Cos²(60^o+A)+Cos²(60^o-A) এর মান-

Explanation:

Another Explanation (5): Cos²(60^o+A)+Cos²(60^o-A) এর মান নির্ণয়: আমরা জানি, Cos(A+B) = CosA CosB - SinA SinB এবং Cos(A-B) = CosA CosB + SinA SinB সুতরাং, Cos(60^o+A) = Cos60^o CosA - Sin60^o SinA = \(\frac{1}{2}\)CosA - \(\frac{\sqrt{3}}{2}\)SinA Cos(60^o-A) = Cos60^o CosA + Sin60^o SinA = \(\frac{1}{2}\)CosA + \(\frac{\sqrt{3}}{2}\)SinA এখন, Cos²(60^o+A)+Cos²(60^o-A) = (\(\frac{1}{2}\)CosA - \(\frac{\sqrt{3}}{2}\)SinA)² + (\(\frac{1}{2}\)CosA + \(\frac{\sqrt{3}}{2}\)SinA)² = \(\frac{1}{4}\)Cos²A - \(\frac{\sqrt{3}}{2}\)CosA SinA + \(\frac{3}{4}\)Sin²A + \(\frac{1}{4}\)Cos²A + \(\frac{\sqrt{3}}{2}\)CosA SinA + \(\frac{3}{4}\)Sin²A = \(\frac{1}{4}\)Cos²A + \(\frac{3}{4}\)Sin²A + \(\frac{1}{4}\)Cos²A + \(\frac{3}{4}\)Sin²A = \(\frac{1}{2}\)Cos²A + \(\frac{3}{2}\)Sin²A = \(\frac{1}{2}\)Cos²A + \(\frac{1}{2}\)Sin²A + Sin²A = \(\frac{1}{2}\)(Cos²A + Sin²A) + Sin²A = \(\frac{1}{2}\) + Sin²A = \(\frac{1}{2}\) + \(\frac{1-Cos2A}{2}\) [ যেহেতু Sin²A = \(\frac{1-Cos2A}{2}\) ] = \(\frac{1}{2}\) + \(\frac{1}{2}\) - \(\frac{1}{2}\)Cos2A = 1 - \(\frac{1}{2}\)Cos2A 🎉 উত্তর: 1 - \(\frac{1}{2}\)Cos2A 🥳