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De Moivre's theorem states that, for real $\theta$ and integer $n$,It follows from $\cos \theta +i\sin \theta =e^{i\theta }$ and exponent multiplication. Negative integers are allowed because this unit-modulus complex number is nonzero. For a general polar number $r{e}^{i\theta }$, its integer power also has the factor $r^n$.

Take the imaginary part of the eighth-power binomial expansion in De Moivre's theorem. With $s=\sin \theta$ and $c=\cos \theta$,Substitute $c^2=1-s^2$ and collect powers:Therefore the requested leading coefficient isThe other coefficients, if retained as an algebraic check, are $b=192,c=-80,d=8$. No division by $s$ or $c$ is needed, so the identity remains valid when either vanishes.

Take the imaginary part of De Moivre's theorem at exponent five. Writing $s=\sin \theta$ and $c=\cos \theta$ givesFor real $x\in [-1,1]$, put $x=\sin \theta$. The equation becomes $\sin 5\theta =1$, so $\theta =\pi /10+2\pi k/5$. The distinct sine values are $1$, $(\sqrt{5}-1)/4$, and $-(1+\sqrt{5})/4$. To establish completeness over the complex numbers as well, factor the polynomial:Thus all polynomial roots, including their multiplicities, areThe repeated sine values account for the double roots; there are no additional nonreal roots hidden by the trigonometric parametrization.

De Moivre's theorem states $(\cos x+i\sin x{)}^n=\cos (nx)+i\sin (nx)$ for integer $n$. For $n=3$, expanding the cube and taking its real part givesRearranging gives the required identity withNo division by $\cos x$ is needed, so the proof also covers its zeros.

By De Moivre's theorem, $(\cos \theta +i\sin \theta {)}^3=\cos 3\theta +i\sin 3\theta$. Expanding the cube and equating its real part givesThe last form uses ${\sin }^2\theta +{\cos }^2\theta =1$; comparing the imaginary parts similarly gives the corresponding sine identity.

Expand $(\cos \theta +i\sin \theta {)}^4$ using De Moivre's theorem, then divide its imaginary and real parts by ${\cos }^4\theta$. With $u=\tan \theta$, this givesThe quotient holds where its denominator and $\cos \theta$ are nonzero. At $\theta =\pi /8$, use the undivided real-part identity: $\cos 4\theta =0$ implies $u^4-6u^2+1=0$. Thus $u^2=3\pm 2\sqrt{2}$. Since $0<\tan (\pi /8)<1$, the smaller positive value applies:This limiting step avoids treating an undefined value of $\tan 4\theta$ as an ordinary finite number.

Put $w=e^{i\theta }$. By Euler's formula, $w-w^{-1}=2i\sin \theta$. The binomial theorem givesGroup opposite powers and use $w^j-w^{-j}=2i\sin (j\theta )$:Since $(2i{)}^7=-128i$, division by $-2i$ yieldsThis derivation is an application of the same exponential representation underlying De Moivre's theorem.

The polar form is $1+i\sqrt{3}=2e^{i\pi /3}$. By De Moivre's theorem,Therefore the real part and imaginary part are

For every integer $m$, De Moivre's theorem states $(\cos \theta +i\sin \theta {)}^m=\cos (m\theta )+i\sin (m\theta )$. Taking the real part for $m=5$ givesReplace ${\sin }^2\theta$ by $1-{\cos }^2\theta$ and expand:The resulting polynomial is the fifth Chebyshev polynomial.

De Moivre's theorem says that, for a real angle $\theta$ and an integer $m$,For a complex number $z=\rho (\cos \theta +i\sin \theta )$, the corresponding formula is $z^m={\rho }^m[\cos (m\theta )+i\sin (m\theta )]$, with $z\ne 0$ required for negative powers.

Put $q=e^{i\theta }$. If $\theta \notin 2\pi \mathbb{Z}$, the finite geometric series givesTo extract the real part, multiply numerator and denominator by $1-q^{-1}$. The denominator becomes $2(1-\cos \theta )$, and the numerator is $q-1-q^{n+1}+q^n$. Thus this finite trigonometric sum isAt $\theta \in 2\pi \mathbb{Z}$, the sum equals $n$; the printed quotient is undefined there, though its continuous limiting value is $n$.

