Question

Show that the difference quotient for $f\left(x\right)=\cos x$is given by

role="math" localid="1646431168099" $$\begin{gathered} \frac{f(x+h)-f\left(x\right)}{h}=\frac{\cos (x+h)-\cos \left(x\right)}{h} \\ =-\sin x·\frac{\sin h}{h}-\cos x·\frac{1-\cos h}{h} \end{gathered}$$

Short answer

The difference quotient for $f\left(x\right)=\cos \left(x\right)$is determined by using the Sum formula for the cosine function as

$$\begin{gathered} \frac{f(x+h)-f\left(x\right)}{h}=\frac{\cos (x+h)-\cos \left(x\right)}{h} \\ =\frac{(\cos x)(\cos h)-(\sin x)(\sin h)-\cos \left(x\right)}{h} \\ =-\sin x·\frac{\sin h}{h}-\cos x·\frac{1-\cos h}{h} \end{gathered}$$

Step-by-step solution

Step 1. Given data 

The given function is

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

the equation that needs to prove is

$$\begin{gathered} \frac{f(x+h)-f\left(x\right)}{h}=\frac{\cos (x+h)-\cos \left(x\right)}{h} \\ =-\sin x·\frac{\sin h}{h}-\cos x·\frac{1-\cos h}{h} \end{gathered}$$

Step 2. Derivation 

Use Sum formula for the cosine function

$$\begin{gathered} \frac{f(x+h)-f\left(x\right)}{h}=\frac{\cos (x+h)-\cos \left(x\right)}{h} \\ =\frac{(\cos x)(\cos h)-(\sin x)(\sin h)-\cos \left(x\right)}{h} \\ =\frac{-(\sin x)(\sin h)}{h}-\frac{\cos \left(x\right)-(\cos x)(\cos h)}{h} \\ =-(\sin x)\frac{(\sin h)}{h}-\frac{(\cos x)(1-(\cos h\left)\right)}{h} \\ =-\sin x·\frac{\sin h}{h}-\cos x·\frac{1-\cos h}{h} \end{gathered}$$