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Apply $R_1$ to a column coordinate vector, then apply $R_2$ in the same fixed basis. The function composition isThus the rightmost matrix acts first. The product is again a rotation matrix, since $(R_2{R}_1{)}^T(R_2{R}_1)=I$ and $\det (R_2{R}_1)=1$. For planar rotations about the same origin, the sine addition formula and cosine addition formula giveFor a concrete composition of rotation matrices, first rotate $(2,0{)}^T$ by $\pi /3$ radians, reaching $(1,\sqrt{3}{)}^T$, then by $\pi /6$ radians:The total angle is $\pi /2$. Reversing these two planar rotations gives the same answer, but noncommuting rotations about fixed axes show why one cannot assume that property for rotations in three dimensions.^[ref]

The cosine addition formula gives the horizontal coordinate after two successive rotations:The minus sign comes from the way a counterclockwise rotation mixes horizontal and vertical coordinates:The two rows recover the cosine addition formula and sine addition formula together.

With $x=\pi /6$ and $y=\pi /4$, the resulting angle is $5\pi /12$, andAlthough the horizontal coordinate becomes small, the vertical one approaches one; their squared sum still obeys the Pythagorean trigonometric identity. Taking $y=-x$ in the cosine addition formula, together with the even function property of cosine and odd function property of sine, also recovers that identity.

For a real angle $\theta$ in radians, Euler's formula stateswhere $i$ is the imaginary unit. It connects the complex exponential to the sine and cosine coordinates of the unit circle. In particular, $|e^{i\theta }|=1$, and multiplying by $e^{i\theta }$ rotates a complex number through angle $\theta$ without changing its modulus. At $\theta =\pi /3$ it gives $e^{i\pi /3}=1/2+i\sqrt{3}/2$; at $\theta =\pi$, it gives $e^{i\pi }=-1$.

The identity makes angle addition algebraic: $e^{i\alpha }{e}^{i\beta }=e^{i(\alpha +\beta )}$. Expanding the product and comparing real parts and imaginary parts recovers the sine addition formula and cosine addition formula. Differentiating $e^{i\omega t}$ gives $i\omega e^{i\omega t}$, and differentiating again gives $-{\omega }^2{e}^{i\omega t}$. Its real and imaginary coordinates therefore solve the harmonic oscillator equation. This is why complex exponentials are useful building blocks in Fourier analysis: differentiation acts on each frequency by simple multiplication.

A short justification separates the even and odd powers in the absolutely convergent power series for $e^{i\theta }$. The even powers give ${\sum }_{n\ge 0}(-1{)}^n{\theta }^{2n}/(2n)!=\cos \theta$, and the odd powers give $i{\sum }_{n\ge 0}(-1{)}^n{\theta }^{2n+1}/(2n+1)!=i\sin \theta$. The absolute convergence permits this regrouping. The formula thus links an analytic definition of the complex exponential to a geometric motion rather than merely introducing a convenient notation.

Leonhard Euler presents the relation between imaginary exponentials and sines and cosines in his 1748 Introductio in analysin infinitorum, volume 1, chapter VIII, section 138. His derivation uses limiting powers after connecting trigonometric addition to algebra. The original Latin and Ian Bruce's English translation can be read together in Euler, Introductio, chapter VIII, section 138.

Let ${\theta }_k=k\pi h=k\pi /(m+1)$ and define $q_j^{(k)}=\sin (j{\theta }_k)$. The artificial boundary values are $q_0^{(k)}=q_{m+1}^{(k)}=0$. For the symmetric tridiagonal Toeplitz matrix $H=[\beta ,\alpha ,\beta ]$, the sine addition formula givesThus all such matrices have the same discrete sine transform eigenvectors, with eigenvaluesThe sine vectors are mutually orthogonal vectors and nonzero, so the matrix $Q=(\sin (ij\pi h){)}_{i,j=1}^m$ is invertible. If $D=\mathrm{diag}({\lambda }_1,\dots ,{\lambda }_m)$, then $HQ=QD$ and therefore

The constant from part (i) is found by setting $x=0$:Taking $a=x+y$ therefore gives the sine addition formula

The sine addition formula gives $(\sin x+\cos x)/\sqrt{2}=\sin (x+\pi /4)$. On the requested interval the shifted angle runs from $0$ to $\pi /2$, so the graph rises from $(-\pi /4,0)$ through $(0,1/\sqrt{2})$ to $(\pi /4,1)$.

