If sinA + cos2A = \(\frac{1}{3}\)
cosA + sin2A = \(\frac{1}{4}\), then
find the value of (sin3A).
Answer & explanation
Correct answer: option 3
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 = \(\frac{-263}{288}\)