Category Archives: Identities

November 26, 2024 · 7:07 am

Infinite Product Expansion of cos (x)

Remember

(1)   \begin{equation*}cos(x)=1-\frac{1}{2!}x^2+\frac{1}{4!}x^4-\frac{1}{6!}x^6+...\end{equation*}

We know that cos(x)=0 for odd integer multiples of \frac{\pi}{2}, i.e. \frac{\pi}{2}, \frac{3\pi}{2}, ..., which is \frac{(2n-1)\pi}{2} for n\neq 0

Hence,

    \begin{equation*}0=1-\frac{1}{2!}x^2+\frac{1}{4!}x^4-\frac{1}{6!}x^6+...\end{equation}

for x=\frac{(2n-1)\pi}{2}, n>0

We can factorise our cos(x) expansion

    \begin{equation*}(1-\frac{x^2}{r_1})(1-\frac{x^2}{r_2})...\end{equation}

We know r_1=\frac{\pi}{2}, r_2=\frac{3\pi}{2}, ...

    \begin{equation*}cos(x)=(1-\frac{x^2}{(\frac{\pi}{2})^2})(1-\frac{x^2}{(\frac{3\pi}{2})^2})...(1-\frac{x^2}{(\frac{(2n-1)\pi}{2})^2})\end{equation}

    \begin{equation*}cos(x)=\Pi_{n=1}^{\infty}(1-\frac{x^2}{(\frac{(2n-1)\pi}{2})^2})\end{equation}

    \begin{equation*}cos(x)=\Pi_{n=1}^{\infty}(1-\frac{4x^2}{(2n-1)^2\pi^2})\end{equation}

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October 1, 2024 · 7:31 am

Using De Moivre’s Theorem for Trigonometric Identities

We are going to use De Moivre’s theorem to prove trigonometric identities.

Remember, De Moivre’s Theorem

If z=r(cos(\theta)+isin(\theta)), then z^n=r^n(cos(n\theta)+isin(n\theta))

Or a shorter version z=rcis(\theta), then z^n=r^ncis(n\theta)

Now, let z=cos(\theta)+isin(\theta), find z+\frac{1}{z}

z+z^{-1}=cos(\theta)+isin(\theta)+cos(-\theta)+isin(-\theta)

Remember cos(\theta)=cos(\theta) and sin(-\theta)=-sin(\theta)

z+\frac{1}{z}=cos(\theta)+isin(\theta)+cos(\theta)-isin(\theta)

z+\frac{1}{z}=2cos(\theta)

It is the same for z^n+\frac{1}{z^n}

z^n+z^{-n}=cos(n\theta)+isin(n\theta)+cos(-n\theta)+isin(-n\theta)

z^n+\frac{1}{z^n}=2cos(n\theta)

Prove cos(2\theta)=2cos^2(\theta)-1
LHS=\frac{1}{2}(z^2+\frac{1}{z^2})
LHS=\frac{1}{2}(z^2+\frac{1}{z^2})+z\times\frac{1}{z}-z\times\frac{1}{z}
LHS=\frac{1}{2}(z^2+2z\times\frac{1}{z}+\frac{1}{z^2})-z\times\frac{1}{z}
LHS=\frac{1}{2}(z+\frac{1}{z})^2-1
LHS=\frac{1}{2}(2cos(\theta))^2-1
LHS=\frac{1}{2}(4cos^2(\theta))-1
LHS=2cos^2(\theta)-1
LHS=RHS

We can do something similar with sine.

z-\frac{1}{z}=cos(\theta)+isin(\theta)-(cos(-\theta)+isin(-\theta))

z-\frac{1}{z}=cos(\theta)+isin(\theta)-(cos(-\theta)+isin(-\theta))

z-\frac{1}{z}=cos(\theta)+isin(\theta)-(cos(\theta)-isin(\theta))

z-\frac{1}{z}=cos(\theta)+isin(\theta)-cos(\theta)+isin(\theta)

z-\frac{1}{z}=2isin(\theta)

Hence z^n-\frac{1}{z^n}=2isin(n\theta)

Prove sin(2\theta)=2sin(\theta)cos(\theta)
LHS=sin(2\theta)
LHS=\frac{1}{2i}(z^2-\frac{1}{z^2})
LHS=\frac{1}{2i}(z-\frac{1}{z})(z+\frac{1}{z})
LHS=\frac{1}{2i}(2isin(\theta)(2cos(\theta))
LHS=sin(\theta)2cos(\theta)
LHS=2sin(\theta)cos(\theta)
LHS=RHS

Let’s find an identity for cos(3\theta)

cos(3\theta)=\frac{1}{2}(z^3+\frac{1}{z^3})

=\frac{1}{2}(z^3+\frac{1}{z^3}+3z^2\times\frac{1}{z}+3z\times\frac{1}{z^2}-3z^2\times\frac{1}{z}-3z\times\frac{1}{z^2})

=\frac{1}{2}((z+\frac{1}{z})^3-3z-\frac{3}{z})

=\frac{1}{2}((z+\frac{1}{z})^3-3(z+\frac{1}{z}))

=\frac{1}{2}(2cos(\theta))^3-3(2cos(\theta)))

=\frac{1}{2}(8cos^3(\theta)-6cos(\theta))

=4cos^3(\theta)-3cos(\theta)

\therefore cos(3\theta)=4cos^3(\theta)-3cos(\theta)

And sin(3\theta)?

sin(3\theta)=\frac{1}{2i}(z^3-\frac{1}{z^3})

=\frac{1}{2i}(z^3-\frac{1}{z^3}-3z^2\times\frac{1}{z}+3z\times\frac{1}{z^2}+3z^2\times\frac{1}{z}-3z\times\frac{1}{z^2}

=\frac{1}{2i}((z-\frac{1}{z})^3+3z-\frac{3}{z})

=\frac{1}{2i}(2isin(\theta))^3+3(z-\frac{1}{z}))

=\frac{1}{2i}(-8isin^3(\theta)+6isin(\theta))

=-4sin^3(\theta)+3sin(\theta)

\therefore sin(3\theta)=3sin(\theta)-4sin^3(\theta)

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