Now take $\theta =2p\pi /(n+1)$ with $1\le p\le n$. This is not a multiple of $2\pi$, while $(n+1)\theta =2p\pi$ and $\cos (n\theta )=\cos (2p\pi -\theta )=\cos \theta$. Substitution gives

De Moivre's theorem states that $(\cos \theta +i\sin \theta {)}^n=\cos (n\theta )+i\sin (n\theta )$ for integer $n$. For $z\ne 0$, write its polar form as $z=r{e}^{i\theta }$, where $r=|z|>0$. If $w=s{e}^{i\phi }$ and $w^n=z$, equality of moduli gives $s=r^{1/n}$ and equality of arguments gives $n\phi =\theta +2\pi k$. Thus the distinct roots of a complex number areChanging the chosen argument of $z$ only permutes these roots. For $z=0$ the only root is zero. Taking $r=8$ and $\theta =\pi$ gives the cube rootsCubing any of them returns $-8$; geometrically they are equally spaced on the circle of radius two.

For $z\ne 0$, the principal value of complex exponentiation is $z^w=\exp (w\mathrm{Log}z)$, where the principal complex logarithm is $\mathrm{Log}z=\log |z|+i\mathrm{Arg}z$ with $-\pi <\mathrm{Arg}z\le \pi$. De Moivre's theorem states that $(\cos \theta +i\sin \theta {)}^n=\cos (n\theta )+i\sin (n\theta )$ for every integer $n$.

Since $\sqrt{3}+i=2e^{i\pi /6}$, the six roots are

Take the real part of De Moivre's theorem. Put $c=\cos \theta$ and use ${\sin }^2\theta =1-c^2$:

Take the imaginary part of De Moivre's theorem, $(\cos \theta +i\sin \theta {)}^5=\cos 5\theta +i\sin 5\theta$. With $s=\sin \theta$ and ${\cos }^2\theta =1-s^2$,The required equation cancels the fifth powers and becomes $5s(1-4s^2)=0$. Thus $\sin \theta =0,\pm 1/2$, giving all real solutions

In polar form of a complex number, $1+i=\sqrt{2}{e}^{i\pi /4}$. By De Moivre's theorem,

By De Moivre's theorem, $(\cos \theta +i\sin \theta {)}^4=\cos 4\theta +i\sin 4\theta$. Take real parts of the binomial expansion:Eliminate ${\sin }^2\theta$ using $1-{\cos }^2\theta$ and collect powers:

By De Moivre's theorem, $(e^{i\alpha }{)}^j=e^{ij\alpha }$. The complex sum whose real part is wanted is therefore a finite geometric series:Use $1-e^{i\beta }=-2i{e}^{i\beta /2}\sin (\beta /2)$ in the numerator and denominator. This yieldsTaking its real part proves the finite trigonometric sumWhen $\alpha \in 2\pi \mathbb{Z}$, the printed quotient is undefined; the original sum is directly $(n+1)\cos \theta$, also the limiting value of the displayed identity. The denominator restriction cannot be omitted from a literal use of the quotient.

By De Moivre's theorem, $\cos 5\theta$ is the real part of $(\cos \theta +i\sin \theta {)}^5$. Writing $c=\cos \theta$ and $s=\sin \theta$, the binomial theorem givesUse the Pythagorean trigonometric identity $s^2=1-c^2$ and expand:Therefore

Put $c=\cos \theta$ and $s=\sin \theta$. Taking the real part of $(c+is{)}^6$ in De Moivre's theorem givesTaking the imaginary parts of the second and fourth powers gives $\sin 2\theta =2cs$ and $\sin 4\theta =4cs(c^2-s^2)$. HenceThe mixed powers cancel exactly.

By De Moivre's theorem, $\sin 5\theta$ is the imaginary part of $(\cos \theta +i\sin \theta {)}^5$. The binomial theorem gives, with $s=\sin \theta$,Use ${\cos }^2\theta =1-s^2$ and simplify:The requested equation cancels the $16s^5$ term, leaving $5s(1-4s^2)=0$. Thus $s=0$ or $s=\pm 1/2$. The complete list in $0\le \theta <2\pi$ is