Previously we showed that:
\(e^x=\sum_{i=0}^\infty \dfrac{x^i}{i!}\)
Consider:
\(e^{i\theta }\)
\(e^{i\theta }=\sum_{j=0}^\infty \dfrac{(i\theta )^j}{j!}\)
\(e^{i\theta }=[\sum_{j=0}^\infty \dfrac{(\theta )^{4j}}{(4j)!}-\sum_{j=0}^\infty \dfrac{(\theta )^{4j+2}}{(4j+2)!}]+i[\sum_{j=0}^\infty \dfrac{(\theta )^{4j+1}}{(4j+1)!}-\sum_{j=0}^\infty \dfrac{(\theta )^{4j+3}}{(4j+3)!}]\)
We then use this to define \(\sin\) and \(\cos\) functions.
\(\cos (\theta ):=\sum_{j=0}^\infty \dfrac{(\theta )^{4j}}{(4j)!}-\sum_{j=0}^\infty \dfrac{(\theta )^{4j+2}}{(4j+2)!}\)
\(\sin (\theta ):=\sum_{j=0}^\infty \dfrac{(\theta )^{4j+1}}{(4j+1)!}-\sum_{j=0}^\infty \dfrac{(\theta )^{4j+3}}{(4j+3)!}\)
So:
\(e^{i\theta }=\cos (\theta )+i\sin (\theta )\)
We know
\(e^{i\theta }=\cos (\theta )+i\sin (\theta )\)
\(e^{-i\theta }=\cos (\theta )-i\sin (\theta )\)
So
\(e^{i\theta }+e^{-i\theta }=cos (\theta )+i\sin (\theta )+\cos (\theta )-i\sin (\theta )\)
\(\cos (\theta )=\dfrac{e^{i\theta }+e^{-i\theta }}{2}\)
And
\(e^{i\theta }-e^{-i\theta }=cos (\theta )+i\sin (\theta )-\cos (\theta )+i\sin (\theta )\)
\(\sin (\theta )=\dfrac{e^{i\theta }-e^{-i\theta }}{2i}\)
Sine is an odd function.
\(\sin (-\theta )=-\sin (\theta )\)
Cosine is an even function.
\(\cos (-\theta )=\cos (\theta )\)
\(e^{i\theta }=\cos (\theta )+i\sin (\theta )\)
Let \(\theta = nx\):
\(e^{inx }=\cos (nx )+i\sin (nx )\)
\((e^{ix})^n=\cos (nx )+i\sin (nx )\)
\((\cos (x)+i\sin (x))^n=\cos (nx )+i\sin (nx )\)
\(\sin (\alpha +\beta )=\sin (\alpha )\cos(\beta )+\cos(\alpha )\sin (\beta )\)
\(\cos (\alpha +\beta )=\cos (\alpha )\cos(\beta )-\sin(\alpha )\sin (\beta )\)
We know that:
\(a\sin(bx+c)=a\sin(bx)\cos(c)+a\sin(c)\cos(bx)\)
So:
\(a\sin(bx+c)+d\sin(bx+e)=a\sin(bx)\cos(c)+a\sin(c)\cos(bx)+ d\sin(bx)\cos(e)+d\sin(e)\cos(bx)\)
We know that:
\(\sin(\theta )=\dfrac{e^{i\theta }-e^{-i\theta }}{2i}\)
So:
\(a\sin(bx+c)+d\sin(bx+f)=a\dfrac{e^{i(bx+c)}-e^{-i(bx+c)}}{2i}+d\dfrac{e^{i(bx+f)}-e^{-i(bx+f)}}{2i}\)
\(a\sin(bx+c)+d\sin(bx+f)=\dfrac{a(e^{i(bx+c)}-e^{-i(bx+c)})+d(e^{i(bx+f)}-e^{-i(bx+f)})}{2i}\)
\(a\sin(bx+c)+d\sin(bx+f)=\dfrac{a(e^{ibx}e^{ic}-e^{-ibx}e^{-ic})+d(e^{ibx}e^{if}-e^{-ibx}e^{-if)})}{2i}\)
\(a\sin(bx+c)+d\sin(bx+f)=\dfrac{(e^{ibx}(ae^{ic}+de^{if})-e^{-ibx}(ae^{-c}+d^{-if})}{2i}\)
\(a_i\sin(b_ix+c_i)+a_j\sin(b_jx+c_j)=a_i\sin(b_ix+c_i)+a_j\sin(b_ix+c_j)\)
\(a_i\sin(b_ix+c_i)+a_j\sin(b_jx+c_j)=a_i\sin(b_ix)\cos(c_i)+a_i\sin(c_i)\cos(b_ix)+a_j\sin(b_ix)\cos(c_j)+a_j\sin(c_j)\cos(b_ix)\)
Note that with imaginary numbers we can reverse all \(i\)s. So:
\(e^{i\theta }=\cos (\theta )+i\sin (\theta )\)
\(e^{-i\theta }=\cos (\theta )-i\sin (\theta )\)
\(e^{i\theta }e^{-i\theta }=(\cos (\theta )+i\sin (\theta ))(\cos (\theta )-i\sin (\theta ))\)
\(e^{i\theta }e^{-i\theta }=\cos (\theta )^2+\sin (\theta )^2\)
\(e^{i\theta }e^{-i\theta }=e^{i\theta -i\theta }=e^0=1\)
So:
\(\cos (\theta )^2+\sin (\theta )^2=1\)
Note that if \(\cos (\theta )^2=0\), then \(\sin (\theta )^2=\pm 1\)
That is, if the real part of \(e^{i\theta }\) is \(0\), the imaginary part is \(\pm 1\). And visa versa.
