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

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:

$\frac{d}{\mathrm{dx}} \left(\sin \left(x\right)\right) = \cos \left(x\right) \text{ "=>" } \frac{d}{\mathrm{dx}} \left(\sin \left(g \left(x\right)\right)\right) = \cos \left(g \left(x\right)\right) \cdot g ' \left(x\right)$

$\frac{d}{\mathrm{dx}} \left(\cos \left(x\right)\right) = - \sin \left(x\right) \text{ "=>" } \frac{d}{\mathrm{dx}} \left(\cos \left(h \left(x\right)\right)\right) = - \sin \left(h \left(x\right)\right) \cdot h ' \left(x\right)$

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

$f ' \left(x\right) = \cos \left(\frac{\pi}{2} - x\right) \frac{d}{\mathrm{dx}} \left(\frac{\pi}{2} - x\right) - \left(- \sin \left(\pi - x\right)\right) \frac{d}{\mathrm{dx}} \left(\pi - x\right)$

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

$f ' \left(x\right) = - \cos \left(\frac{\pi}{2} - x\right) - \sin \left(\pi - x\right)$

Now, to see if the function is increasing or decreasing, we must find the value of the derivative at $x = \frac{\pi}{3}$.

• If $f ' \left(\frac{\pi}{3}\right) < 0$, then $f \left(x\right)$ is decreasing at $x = \frac{\pi}{3}$.
• If $f ' \left(\frac{\pi}{3}\right) > 0$, then $f \left(x\right)$ is increasing at $x = \frac{\pi}{3}$.

The value of the derivative at $x = \frac{\pi}{3}$ is:

$f ' \left(\frac{\pi}{3}\right) = - \cos \left(\frac{\pi}{2} - \frac{\pi}{3}\right) - \sin \left(\pi - \frac{\pi}{3}\right) = - \cos \left(\frac{\pi}{6}\right) - \sin \left(\frac{2 \pi}{3}\right)$

We could continue and find the exact value, but we know that cosine is positive in the first quadrant, where $\frac{\pi}{6}$ is, and that sine is positive in the second quadrant, where $\frac{2 \pi}{3}$ is, so we will have a negative result.

Since $f ' \left(\frac{\pi}{3}\right)$ is negative, the function must be decreasing at $x = \frac{\pi}{3}$.

Feb 14, 2016

Decreasing.

#### Explanation:

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

• $\sin \left(a - b\right) = \sin \left(a\right) \cos \left(b\right) - \cos \left(a\right) \sin \left(b\right)$
• $\cos \left(a - b\right) = \cos \left(a\right) \cos \left(b\right) + \sin \left(a\right) \sin \left(b\right)$

The function can then be written as

$f \left(x\right) = \sin \left(\frac{\pi}{2}\right) \cos \left(x\right) - \cos \left(\frac{\pi}{2}\right) \sin \left(x\right) - \left(\cos \left(\pi\right) \cos \left(x\right) + \sin \left(\pi\right) \sin \left(x\right)\right)$

Simplify.

$f \left(x\right) = \left(1\right) \cos \left(x\right) - \left(0\right) \sin \left(x\right) - \left(\left(- 1\right) \cos \left(x\right) + \left(0\right) \sin \left(x\right)\right)$

$f \left(x\right) = \cos \left(x\right) + \cos \left(x\right) = 2 \cos \left(x\right)$

Differentiation is now simple if you know that $\frac{d}{\mathrm{dx}} \left(\cos \left(x\right)\right) = - \sin \left(x\right)$.

$f ' \left(x\right) = - 2 \sin \left(x\right)$

The value of the derivative at $x = \frac{\pi}{3}$ is

$f ' \left(\frac{\pi}{3}\right) = - 2 \sin \left(\frac{\pi}{3}\right) = - \sqrt{3}$

Since this is negative, the function is decreasing at $x = \frac{\pi}{3}$.