Question

If f(x) = sin x and g(x) = cos x, determine the following.

f (4n + 1)(x)=
g(4n + 1)(x)=
f (4n + 2)(x)=
g(4n + 2)(x)=
f (4n + 3)(x)=
g(4n + 3)(x)=
f (4n + 4)(x)=
g(4n + 4)(x)=

Now, use your answers to calculate F^(157)(x) when
F(x) = 4 sin x − 3 cos x.

F(157)(x)=

Answer 1

what does f(4n+1)(x) mean?

Answer 2

is this meant to be \[f^{(4n+1)}(x) \text{ where this means the } (4n+1)th \text { derivative of} f\]

Answer 3

\[f(x)=\sin(x)\\f'(x)=\cos(x)\\f''(x)=-\sin(x)\\f'''(x)=-\cos(x)\\f''''(x)=\sin(x)=f(x)\]
Notice how it loops, right?

So here's the answer to first one: \(f^{(4n+1)}(x) = f'(x) = \cos(x)\)

Is it clear? Can you handle the rest?