can someone help me out with this question? I am at a complete loss...
define these two functions: \begin{align} f(x, \delta) &= \cos \left(x + \delta\right) - \cos(x) \\[.5em] g(x, \delta) &= -2\sin\left(x + \left( \frac{\delta}{2}\right)\right) \sin\left(\frac{\delta}{2}\right) \end{align} How are $f(x,\delta)/\delta$ and $g(x, \delta)/\delta$ good approximations of $-\sin(x)$ as $\delta$ becomes smaller and smaller?
Secondly, how can we prove, analytically, that $f = g$ using trig identities only?
Would highly appreciate any insight into this question.
You should use two kind of trigonometric identities. First one is $$ \cos(a+b)=\cos(a)\cos(b) -\sin(a)\sin(b)$$ and the second is $$\cos(a) - \cos(b)= -2\sin\left(\frac{a+b}{2}\right)\sin\left(\frac{a-b}{2}\right)$$
The first one is able to prove that $f(x,\delta)$ tends to $\sin(x)$ when $\delta$ tends to $0$ by using the small angles approach: $\sin(\delta)\approx\delta$ and $\cos(\delta)\approx1$
$$f(x,\delta)=\cos(x+\delta)-\cos(x)=\cos(x)\cos(\delta) -\sin(x) \sin(\delta)-\cos(x) \approx -\delta \sin(x)$$
The second one is able to prove that $f$ and $g$ are equivalent by this replacement: $a=x+\delta$ and $b=\delta$ \begin{align} f(x,\delta)=\cos(x+\delta)-\cos(\delta) &=-2\sin\left(\frac{x+\delta+x}{\delta}\right)\sin\left(\frac{x+\delta-x}{2}\right) \\ &=-2\sin\left(x+\frac{\delta}{2}\right)\sin\left(\frac{\delta}{2}\right) =g(x,\delta) \end{align}