#### To determine

**To find:** The equation of the tangent line to the curve at the point.

#### Answer

The equation of the tangent line to the curve
y=sin2xcosx at
(π2,0) is
y=x−π2.

#### Explanation

**Given:**

The function is
y=sin2xcosx.

The point is
(π2,0).

**Result used:**

**Chain Rule**

If *h* is differentiable at *x* and *g* is differentiable at
h(x), then the composite function
F=g∘h defined by
F(x)=g(h(x)) is differentiable at *x* and
F′ is given by the product

F′(x)=g′(h(x))⋅h′(x) (1)

**Product Rule:**

If
f(x).and
g(x) are both differentiable function, then

ddx[f(x)g(x)]=f(x)ddx[g(x)]+g(x)ddx[f(x)] (2)

**Formula used:**

The equation of the tangent line at
(x1,y1) is,
y−y1=m(x−x1). (3)

where, *m* is the slope of the tangent line at
(x1,y1) and
m=dydx|x=x1.

**Calculation:**

The derivative of
y is
dydx, which is obtained as follows,

dydx=ddx(y)=ddx(sin2xcosx)

Apply the product rule as shown in equation (2).

dydx=sin2xddx[cosx]+cosxddx[sin2x]=sin2x(−sinx)+cosxddx[sin2x]

Use the chain rule as shown in equation (1).

dydx=sin2x(−sinx)+cosx(2sinx)ddx[sinx]=−sin3x+cosx(2sinx)(cosx)=−sin3x+2sinxcos2x

Therefore, the derivative of
y is
dydx=−sin3x+2sinxcos2x.

The slope of the tangent line at
(π2,0) is computed as follows,

m=dydx|x=π2=−sin3(π2)+2sin(π2)cos2(π2) =1+2(1)(0) =1

Thus, the slope of the tangent line at
(π2,0) is
m=1.

Substitute
(π2,0) for
(x1,y1) and
m=1 in equation (3),

y−0=1(x−π2)y=x−π2

Therefore, the equation of the tangent line to the curve
y at
(π2,0) is
y=x−π2.