Suppose $f$ and $g$ are non-constant, differentiable, real- valued functions defined on $(-\infty, \infty) .$ Furthermore, suppose that for each pair of real numbers $x$ and $y$

$f(x+y)=f(x) f(y)-g(x) g(y)$ and

$g(x+y)=f(x) g(y)+g(x) f(y)$

If $$f^{\prime}(0)=0,$ prove that $(f(x))^{2}+(g(x))^{2}=1$ for all $x$$

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## Recommended Questions

Show that if $f(x)$ and $g(x)$ are linear functions such that $f(0)=g(0)$ and $f(1)=g(1),$ then $f(x)=g(x).$

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$$E(x)=\frac{f(x)+f(-x)}{2}$$

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$$ \int_0^a f(x) g^{\prime\prime} (x) dx = f(a)g^\prime(a) - f^\prime(a)g(a) + \int_0^a f^{\prime\prime} (x) g(x) dx $$

Proof Let f and g be one-to-one functions. Prove that

\begin{equation}\begin{array}{l}{\text { (a) } f\circ g \text { is one-to-one. }} \\ {\text { (b) }(f \circ g)^{-1}(x)=\left(g^{-1} \circ f^{-1}\right)(x)}\end{array}\end{equation}

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$$f(x)=4-2 x$$

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$$[f(g(h(x)))]^{\prime}=f^{\prime}(g(h(x))) g^{\prime}(h(x)) h^{\prime}(x)$$

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$$

f(x)=\cos x \cdot \cot x \quad g(x)=\csc x-\sin x

$$

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$$

\int_{0}^{1} f(x) d x=\int_{0}^{1} f(1-x) d x

$$

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$$\int_{0}^{1} f(x) d x=\int_{0}^{1} f(1-x) d x$$

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Hint: Show that

$$

\lim _{(\Delta r, \Delta y) \rightarrow(0,0)} f(a+\Delta x, b+\Delta y)=f(a, b)

$$

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