If x = s i n θ , y = s i n 3 θ then d 2 y d x 2 at θ = π 2 is ….

Step-by-step solution

We have , x = s i n θ , y = s i n 3 θ ∴ d y d x = d y d θ d x d θ = 3 s i n 2 θ c o s θ c o s θ ⇒ d y d x = 3 s i n 2 θ Now, d 2 y d x 2 = 3 2 s i n θ c o s θ d θ d x = 3 2 s i n θ c o s θ 1 c o s θ = 6 s i n θ at θ = π 2 , d 2 y d x 2 = 6 sin ⁡ π 2 = 6 × 1 = 6