How do you find the linearization at a=pi/6 of #f(x)=sin2x#? Calculus Applications of Derivatives Using the Tangent Line to Approximate Function Values 1 Answer Jim H Jan 17, 2018 #L(x) = sqrt3/2+(x-pi/6)# Explanation: #L(x) = f(a)+f'(a)(x-a)# #f(pi/6) = sin(pi/3) = sqrt3/2# #f'(x) = 2cos2x#, so #f'(pi/6) = 2cos(pi/3) = 1# #L(x) = sqrt3/2+(x-pi/6)# Answer link Related questions How do you find the linear approximation of #(1.999)^4# ? How do you find the linear approximation of a function? How do you find the linear approximation of #f(x)=ln(x)# at #x=1# ? How do you find the tangent line approximation for #f(x)=sqrt(1+x)# near #x=0# ? How do you find the tangent line approximation to #f(x)=1/x# near #x=1# ? How do you find the tangent line approximation to #f(x)=cos(x)# at #x=pi/4# ? How do you find the tangent line approximation to #f(x)=e^x# near #x=0# ? How do you use the tangent line approximation to approximate the value of #ln(1003)# ? How do you use the tangent line approximation to approximate the value of #ln(1.006)# ? How do you use the tangent line approximation to approximate the value of #ln(1004)# ? See all questions in Using the Tangent Line to Approximate Function Values Impact of this question 2204 views around the world You can reuse this answer Creative Commons License