sin A + cos A = sin B + cos B এবং A = 2π/9 হলে B =?

প্রশ্ন: sin A + cos A = sin B + cos B এবং A = 2π/9 হলে B =?

সমাধান:

দেওয়া আছে, sin A + cos A = sin B + cos B এবং A = 2π/9। সুতরাং,

sin (2π/9) + cos (2π/9) = sin B + cos B

আমরা জানি, sin θ + cos θ = √2 sin(θ + π/4)

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

⇒ sin(2π/9 + π/4) = sin(B + π/4)

⇒ sin((8π + 9π)/36) = sin(B + π/4)

⇒ sin(17π/36) = sin(B + π/4)

এখন, sin x = sin y হলে, x = y অথবা x = π - y হয়।

সুতরাং,

1) B + π/4 = 17π/36

⇒ B = 17π/36 - π/4

⇒ B = (17π - 9π)/36

⇒ B = 8π/36 = 2π/9 = A (যা গ্রহণযোগ্য নয়, কারণ \(A \neq B\))

2) B + π/4 = π - 17π/36

⇒ B + π/4 = (36π - 17π)/36

⇒ B + π/4 = 19π/36

⇒ B = 19π/36 - π/4

⇒ B = (19π - 9π)/36

⇒ B = 10π/36 = 5π/18

অতএব, B = 5π/18। 🎉