Equivalence of Definitions of Sine and Cosine

From ProofWiki
Jump to navigation Jump to search


Theorem

The definitions for sine and cosine are equivalent.

That is:

$\displaystyle \sin x = \sum_{n \mathop = 0}^\infty \left({-1}\right)^n \frac {x^{2 n + 1} } {\left({2 n + 1}\right)!} \iff \sin x = \frac {\text{Opposite}} {\text{Hypotenuse}}$
$\displaystyle \cos x = \sum_{n \mathop = 0}^\infty \left({-1}\right)^n \frac {x^{2 n} } {\left({2 n}\right)!} \iff \cos x = \frac {\text{Adjacent}} {\text{Hypotenuse}}$


Proof

Let $s \left({x}\right): \R \to \R$, $c \left({x}\right): \R \to \R$ be two functions that satisfy:

$(1): \quad s' \left({x}\right) = c \left({x}\right)$
$(2): \quad c' \left({x}\right) = -s \left({x}\right)$
$(3): \quad s \left({0}\right) = 0$
$(4): \quad c \left({0}\right) = 1$
$(5): \quad \forall x: s^2 \left({x}\right) + c^2 \left({x}\right) = 1$

where $s'$ denotes the derivative with respect to $x$.


Let $f \left({x}\right): \R \to \R$, $g \left({x}\right): \R \to \R$ also be two functions that satisfy:

$(1): \quad f' \left({x}\right) = g \left({x}\right)$
$(2): \quad g' \left({x}\right) = -f \left({x}\right)$
$(3): \quad f \left({0}\right) = 0$
$(4): \quad g \left({0}\right) = 1$
$(5): \quad \forall x: f^2 \left({x}\right) + g^2 \left({x}\right) = 1$


It will be shown that:

$f \left({x}\right) = s \left({x}\right)$

and:

$c \left({x}\right) = g \left({x}\right)$


Define:

$h \left({x}\right) = \left({c \left({x}\right) - g \left({x}\right)}\right)^2 + \left({s \left({x}\right) - f \left({x}\right)}\right)^2$

Notice that:

$\left({\forall x: h \left({x}\right) = 0}\right) \iff \left({\forall x: c \left({x}\right) = g \left({x}\right), s \left({x}\right) = f \left({x}\right)}\right)$


Then:

\(\displaystyle h \left({x}\right)\) \(=\) \(\displaystyle c^2 \left({x}\right) - 2 c \left({x}\right) g \left({x}\right) + g^2 \left({x}\right) + s^2 \left({x}\right) - 2 s \left({x}\right) f \left({x}\right) + f^2 \left({x}\right)\)
\(\displaystyle \) \(=\) \(\displaystyle 2 - 2 \left({c \left({x}\right) g \left({x}\right) + s \left({x}\right) f \left({x}\right)}\right)\) Property $(5)$


By taking $h' \left({x}\right)$:

\(\displaystyle h' \left({x}\right)\) \(=\) \(\displaystyle -2 \left({c \left({x}\right) \left({-f \left({x}\right)}\right) + g \left({x}\right) \left({-s \left({x}\right)}\right) + s \left({x}\right) g \left({x}\right) + f \left({x}\right) c \left({x}\right)}\right)\) Properties $(1)$ and $(2)$ and Product Rule
\(\displaystyle \) \(=\) \(\displaystyle 0\)

So $h \left({x}\right)$ is a constant function:

$h \left({x}\right) = k$



Also:

\(\displaystyle h \left({0}\right)\) \(=\) \(\displaystyle \left({1 - 1}\right)^2 + \left({0 - 0}\right)^2\) Properties $(3)$ and $(4)$
\(\displaystyle \) \(=\) \(\displaystyle 0\)

Since $h \left({x}\right)$ is constant, then:

$\forall x: h \left({x}\right) = 0$

Then:

$c \left({x}\right) = g \left({x}\right)$

and:

$s \left({x}\right) = f \left({x}\right)$

By:

Derivative of Sine Function
Derivative of Cosine Function
Sine of Zero is Zero
Cosine of Zero is One
Sum of Squares of Sine and Cosine

both definitions satisfy all these properties.

Therefore they must be the same.

$\blacksquare$