Is #f(x)=sin(pi/2-x)-cos(pi-x)# increasing or decreasing at #x=pi/3#?

2 Answers
Feb 14, 2016

Decreasing.

Explanation:

We could use the chain rule to differentiate this. The chain rule, when specifically applied to the cosine and sine functions, is as follows:

#d/dx(sin(x))=cos(x)" "=>" "d/dx(sin(g(x)))=cos(g(x))*g'(x)#

#d/dx(cos(x))=-sin(x)" "=>" "d/dx(cos(h(x)))=-sin(h(x))*h'(x)#

When we apply this to the given function, we see that

#f'(x)=cos(pi/2-x)d/dx(pi/2-x)-(-sin(pi-x))d/dx(pi-x)#

Note that the derivative of each of these terms is #-1#.

#f'(x)=-cos(pi/2-x)-sin(pi-x)#

Now, to see if the function is increasing or decreasing, we must find the value of the derivative at #x=pi/3#.

  • If #f'(pi/3)<0#, then #f(x)# is decreasing at #x=pi/3#.
  • If #f'(pi/3)>0#, then #f(x)# is increasing at #x=pi/3#.

The value of the derivative at #x=pi/3# is:

#f'(pi/3)=-cos(pi/2-pi/3)-sin(pi-pi/3)=-cos(pi/6)-sin((2pi)/3)#

We could continue and find the exact value, but we know that cosine is positive in the first quadrant, where #pi/6# is, and that sine is positive in the second quadrant, where #(2pi)/3# is, so we will have a negative result.

Since #f'(pi/3)# is negative, the function must be decreasing at #x=pi/3#.

Feb 14, 2016

Decreasing.

Explanation:

We could also expand these functions with the sine and cosine angle subtraction formulas.

  • #sin(a-b)=sin(a)cos(b)-cos(a)sin(b)#
  • #cos(a-b)=cos(a)cos(b)+sin(a)sin(b)#

The function can then be written as

#f(x)=sin(pi/2)cos(x)-cos(pi/2)sin(x)-(cos(pi)cos(x)+sin(pi)sin(x))#

Simplify.

#f(x)=(1)cos(x)-(0)sin(x)-((-1)cos(x)+(0)sin(x))#

#f(x)=cos(x)+cos(x)=2cos(x)#

Differentiation is now simple if you know that #d/dx(cos(x))=-sin(x)#.

#f'(x)=-2sin(x)#

The value of the derivative at #x=pi/3# is

#f'(pi/3)=-2sin(pi/3)=-sqrt3#

Since this is negative, the function is decreasing at #x=pi/3#.