Given, y=sin‌3‌x . . . (i) At x=π∕4,y=1∕√2 ∴ Point is (π∕4,1∕√2) Now, differentiating Eq. (i) w.r.t x, we get ‌y′=3‌cos‌3‌x ⇒‌‌(y′)x=‌
Ï€
4
=3‌cos‌
3Ï€
4
=−‌
3
√2
∴‌ Slope of normal ‌=√2∕3‌‌[m1⋅m2=−1] ‌ Equation of normal is ‌(y−‌