Its derivative $\cos (x+\pi /4)$ is nonnegative and reaches zero only at the right endpoint. Its second derivative is $-\sin (x+\pi /4)\le 0$, so the rising arc is concave downward and ends with a horizontal tangent.

Using the sine addition formula, combine the surviving modes:It has amplitude $2$ and a phase delay $\pi /3$. The full general solution would add $C{e}^{-x/\sqrt{3}}$, but this exponential is periodic only for $C=0$, so the displayed periodic solution is unique.

Write the sine argument as $\pi /2+\delta$, with $\delta =\frac{\pi }{2}(e^x-1)=\frac{\pi }{2}x+\frac{\pi }{4}{x}^2+O(x^3)$. The sine addition formula gives $\sin (\pi /2+\delta )=\cos \delta$. Its Maclaurin series is $1-{\delta }^2/2+O({\delta }^4)$, soThere is no linear term because the outer sine has zero derivative at $\pi /2$.

The Pythagorean trigonometric identity makes the second factor $1+{\cos }^2\theta$, which is strictly positive. It is therefore enough to solve $\sin \theta -\cos \theta \le 0$. The sine addition formula givesThe shifted angle lies between $\pi /4$ and $5\pi /4$. On this interval the sine is nonpositive precisely from $\pi$ onwards. Restoring the original angle and retaining its strict upper bound gives

A standing wave results from the superposition principle applied to equal-frequency, equal-amplitude waves travelling in opposite directions. Its spatial nodes of a standing wave remain fixed while its displacement oscillates in time.

For a pure standing wave, the two travelling waves also have equal amplitudes: their interference leaves a fixed mode shape, rather than a profile moving along the string. Choose the position origin and time origin so that their transverse displacements areHere $a>0$ is each constituent's amplitude, $k=2\pi /\lambda$ is the wavenumber, $\lambda$ is the wavelength, and $\omega =2\pi f$ is the angular frequency corresponding to frequency $f$. Their wave speed is $c=\omega /k$. The superposition principle and the sine addition formula giveThe spatial factor stays put; the temporal factor changes its size and sign. Other constant wave phases shift the position or time origin. Unequal constituent amplitudes generally leave moving displacement at the would-be nodes of a standing wave, rather than this exact factorization.^[ref]

For an original ideal-string example, set $L=1\mathrm{m}$, $c=20\mathrm{m}{\mathrm{s}}^{-1}$, and choose the third normal mode. Then $k=3\pi {\mathrm{m}}^{-1}$, $\omega =60\pi {\mathrm{s}}^{-1}$, and $f=30\mathrm{Hz}$ by fixed-end standing wave frequencies. With each constituent amplitude $a=5\mathrm{mm}$,Its nodes of a standing wave are at $x=0,1/3,2/3,1\mathrm{m}$ and its antinodes at $x=1/6,1/2,5/6\mathrm{m}$. These are constructed model parameters, not experimental readings.

At $t=1/120\mathrm{s}$, every point has zero displacement, but most have nonzero transverse velocity: the string is crossing its straight configuration. The black dots alone mark permanent nodes of a standing wave. At an antinode the greatest transverse speed is $2a\omega =0.6\pi \mathrm{m}{\mathrm{s}}^{-1}$, much smaller than the chosen $20\mathrm{m}{\mathrm{s}}^{-1}$ wave speed. One is motion of string material; the other is propagation of the constituent profiles.

The beauty of a standing wave is that motion reveals geometry: fixed nodes of a standing wave turn a moving disturbance into a spatial ruler. Their separation measures wavelength, and boundary conditions turn the continuous choice of wavenumber into a discrete set of normal modes. This is useful both for interpreting string vibrations and for understanding how a bounded wave equation selects its eigenfrequencies.