tan θ = 1/2 হলে sin 2θ এর মান কোনটি?

সমাধান:

প্রদত্ত: \(\tan \theta = \frac{1}{2}\)

প্রথমে, \(\sin \theta\) এবং \(\cos \theta\) খুঁজে নিতে হবে।

তাই, \(\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{1}{2}\)

এখন, ধরুন:

\(\sin \theta = y\)

\(\cos \theta = x\)

তাহলে, \(\frac{y}{x} = \frac{1}{2}\) অর্থাৎ, \(y = \frac{1}{2}x\)

তবে, \(\sin^2 \theta + \cos^2 \theta = 1\)

অর্থাৎ, \(y^2 + x^2 = 1\)

এখন, \(y = \frac{1}{2}x\) বসিয়ে নিই:

\(\left(\frac{1}{2}x\right)^2 + x^2 = 1\)

\(\frac{1}{4}x^2 + x^2 = 1\)

\(\frac{1}{4}x^2 + \frac{4}{4}x^2 = 1\)

\(\frac{5}{4}x^2 = 1\)

অতএব,

\(x^2 = \frac{4}{5}\)

এবং,

\(x = \pm \sqrt{\frac{4}{5}} = \pm \frac{2}{\sqrt{5}}\)

এখন, \(y = \frac{1}{2}x\), তাই:

\(y = \pm \frac{1}{2} \times \frac{2}{\sqrt{5}} = \pm \frac{1}{\sqrt{5}}\)

তাহলে, \(\sin \theta = y = \pm \frac{1}{\sqrt{5}}\)

এবং, \(\cos \theta = x = \pm \frac{2}{\sqrt{5}}\)

এখন, \(\sin 2\theta = 2 \sin \theta \cos \theta\)

সুতরাং,

\(\sin 2\theta = 2 \times \frac{1}{\sqrt{5}} \times \frac{2}{\sqrt{5}} = 2 \times \frac{2}{5} = \frac{4}{5}\)

অতএব,

উত্তর: \(\boxed{\frac{4}{5}}\)