How do you find the equation of the tangent line #y=sinx# at #(pi/6, 1/2)#?

Question

Isabella Ackerman · Accepted Answer

#y-1/2=sqrt3/2(x-pi/6)#

The slope of the tangent line to a function #y# at #x=a# is found by calculating the value of #dy/dx#, the derivative of #y#, at #x=a#.

Where

#y=sinx#

the derivative is given by

#dy/dx=cosx#

The slope of the tangent line to #y# at #x=pi/6# is found by evaluating the derivative of #y# at #x=pi/6#:

#m=dy/dx|_(x=pi/6)=cos(pi/6)=sqrt3/2#

The slope of the tangent line is #sqrt3/2#. Writing the equation of the line that passes through #(pi/6,1/2)# with slope #sqrt3/2# in point-slope form, we get:

#y-y_1=m(x-x_1)#

#y-1/2=sqrt3/2(x-pi/6)#

Cameron Alexander · Answer

Differentiate y and evaluate #dy/dx# at #x=pi/6#

The equation of the tangent line would then be #y=ax+b#, where #a# is the derivative at #x=pi/6# and #b# can be solved by setting #y=1/2, x=pi/6#

The equation would be #y=sqrt(3)/2x+1/2-(sqrt(3)pi)/12# Let the equation of the tangent line be #y=ax+b# #dy/dx=cos x# #a=cos pi/6=sqrt(3)/2# #:. y=sqrt(3)/2x+b# #1/2=sqrt(3)/2*pi/6+b# #b=1/2-(sqrt(3)pi)/12# Hence the equation of the tangent line is #y=sqrt(3)/2x+1/2-(sqrt(3)pi)/12# You can verify this answer visually too graph{(y-sqrt(3)/2x-1/2+(sqrt(3)pi)/12)(y-sin(x))=0 [-1.259, 1.781, -0.477, 1.04]} The reason the equation of a tangent line is as shown above is because in a linear function, #y=ax+b#, a represents the gradient / slope of the line and b represents the y-intercept. By definition, the gradient of a tangent line is equal to the slope of a curve at the point where the tangent line meets the curve. Hence, #a=dy/dx# when #x=pi/6# and the rest of the equation can be derived through algebra

Samantha Adkison · Answer

undefined To find the equation of the tangent line to the curve y = sin(x) at the point (π/6, 1/2), we need to determine the slope of the tangent line at that point.

The slope of the tangent line can be found by taking the derivative of the function y = sin(x) with respect to x.

The derivative of sin(x) is cos(x).

Evaluating cos(π/6), we get √3/2.

Therefore, the slope of the tangent line at (π/6, 1/2) is √3/2.

Using the point-slope form of a linear equation, the equation of the tangent line is y - 1/2 = (√3/2)(x - π/6).