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

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)

অতএব, sin(πCosθ) = sin(π/2 - πsinθ)

সুতরাং, πCosθ = π/2 - πsinθ অথবা πCosθ = π - (π/2 - πsinθ)

কেস ১:

\(\pi\)Cosθ = \(\frac{\pi}{2}\) - \(\pi\)sinθ

Cosθ = \(\frac{1}{2}\) - sinθ

Cosθ + sinθ = \(\frac{1}{2}\)

\(\frac{1}{\sqrt{2}}\)Cosθ + \(\frac{1}{\sqrt{2}}\)sinθ = \(\frac{1}{2\sqrt{2}}\)

Cos(\(\frac{\pi}{4}\))Cosθ + Sin(\(\frac{\pi}{4}\))sinθ = \(\frac{1}{2\sqrt{2}}\)

Cos(θ - \(\frac{\pi}{4}\)) = \(\frac{1}{2\sqrt{2}}\)

θ - \(\frac{\pi}{4}\) = Cos^-1(\(\frac{1}{2\sqrt{2}}\))

θ = \(\frac{\pi}{4}\) + Cos^-1(\(\frac{1}{2\sqrt{2}}\))

কেস ২:

\(\pi\)Cosθ = \(\pi\) - (\(\frac{\pi}{2}\) - \(\pi\)sinθ)

\(\pi\)Cosθ = \(\frac{\pi}{2}\) + \(\pi\)sinθ

Cosθ = \(\frac{1}{2}\) + sinθ

Cosθ - sinθ = \(\frac{1}{2}\)

\(\frac{1}{\sqrt{2}}\)Cosθ - \(\frac{1}{\sqrt{2}}\)sinθ = \(\frac{1}{2\sqrt{2}}\)

Cos(\(\frac{\pi}{4}\))Cosθ + Sin(-\(\frac{\pi}{4}\))sinθ = \(\frac{1}{2\sqrt{2}}\)

Cos(θ + \(\frac{\pi}{4}\)) = \(\frac{1}{2\sqrt{2}}\)

θ + \(\frac{\pi}{4}\) = Cos^-1(\(\frac{1}{2\sqrt{2}}\))

θ = -\(\frac{\pi}{4}\) + Cos^-1(\(\frac{1}{2\sqrt{2}}\))

সুতরাং, θ = ± \(\frac{\pi}{4}\) + Cos^-1(\(\frac{1}{2\sqrt{2}}\)) 😃

অতএব, উত্তর: ± \(\frac{\pi}{4}\) + Cos^-1(\(\frac{1}{2\sqrt{2}}\))