The sine and cosine functions, and identifying \(\pi\)

Sine and cosine

Defing sine and cosine using Euler’s formula

Euler’s formula

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:

\(e^{i\theta }=\cos (\theta )+i\sin (\theta )\)

Alternative formulae for sine and cosine

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 and cosine are odd and even functions

Sine is an odd function.

\(\sin (-\theta )=-\sin (\theta )\)

Cosine is an even function.

\(\cos (-\theta )=\cos (\theta )\)

De Moive’s formula

\(e^{i\theta }=\cos (\theta )+i\sin (\theta )\)

Let \(\theta = nx\):

\(e^{inx }=\cos (nx )+i\sin (nx )\)

\((e^{ix})^n=\cos (nx )+i\sin (nx )\)

\((\cos (x)+i\sin (x))^n=\cos (nx )+i\sin (nx )\)

Expanding sine and cosine

Expansion

\(\sin (\alpha +\beta )=\sin (\alpha )\cos(\beta )+\cos(\alpha )\sin (\beta )\)

\(\cos (\alpha +\beta )=\cos (\alpha )\cos(\beta )-\sin(\alpha )\sin (\beta )\)

Addition of sine and cosine

Adding waves with same frequency

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:

\(a\sin(bx+c)+d\sin(bx+f)=a\dfrac{e^{i(bx+c)}-e^{-i(bx+c)}}{2i}+d\dfrac{e^{i(bx+f)}-e^{-i(bx+f)}}{2i}\)

\(a\sin(bx+c)+d\sin(bx+f)=\dfrac{a(e^{i(bx+c)}-e^{-i(bx+c)})+d(e^{i(bx+f)}-e^{-i(bx+f)})}{2i}\)

\(a\sin(bx+c)+d\sin(bx+f)=\dfrac{a(e^{ibx}e^{ic}-e^{-ibx}e^{-ic})+d(e^{ibx}e^{if}-e^{-ibx}e^{-if)})}{2i}\)

\(a\sin(bx+c)+d\sin(bx+f)=\dfrac{(e^{ibx}(ae^{ic}+de^{if})-e^{-ibx}(ae^{-c}+d^{-if})}{2i}\)

\(a_i\sin(b_ix+c_i)+a_j\sin(b_jx+c_j)=a_i\sin(b_ix+c_i)+a_j\sin(b_ix+c_j)\)

\(a_i\sin(b_ix+c_i)+a_j\sin(b_jx+c_j)=a_i\sin(b_ix)\cos(c_i)+a_i\sin(c_i)\cos(b_ix)+a_j\sin(b_ix)\cos(c_j)+a_j\sin(c_j)\cos(b_ix)\)

Calculus of sine and cosine

Unity

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.

Sine and cosine are linked by their derivatives

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:

\(\dfrac{\delta }{\delta \theta }\sin (\theta )=cos(\theta )\)

Both sine and cosine oscillate

\(\dfrac{\delta^2 }{\delta \theta^2}\sin (\theta )=-\sin (\theta )\)

\(\dfrac{\delta^2 }{\delta \theta^2}\cos (\theta )=-\cos (\theta )\)

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 \(\theta =\tau *\dfrac{3}{4}\)

  • \(\cos (\theta )=0\)

  • \(\sin (\theta )=-1\)

Relationship between\( \cos (\theta )\) and \(\sin(\theta )\)

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.

\(e^{i\theta } = e^{i\theta +n\tau }\)

\(n \in \mathbb{N}\)

\(e^{i\theta } = \cos(\theta )+i\sin(\theta )\)

\(e^{i\theta } = \cos(\theta +n\tau )+i\sin(\theta +n\tau ) \)

\(e^{i\theta } = e^{i(\theta +n\tau )}\)

Calculus of trig

Relationship between cos and sine

\(\sin(x+\dfrac{\pi }{2})=\cos(x)\)

\(\cos(x+\dfrac{\pi }{2})=-\sin(x)\)

\(\sin(x+\pi )=-\sin(x)\)

\(\cos(x+\pi )=-\cos(x)\)

\(\sin(x+\tau )=\sin(x)\)

\(\cos(x+\tau )=\cos(x)\)