sinA + cosA = sinB + cosB ও= π/9,  B=?

প্রশ্ন: sinA + cosA = sinB + cosB এবং A=π/9,  B=?

সমাধান:

দেওয়া আছে, sinA + cosA = sinB + cosB

এবং A = π/9

সুতরাং, sin(π/9) + cos(π/9) = sinB + cosB

উভয়পক্ষে √2 দিয়ে গুণ করে প???ই,

√2sin(π/9) + √2cos(π/9) = √2sinB + √2cosB

আমরা জানি, sin(π/4) = cos(π/4) = 1/√2, সুতরাং √2 = 1/sin(π/4) = 1/cos(π/4)

সুতরাং, √2sin(π/9) + √2cos(π/9) = sin(π/4)cos(π/9) + cos(π/4)sin(π/9)

= sin(π/4 + π/9) = sin((9π + 4π)/36) = sin(13π/36)

আবার, √2sinB + √2cosB = sin(π/4)cosB + cos(π/4)sinB = sin(B + π/4)

অতএব, sin(13π/36) = sin(B + π/4)

সুতরাং, B + π/4 = 13π/36 অথবা B + π/4 = π - 13π/36 = (36π-13π)/36 = 23π/36

যদি B + π/4 = 13π/36 হয়, তবে B = 13π/36 - π/4 = 13π/36 - 9π/36 = 4π/36 = π/9

যদি B + π/4 = 23π/36 হয়, তবে B = 23π/36 - π/4 = 23π/36 - 9π/36 = 14π/36 = 7??/18

যেহেতু A=B হতে পারেনা , তাই B = 7π/18

অতএব, B = 7π/18

বিকল্প পদ্ধতি:

sinA + cosA = sinB + cosB

sinA - sinB = cosB - cosA

2cos((A+B)/2)sin((A-B)/2) = 2sin((A+B)/2)sin((A-B)/2)

2sin((A-B)/2)[cos((A+B)/2) - sin((A+B)/2)] = 0

সুতরাং, sin((A-B)/2) = 0 অথবা cos((A+B)/2) - sin((A+B)/2) = 0

যদি sin((A-B)/2) = 0 হয়, তবে (A-B)/2 = 0, সুতরাং A = B, যা গ্রহণযোগ্য নয়।

যদি cos((A+B)/2) - sin((A+B)/2) = 0 হয়, তবে cos((A+B)/2) = sin((A+B)/2)

সুতরাং, tan((A+B)/2) = 1

(A+B)/2 = π/4

A+B = π/2

B = π/2 - A = π/2 - π/9 = (9π - 2π)/18 = 7π/18

অতএব, B = 7π/18 🥳