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The double-angle formula givesFor $\sigma >1$, the logarithm of the absolutely convergent Euler product givesExponentiation provesSuppose $\zeta (1+it)=0$ for $t\ne 0$, of order $r\ge 1$. As $\sigma \downarrow 1$, its middle factor is $O((\sigma -1{)}^{4r})$, the first is $O((\sigma -1{)}^{-3})$ because the pole at one is simple, and the final factor is bounded because $1+2it\ne 1$. The product is $O((\sigma -1{)}^{4r-3})\to 0$, contradicting the lower bound one. ThereforeThis is the three-four-one zero-free-region argument specialized to the line with real part one.

The homogeneous linear differential equation $y^{\prime \prime }+y=0$ has general solution $C_1\cos x+C_2\sin x$. For a particular solution, use variation of parameters with $y_1=\cos x$, $y_2=\sin x$ and Wronskian one:The double-angle formulas and integration by parts giveSubstitution simplifies to $-x^2\cos x/4+x\sin x/4+\cos x/8$. Absorb the last term into the homogeneous linear differential equation solution. ThereforeIndeed, applying $d^2/d{x}^2+1$ to $x\sin x$ gives $2\cos x$, and applying it to $x^2\cos x$ gives $2\cos x-4x\sin x$; their difference divided by four is the required $x\sin x$.

Euler's formula and its version with $-\theta$ giveSquaring $e^{i\theta }=\cos \theta +i\sin \theta$ and comparing real and imaginary parts with $e^{2i\theta }$ yields the double-angle formulas

The double-angle formula gives $r^2\sin 2\varphi =2(r\cos \varphi )(r\sin \varphi )=2xy$. Therefore

The complex conjugate is $z^{\ast }=x-iy$. From the complex sine and cosine identities,Divide these expressions and multiply by the complex conjugate of the denominator. Its squared modulus is ${\cos }^{2}x{\cosh }^{2}y+{\sin }^{2}x{\sinh }^{2}y=(\cos 2x+\cosh 2y)/2$. The numerator simplifies using ${\cosh }^{2}y-{\sinh }^{2}y=1$ and the double-angle formulas. ThusThese expressions apply except at the poles $y=0$, $x=\pi /2+k\pi$ with integer $k$, where the original tangent function is undefined.

The double-angle formula gives $\cos 2x=1-2{\sin }^{2}x$. Setting $s=\sin x$ reduces the equation to $2s^2+s-1=(2s-1)(s+1)=0$. On $0\le x\le \pi$, the sine is nonnegative, so only $s=1/2$ is allowed. Therefore

The cardioid has a nonnegative radial boundary, so each ray from the origin contributes $0\le r\le a(1+\cos \theta )$. Integrating the polar coordinates area element yieldsThe three angular integrals are $2\pi$, zero and $\pi$, respectively, using the double-angle formula for the last. Thus

Choose the trigonometric substitution $x=\sin u$ with $-\pi /2\le u\le \pi /2$. The cosine is nonnegative there, so $\sqrt{1-x^2}=\cos u$ and $dx=\cos udu$. The double-angle formula then givesSince $u=\arcsin x$ and $\sin 2u=2x\sqrt{1-x^2}$,The formula is an antiderivative on $(-1,1)$ and has continuous endpoint values at $\pm 1$.

The double-angle formula for the tangent function gives $1=2t/(1-t^2)$. Here $0<t<1$, because $0<\pi /8<\pi /4$, so the denominator is nonzero. Rearranging yields the quadratic equation

Using the Pythagorean trigonometric identity and the double-angle formula,The required value $1/2$ therefore occurs when ${\sin }^2(2\theta )=1$. Within $0\le \theta \le \pi$, the only possibilities are $2\theta =\pi /2$ and $3\pi /2$. HenceEquivalently equality requires ${\sin }^2\theta ={\cos }^2\theta =1/2$.