141111AAq2MGn