sin ^2 ( π/4 - A) + sin^2 ( π/4 + A ) এর মান কোনটি?

Explanation:  

Another Explanation (5): sin^2 (π/4 - A) + sin^2 (π/4 + A) এর মান নির্ণয়: আমরা জানি, sin(a - b) = sin a cos b - cos a sin b এবং sin(a + b) = sin a cos b + cos a sin b সুতরাং, sin(π/4 - A) = sin(π/4)cos(A) - cos(π/4)sin(A) = \(\frac{1}{\sqrt{2}}\)cos(A) - \(\frac{1}{\sqrt{2}}\)sin(A) sin(π/4 + A) = sin(π/4)cos(A) + cos(π/4)sin(A) = \(\frac{1}{\sqrt{2}}\)cos(A) + \(\frac{1}{\sqrt{2}}\)sin(A) এখন, sin^2(π/4 - A) + sin^2(π/4 + A) = [ \(\frac{1}{\sqrt{2}}\)cos(A) - \(\frac{1}{\sqrt{2}}\)sin(A) ]^2 + [ \(\frac{1}{\sqrt{2}}\)cos(A) + \(\frac{1}{\sqrt{2}}\)sin(A) ]^2 = \(\frac{1}{2}\)[cos(A) - sin(A)]^2 + \(\frac{1}{2}\)[cos(A) + sin(A)]^2 = \(\frac{1}{2}\)[cos^2(A) - 2cos(A)sin(A) + sin^2(A)] + \(\frac{1}{2}\)[cos^2(A) + 2cos(A)sin(A) + sin^2(A)] = \(\frac{1}{2}\)[cos^2(A) + sin^2(A) - 2cos(A)sin(A) + cos^2(A) + sin^2(A) + 2cos(A)sin(A)] = \(\frac{1}{2}\)[2cos^2(A) + 2sin^2(A)] = \(\frac{1}{2}\)*2[cos^2(A) + sin^2(A)] = cos^2(A) + sin^2(A) আমরা জানি, cos^2(A) + sin^2(A) = 1 সুতরাং, sin^2 (π/4 - A) + sin^2 (π/4 + A) = 1 🎉