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

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

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

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

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

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

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

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

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

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