If sinA + cos2A = \(\frac{1}{3}\)

cosA + sin2A = \(\frac{1}{4}\), then

find the value of (sin3A).

\(\frac{-263}{288}\)

Square both the equation:

⇒ sin²A + cos²2A + 2sinA cos2A =\(\frac{1}{9}\)

⇒ cos²A + sin²2A + 2cosA sin2A =\(\frac{1}{16}\)

Now, add the equations:

⇒ 1 + 1 + 2sin(2A + A) = \(\frac{1}{9}\)+\(\frac{1}{16}\)

⇒ 2 + 2sin3A =\(\frac{25}{144}\)

⇒ 2(sin3A) = \(\frac{25}{144}\)-2

⇒ sin3A = \(\frac{-263}{288}\)