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Put $h=x-\pi /4$. The trigonometric identities give $\cos x\sin x=\frac{1}{2}\sin 2x=\frac{1}{2}\cos 2h$. Expanding cosine at zero givesThe linear and cubic terms vanish, so these are the first two nonzero Taylor series terms.

The double-angle trigonometric identity $\sin 2x=2\sin x\cos x$ makes the integrand $2\sin x{\cos }^{2}x$. The integration by substitution $u=\cos x$, $du=-\sin xdx$, gives

Put $\theta =\pi /12$. The double-angle trigonometric identities imply ${\cos }^2(2\theta )=1-{\sin }^2(2\theta )$. Since $2\theta =\pi /6$ is in the first quadrant, the positive square root is required. The half-angle identities then give the requested expression in terms of $\sin (\pi /6)$:

The double-angle trigonometric identity gives $\cos x=2\sin x\cos x$, or $\cos x(1-2\sin x)=0$. Preserve both factors: $\cos x=0$ contributes the endpoint, while $\sin x=1/2$ contributes the interior solution. In the specified interval,

Use the product-to-sum formula and the double-angle trigonometric identities:By Fourier orthogonality, these are already its only nonzero Fourier series coefficients:

The trigonometric identity gives $\sin x\cos x=\frac{1}{2}\sin 2x$, and the complementary-angle trigonometric identity gives $\cos (\pi /2-x)=\sin x$. Thus the Fourier series is already the finite expressionThe only nonzero Fourier coefficients are $b_1=1$ and $b_2=1/2$.

The equation as printed requires $v\ne 0$; its sign does not affect the result. Seek separated solutions $\psi =X(x)T(t)$. Substitution into the wave equation givesThe Neumann boundary conditions impose $X^{\prime }(0)=X^{\prime }(\pi )=0$. Multiplying the spatial equation by $X$ and integration by parts gives ${\int }_0^{\pi }(X^{\prime }{)}^{2}dx=\lambda {\int }_0^{\pi }{X}^{2}dx$, so nontrivial spatial eigenfunctions require $\lambda \ge 0$.

For $\lambda =0$, $X$ is constant and $T=A_0+B_{0}t$. For $\lambda =k^2>0$, $X=C\cos kx+D\sin kx$. The first boundary condition gives $D=0$, and the second gives $\sin k\pi =0$. Thus $k=n$ is a positive integer and the spatial eigenfunctions are $\cos nx$. Their time dependence is $A_n\cos (nvt)+B_n\sin (nvt)$. By superposition,The zero initial velocity sets $B_0=B_n=0$. The trigonometric identity gives the initial profile ${\cos }^2(x/2)=(1+\cos x)/2$, so only the constant mode and the first cosine mode occur. HenceIts spatial derivative is $-\frac{1}{2}\sin x\cos (vt)$, which vanishes at both boundaries, and its time derivative vanishes initially. Thus it satisfies all the required data.

Use the trigonometric identity $2{\cos }^2(ax)=1+\cos (2ax)$ and express the cosine as a sum of two complex exponentials. The Fourier transform of each exponential is a shifted Dirac delta distribution:By linearity,The original function does not decay and is not absolutely integrable on the real line, so this answer is a tempered distribution, rather than an ordinary convergent Fourier integral. Its three lines represent the constant part and the two opposite spatial frequencies.

The relevant comparison is between gravity and surface tension. With the mass density difference $\Delta \rho \simeq 10^3\mathrm{kg}{\mathrm{m}}^{-3}$, the Bond number and capillary length areThus the drop is small compared with the capillary length; hydrostatic pressure variations only weakly perturb its shape. The Young–Laplace equation then gives approximately constant interfacial mean curvature, and an axisymmetric drop on a uniform flat substrate is a spherical cap. The substrate selects the contact angle, measured through the water.

Write $R$ for the radius of the generating sphere, $s$ for the contact-circle radius, and $h$ for the apex height. Elementary trigonometric identities giveTaking the stated droplet radius to mean its radius of curvature, $R=r$, givesThere is no diagram or explicit radius convention in the original PDF. If $r$ instead denotes the contact-circle radius, the answer is $h=r\tan 27.5^{\circ }=0.260\mathrm{mm}$. If it denotes the radius of an equal-volume free sphere, use the volume of a spherical cap, $\pi h^2(3R-h)/3=4\pi r^3/3$, to obtain $h=r[4(1-\cos \theta )/(2+\cos \theta ){]}^{1/3}=0.436\mathrm{mm}$. The spherical cap approximation is justified in all three conventions; a unique numerical height requires specifying which radius is meant.

As an original stipulated low-field model, take the axial magnetic field intensity $H$ in $A/m$ and prescribeAll strain components are dimensionless, and their trace vanishes. This is an even, quasistatic, unsaturated toy law with a fixed measuring axis. It is distinct from the fixed-magnitude directional model in saturated isotropic magnetostrictive strain. In particular, one must not replace local unit magnetization by normalized net magnetization in a many-domain specimen and assume this law follows.

For $H(t)=H_0\sin (2\pi ft)$, $H_0=20kA/m$, and $f=50\mathrm{Hz}$, a trigonometric identity givesThe axial strain ranges from zero to $20\times 10^{-6}$, with mean $10\times 10^{-6}$ and a $100\mathrm{Hz}$ component of amplitude $10\times 10^{-6}$. The cosine coefficient is negative with the chosen time origin. For $L=100\mathrm{mm}$ the length ranges from zero to $2\mu \mathrm{m}$ relative to $H=0$. There is no $50\mathrm{Hz}$ term in this unbiased model.

A static bias $H_b$ changes the conclusion. Expanding $\kappa [H_b+H_0\sin (2\pi ft){]}^2$ produces a fundamental-frequency term $2\kappa H_b{H}_0\sin (2\pi ft)$ as well as the doubled-frequency term. This is an incremental linear response about a state with a chosen magnetic axis; it does not contradict isotropic linear magnetostriction vanishes at an unbiased isotropic reference. Real hysteresis, thermal expansion from Joule heating, and mechanical resonance can add further components or change their phases. Frequency alone cannot identify the cause of motion: Joule heating also contains an even-in-electric current contribution.

A trigonometric identity is an equality involving trigonometric functions that holds for every input in its stated domain.