#### To determine

**To find:** The linearization L(x) of the function at *a*.

#### Answer

The linearization L(x) of the function at a=π6 is L(x)=32x−π43+12.

#### Explanation

**Given:**

The function f(x)=sinx and a=π6.

**Proof:**

The linearization of f(x) at *a* is L(x)=f(a)+f′(a)(x−a) (1)

Substitute a=π6 in f(x)=sinx,

f(π6)=sin(π6)=12

The derivative of the function f(x)=sinx is computed as follows,

f′(x)=ddx(sinx)=cosx

Substitute a=π6 in f′(x),

f′(−2)=cos(π6)=32

Substitute π6 for *a* in the equation (1),

L(x)=f(π6)+f′(π6)(x−π6)

Substitute f(π6)=12 and f′(π6)=32,

L(x)=12+32(x−π6)=12+(32x−32π2×3)=32x−π43+12

Therefore, the linearization L(x) of the function at a=π6 is L(x)=32x−π43+12.