Similarly if the derivative of the real part of \(e^{i\theta }\) is \(0\), the imaginary part is \(\pm 1\). And visa versa.
Note that these functions are linked in their derivatives.
\(\dfrac{\delta }{\delta \theta }\cos (\theta )=\sum_{j=0}^\infty \dfrac{(\theta )^{(4j+3)}}{(4j+3)!}-\sum_{j=0}^\infty \dfrac{(\theta )^{4j+1}}{(4j+1)!}\)
\(\dfrac{\delta }{\delta \theta }\cos (\theta )=-\sin (\theta )\)
Similarly:
\(\dfrac{\delta }{\delta \theta }\sin (\theta )=cos(\theta )\)
\(\dfrac{\delta^2 }{\delta \theta^2}\sin (\theta )=-\sin (\theta )\)
\(\dfrac{\delta^2 }{\delta \theta^2}\cos (\theta )=-\cos (\theta )\)
So for either of:
\(y=\cos (\theta )\)
\(y=\sin (\theta )\)
We know that
\(\dfrac{\delta^2 }{\delta \theta^2}y(\theta )=-y(\theta )\)
Consider \(\theta =0\).
\(e^{i.0}=\cos (0)+i\sin (0)\)
\(1=\cos (0)+i\sin (0)\)
\(\sin (0)=0\)
\(\cos (0)=1\)
Similarly we know that the derivative:
\(\sin'(0)=\cos(0)=1\)
\(\cos'(0)=-\sin(0)=0\)
Consider \(\cos(\theta )\).
As \(\cos (0)\) is static at \(\theta =0\), and is positive, it will fall until \(\cos (\theta )=0\).
While this is happening, \(\sin (\theta )\) is increasing. As:
\(\cos (\theta )^2+\sin (\theta )^2=1\)
\(\sin (\theta )\) will equal \(1\) where \(\cos (\theta )=0\).
Due to symmetry this will repeat \(4\) times.
Let’s call the length of this period \(\tau\).
Where \(\theta =\tau *0\)
\(\cos (\theta )=1\)
\(\sin (\theta )=0\)
Where \(\theta =\tau *\dfrac{1}{4}\)
\(\cos (\theta )=0\)
\(\sin (\theta )=1\)
Where \(\theta =\tau *\dfrac{2}{4}\)
\(\cos (\theta )=-1\)
\(\sin (\theta )=0\)
Where \(\theta =\tau *\dfrac{3}{4}\)
\(\cos (\theta )=0\)
\(\sin (\theta )=-1\)
Note that \(\sin(\theta + \dfrac{\tau }{4})=\cos(\theta )\)
Note that \(\sin (\theta )=\cos (\theta )\) at
\(\tau *\dfrac{1}{8}\)
\(\tau *\dfrac{5}{8}\)
And that all these answers loop. That is, add any integer multiple of \(\tau\) to \(\theta\) and the results hold.
\(e^{i\theta } = e^{i\theta +n\tau }\)
\(n \in \mathbb{N}\)
\(e^{i\theta } = \cos(\theta )+i\sin(\theta )\)
\(e^{i\theta } = \cos(\theta +n\tau )+i\sin(\theta +n\tau )\)
\(e^{i\theta } = e^{i(\theta +n\tau )}\)
Relationship between cos and sine
\(\sin(x+\dfrac{\pi }{2})=\cos(x)\)
\(\cos(x+\dfrac{\pi }{2})=-\sin(x)\)
\(\sin(x+\pi )=-\sin(x)\)
\(\cos(x+\pi )=-\cos(x)\)
\(\sin(x+\tau )=\sin(x)\)
\(\cos(x+\tau )=\cos(x)\)