Note 12.27.

\begin {align*} \sin (x + 2\pi ) & = \sin x \cos 2\pi + \cos x \sin 2\pi \\ & = \sin x \cdot (1) + \cos x \cdot (0)\\ & = \sin x \end {align*}

and

\begin {align*} \cos (x + 2\pi ) & = \cos x\cos 2\pi - \sin x \sin 2\pi \\ & = \cos x \cdot (1) - \sin x \cdot (0)\\ & = \cos x \end {align*}

\begin {align*} \sin (x + 2\pi ) & = \sin x\\ \cos (x + 2\pi ) & = \cos x \end {align*}

Thus the sine function and the cosine function are periodic and both have the period \(2\pi \).

Similarly

\(\sin (x - 2\pi ) = \sin x \)

\(\cos (x - 2\pi ) = \cos x\)

Note that the period of the tangent function is \(\pi \).

Consider the equation \(\, \sin x = \frac {1}{2}.\quad \) Here we see that \(\, x = \frac {\pi }{6}\,\) is a solution to the equation.
But since \(\, \sin ( x - 2\pi ) = \sin x = \frac {1}{2}\,\) we see that \(\, x - 2\pi = \frac {\pi }{6},\,\) which gives \(\, x = \frac {\pi }{6} + 2\pi \).

Also \(\, \sin (x - 4\pi ) = \sin x = \frac {1}{2}\)

\(\implies \quad x - 4\pi = \frac {\pi }{6}\)

\(\implies \quad x = \frac {\pi }{6} + 4\pi .\,\) in fact for any positive integer \(k\).

\(\sin ( x - 2k\pi ) = \sin x = \frac {1}{2}\)

\(\implies \quad x - 2k\pi = \frac {\pi }{6}\)

\(\implies \quad x = \frac {\pi }{6} + 2k\pi \)

Hence \(\, \frac {\pi }{6} + 2k\pi \,\) is also a solution to the equation \(\, \sin x = \frac {1}{2}\).

Similarly, \(\quad \sin ( x + 2\pi ) = \sin x = \frac {1}{2}\)

\(\implies \quad x + 2\pi = \frac {\pi }{6}\)

\(\implies \quad x = \frac {\pi }{6}- 2\pi \)

So that in the same way if it is a positive integer

\(\sin ( x + 2\pi k ) = \sin x = \frac {1}{2}\)

\(\implies \quad x + 2\pi k = \frac {\pi }{6}\)

\(\implies \quad x = \frac {\pi }{6} - 2\pi k\,\) is also a solution to the equation \(\, \sin x = \frac {1}{2}\)

Combining the two solutions i.e \[x = \frac {\pi }{6} + 2k\pi \qquad \text {and}\qquad x = \frac {\pi }{6} - 2k\pi \]

We conclude that for any integer \(n,\quad x = \frac {\pi }{6} + 2n\pi \,\) is a solution to the equation \(\, \sin x = \frac {1}{2}\).

Similarly, for the angle in the second quadrant \(\, x = \frac {5\pi }{6}\,\) is a solution to the equation. Hence by the periodicity of sine functions we see that for any integer \(n,\quad x = \frac {5\pi }{6} + 2n\pi \,\) is also a solution to the equation \(\, \sin x = \frac {1}{2}\).

Therefore the most general solution to the equation \(\sin x=\frac {1}{2}\) is given by

\[ x =\, \left .\begin {aligned} \frac {\pi }{6} + 2n\pi \\\\ \frac {5\pi }{6} + 2n\pi \\ \end {aligned} \right \}\, n \in \mathbb {Z} \]

Note 12.31. Amplitude is a distance, so it is never negative — which is why the absolute value appears. The negative sign in \(-7\sin x\) does not shrink the swing; it flips the curve upside down, so the graph is that of \(7\sin x\) reflected in the \(x\)-axis.

Note too that amplitude is not the same as the maximum value. For \(f(x)=2+3\sin x\) the maximum is \(5\) and the minimum \(-1\), so the amplitude is \(\frac {5-(-1)}{2}=3\), not \(5\): adding \(2\) raises the whole curve without widening its swing. Defining amplitude as “the maximum the function attains” is only correct when the curve is centred on the \(x\)-axis, which is the case in every example here but not in general.

Note 12.33. Each constant does one job, and it is worth seeing why.

\(a\) stretches the curve vertically, so it scales the swing: amplitude \(\left |a\right |\).

\(b\) compresses it horizontally. The parent function completes one cycle as its argument runs through \(2\pi \), so \(bx\) must run through \(2\pi \), which takes \(x\) only \(\frac {2\pi }{b}\) — a larger \(b\) means a shorter period and more cycles in the same space.

\(c\) shifts it sideways. Factorising, \(bx+c=b\left (x+\frac {c}{b}\right )\), and replacing \(x\) by \(x+\frac {c}{b}\) moves a graph left by \(\frac {c}{b}\). That is why the shift is \(-\frac {c}{b}\) rather than \(-c\), and why the sign feels backwards: adding inside the bracket moves the curve the opposite way to what the plus sign suggests.

\(d\) raises the whole curve, moving the centre line to \(y=d\) without touching the amplitude.

The practical rule for the shift is the one used below: set the whole bracket to zero and solve. That locates where the parent curve’s starting point has moved to, and it works without remembering any sign convention.