How do you find the equation of the tangent line to the curve #f(x) = sin cos(x)# at x = pi/2?

Question

Cameron Alexander · Accepted Answer

For this periodic-wave-function, with period #2pi, (pi/2. 0)# is a tangent-crossing-curve point of inflexion. The equation of the tangent is #x + y = pi/2#.

#f'(x) = - sin x cos (cos x)#. #f'(pi/2)=-sin (pi/2) cos (cos( pi/2))=(-1)(1)=-1# #f''(x)=-cos x cos(cos x)-sin x (-sin(cos x)(-sin x)#. #f''(pi/2)=0#. So, for this periodic-wave-function, with period #2pi, (pi/2. 0)# is a tangent-crossing-curve point of inflexion. The slope of the tangent is #-1#. The equation of the tangent is #x + y = pi/2#. An idiosyncrasy of this graph is that the amplitude is not 1. It is sin(1 radian)=0.84147, nearly. This is evident from #|cos x|<=1, and so, |f(x)|<=|sin( 1 radian )|<=0.841471#. One wave of this periodic graph is given for #x in [0, 2pi]#.

Ezra Adams · Answer

To find the equation of a line, we need a point and a slope. #f(x) = sin(cos(x))# Find the #y# coordinate 
of the point on the curve where #x=pi/2#. #y = f(pi/2) = sin(cos(pi/2)) = sin(0)=0#. So the point is #(pi/2,0)# Find the slope 
of the tangent line at #x=pi/2#. Use the derivative.  Use the chain rule to differentiate: #f'(x) = cos(cos(x))(d/dx(cos(x))) = cos(cos(x))(-sin(x))# At #pi/2#, the slope of the tangent line is  #m = f'(pi/2) = cos(cos(pi/2))(-sin(pi/2))# # = cos(0)(-1) = (1)(-1) = -1# Find an equation of the line  with slope #m=-1# through the point #(pi/2,0)# There are various answers possible. #y-0 = -1(x-pi/2)# #y = -x+pi/2# #x+y=pi/2#

Charles Arocha · Answer

undefined To find the equation of the tangent line to the curve f(x) = sin(cos(x)) at x = pi/2, we need to find the derivative of the function at that point and then use the point-slope form of a line.

First, we find the derivative of f(x) using the chain rule. The derivative of sin(cos(x)) with respect to x is -sin(x) * cos(cos(x)).

Next, we substitute x = pi/2 into the derivative to find the slope of the tangent line at that point. The slope is -sin(pi/2) * cos(cos(pi/2)) = -1 * cos(0) = -1.

Now, we have the slope of the tangent line. To find the equation of the line, we use the point-slope form: y - y1 = m(x - x1), where (x1, y1) is the point on the curve.

Since we are given x = pi/2, we substitute this value into the original function to find the corresponding y-coordinate. f(pi/2) = sin(cos(pi/2)) = sin(0) = 0.

Therefore, the point on the curve is (pi/2, 0).

Using the point-slope form, we have y - 0 = -1(x - pi/2).

Simplifying, we get y = -x + pi/2 as the equation of the tangent line to the curve f(x) = sin(cos(x)) at x = pi/2.