sin(πCosθ)= cos(πsinθ) হলে θ এর মান কোনটি?

Another Explanation (5): ```html

প্রশ্ন: sin(πCosθ)= cos(πsinθ) হলে θ এর মান কোনটি?

সমাধান:

আমরা জানি, cos(x) = sin(π/2 - x)

সুতরাং, cos(πsinθ) = sin(π/2 - πsinθ)

প্রশ্নানুসারে, sin(πCosθ) = cos(πsinθ)

অতএব, sin(πCosθ) = sin(π/2 - πsinθ)

সুতরাং, πCosθ = nπ + (-1)^n (π/2 - πsinθ), যেখানে n একটি পূর্ণসংখ্যা। ℤ

Case 1: যখন n জোড় সংখ্যা (n = 2k, k ∈ ℤ)

πCosθ = 2kπ + (π/2 - πsinθ)

Cosθ = 2k + 1/2 - sinθ

Cosθ + sinθ = 2k + 1/2

\(\sqrt{2}\) [1/\(\sqrt{2}\)Cosθ + 1/\(\sqrt{2}\)sinθ] = 2k + 1/2

\(\sqrt{2}\) [Cos(π/4)Cosθ + sin(π/4)sinθ] = 2k + 1/2

\(\sqrt{2}\) Cos(θ - π/4) = 2k + 1/2

Cos(θ - π/4) = (4k + 1) / (2\(\sqrt{2}\))

যদি k = 0 হয়,

Cos(θ - π/4) = 1 / (2\(\sqrt{2}\))

θ - π/4 = ± Cos^-1(1 / (2\(\sqrt{2}\)))

θ = π/4 ± Cos^-1(1 / (2\(\sqrt{2}\)))

Case 2: যখন n বিজোড় সংখ্যা (n = 2k + 1, k ∈ ℤ)

πCosθ = (2k + 1)π - (π/2 - πsinθ)

Cosθ = 2k + 1 - 1/2 + sinθ

Cosθ - sinθ = 2k + 1/2

\(\sqrt{2}\) [1/\(\sqrt{2}\)Cosθ - 1/\(\sqrt{2}\)sinθ] = 2k + 1/2

\(\sqrt{2}\) [Cos(π/4)Cosθ - sin(π/4)sinθ] = 2k + 1/2

\(\sqrt{2}\) Cos(θ + π/4) = 2k + 1/2

Cos(θ + π/4) = (4k + 1) / (2\(\sqrt{2}\))

যদি k = 0 হয়,

Cos(θ + π/4) = 1 / (2\(\sqrt{2}\))

θ + π/4 = ± Cos^-1(1 / (2\(\sqrt{2}\)))

θ = -π/4 ± Cos^-1(1 / (2\(\sqrt{2}\)))

সুতরাং, θ = ± π/4 ± Cos^-1(1 / (2\(\sqrt{2}\)))

ফাইনাল উত্তর:

θ = ± π/4 + Cos^-1(1 / (2\(\sqrt{2}\)))

😊 ```