sin²(60°+A)+sin²A+sin²(60°–A) এর মান কত ?

Explanation:

Another Explanation (5): sin²(60°+A)+sin²A+sin²(60°–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, sin²(60°+A) = [sin(60°)cos(A) + cos(60°)sin(A)]² = [(\(\sqrt{3}\)/2)cos(A) + (1/2)sin(A)]² = (3/4)cos²(A) + (\(\sqrt{3}\)/2)cos(A)sin(A) + (1/4)sin²(A) এবং, sin²(60°–A) = [sin(60°)cos(A) – cos(60°)sin(A)]² = [(\(\sqrt{3}\)/2)cos(A) - (1/2)sin(A)]² = (3/4)cos²(A) - (\(\sqrt{3}\)/2)cos(A)sin(A) + (1/4)sin²(A) অতএব, sin²(60°+A) + sin²(60°–A) = (3/4)cos²(A) + (\(\sqrt{3}\)/2)cos(A)sin(A) + (1/4)sin²(A) + (3/4)cos²(A) - (\(\sqrt{3}\)/2)cos(A)sin(A) + (1/4)sin²(A) = (3/2)cos²(A) + (1/2)sin²(A) এখন, প্রদত্ত রাশিমালাটি হল: sin²(60°+A)+sin²A+sin²(60°–A) = (3/2)cos²(A) + (1/2)sin²(A) + sin²(A) = (3/2)cos²(A) + (3/2)sin²(A) = (3/2)[cos²(A) + sin²(A)] = (3/2) * 1 [∵ cos²(A) + sin²(A) = 1] = 3/2 সুতরাং, sin²(60°+A)+sin²A+sin²(60°–A) = 3/2 😊.