Question

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

Answer 1

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`.

Answer 2

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`.