sin10^o sin50^o sin70^o- এর মান কত? 

Explanation:

Another Explanation (5): sin10^osin50^osin70^o এর মান নির্ণয়: আমরা জানি, \(sin(90^{\circ} - \theta) = cos\theta\) সুতরাং, sin70^o = sin(90^o - 20^o) = cos20^o তাহলে, sin10^osin50^osin70^o = sin10^osin50^ocos20^o এখন, sin50^o = sin(60^o - 10^o) সুতরাং, sin10^osin50^ocos20^o = sin10^osin(60^o - 10^o)cos20^o আমরা জানি, \(sin(A - B) = sinAcosB - cosAsinB\) সুতরাং, sin(60^o - 10^o) = sin60^ocos10^o - cos60^osin10^o = \(\frac{\sqrt{3}}{2}\)cos10^o - \(\frac{1}{2}\)sin10^o তাহলে, sin10^osin(60^o - 10^o)cos20^o = sin10^o(\(\frac{\sqrt{3}}{2}\)cos10^o - \(\frac{1}{2}\)sin10^o)cos20^o = (\(\frac{\sqrt{3}}{2}\)sin10^ocos10^o - \(\frac{1}{2}\)sin²10^o)cos20^o আমরা জানি, 2sinθcosθ = sin2θ, সুতরাং sinθcosθ = \(\frac{1}{2}\)sin2θ = (\(\frac{\sqrt{3}}{4}\)sin20^o - \(\frac{1}{2}\)sin²10^o)cos20^o = \(\frac{\sqrt{3}}{4}\)sin20^ocos20^o - \(\frac{1}{2}\)sin²10^ocos20^o = \(\frac{\sqrt{3}}{8}\)sin40^o - \(\frac{1}{2}\)sin²10^ocos20^o আবার, sin10^osin50^ocos20^o = sin10^ocos40^ocos20^o = \(\frac{2sin10^{\circ}cos10^{\circ}cos20^{\circ}cos40^{\circ}}{2cos10^{\circ}}\) = \(\frac{sin20^{\circ}cos20^{\circ}cos40^{\circ}}{2cos10^{\circ}}\) = \(\frac{2sin20^{\circ}cos20^{\circ}cos40^{\circ}}{2 \cdot 2cos10^{\circ}}\) = \(\frac{sin40^{\circ}cos40^{\circ}}{4cos10^{\circ}}\) = \(\frac{2sin40^{\circ}cos40^{\circ}}{2 \cdot 4cos10^{\circ}}\) = \(\frac{sin80^{\circ}}{8cos10^{\circ}}\) = \(\frac{cos10^{\circ}}{8cos10^{\circ}}\) = \(\frac{1}{8}\) অতএব, sin10^osin50^osin70^o = \(\frac{1}{8}\) 🎉🎉🎉