P1 3.3, P1 1.11, P1 1.12 - Trigonometric graphs

Sine cosine and tangent graphs

How could you solve $\sin x=x/2$ when isolating $x$ by ordinary algebra does not work? Plot both sides as heights: a shared height will reveal a solution. First, we need to know what the trigonometric graphs look like.

Imagine a point moving anticlockwise around a circle of radius $1$, starting at $(1,0)$. After a turn of $x$ radians, its horizontal coordinate is $\cos x$ and its vertical coordinate is $\sin x$. These extend the familiar right-triangle ratios to angles beyond a right angle and to negative turns. A clockwise turn gives a negative input.

One complete turn gives these landmarks:

At $x=\pi/4$, the unit-circle coordinates are $(\sqrt2/2,\sqrt2/2)$ and the slope is $1$, so $\tan(\pi/4)=1$. A tangent branch increases from arbitrarily large negative values to arbitrarily large positive values between its asymptotes. Draw separate branches; never connect across an undefined input.

[DIAGRAM: asset_name: The three trigonometric graphs; asset_slug: trig-base-graphs; caption: Sine and cosine are bounded waves; tangent has separate unbounded branches.; description: The upper panel shows sine in blue and cosine in green from negative two pi to two pi, with pi-over-two landmarks. The lower panel shows tangent in rose with dashed vertical asymptotes at odd multiples of pi over two and zeros at integer multiples of pi.]

All horizontal coordinates in these graphs are radians. For example, $\sin(\pi/2)=1$; typing $\pi/2$ into a calculator in degree mode would evaluate a different angle.

Symmetry and period

A whole repeating graph can be understood from one cycle and its symmetry.

After one full turn, the unit-circle point returns to its starting position, so both sine and cosine repeat after $2\pi$. The radius line already has the same slope after half a turn, so tangent repeats after $\pi$. A repeat need not be the shortest repeat: $2\pi$ is also a period of tangent, but its least positive period is $\pi$.

Cosine has even symmetry: turning clockwise or anticlockwise by the same angle gives the same horizontal coordinate, so $\cos(-x)=\cos x$. Sine and tangent have odd symmetry: their heights or slopes change sign, so $\sin(-x)=-\sin x$ and $\tan(-x)=-\tan x$, wherever tangent is defined.

These properties let you read values outside the first turn. For instance, $9\pi/4=\pi/4+2\pi$, so $\sin(9\pi/4)=\sqrt2/2$. Similarly, $5\pi/4=\pi/4+\pi$, so $\tan(5\pi/4)=1$. By reflection, $\cos(-\pi/4)=\sqrt2/2$ but $\sin(-\pi/4)=-\sqrt2/2$.

The next choices separate a reflection from a repeat.

Single transformations in radians

The same point mappings used for algebraic graphs apply to trigonometric graphs; the horizontal coordinates are now radian angles.

A multiplier outside the function changes its heights. A multiplier inside changes where each old height occurs: to find the image of input $p$ on $y=\sin(kx)$, solve $kx=p$. Consequently, the period of $\sin(kx)$ and $\cos(kx)$ is $2\pi/|k|$, while the period of $\tan(kx)$ is $\pi/|k|$, for $k\ne0$.

For a single translation, $y=\sin(x+\pi/6)$ moves the sine curve left by $\pi/6$. Thus its zero at $0$ moves to $-\pi/6$, and its maximum at $\pi/2$ moves to $\pi/3$. In contrast, $y=\sin x+1$ moves heights up by $1$: its values run from $0$ to $2$, while the period remains $2\pi$.

Change only the position of the multiplier $3$, from outside sine to inside its input. The range now stays fixed and the period changes.

[DIAGRAM: asset_name: Height change input change and translation; asset_slug: trig-single-transformations; caption: Moving the multiplier changes height scaling into horizontal scaling; an added input shifts the landmarks left.; description: Three labelled panels show y equals three sine x, y equals sine three x, and y equals sine of x plus pi over six. Periods, extreme heights and the translated maximum at pi over three are shown.]

Reflections also follow the old rules: $y=-\cos x$ reflects cosine in the $x$-axis, with $(0,1)$ becoming $(0,-1)$. For $y=\tan(2x)$, divide both zeros and asymptote positions by $2$: the period is $\pi/2$, and the closest vertical asymptotes to the origin are $x=-\pi/4$ and $x=\pi/4$.

Reading intersections from trigonometric graphs

An equation is satisfied where the two sides give the same height at the same input, so its solutions are the horizontal coordinates of intersections.

For $\sin x=x/2$, draw $y=\sin x$ and $y=x/2$ on the same axes. Read the $x$-coordinates where they meet, including interval endpoints. These are approximate graphical solutions: the graph's scale determines how many decimal places can be justified. Merely crossing the $x$-axis is relevant only when the other side of the equation is zero.

[DIAGRAM: asset_name: Sine meeting two straight lines; asset_slug: trig-line-intersections; caption: A steeper line changes the nonzero solution while the origin remains an intersection.; description: Two panels show sine from zero to two pi against the lines y equals x over four and y equals x over two respectively. Intersections are marked, with dashed vertical guides to the horizontal axis. The vertical display is clipped above 1.6; each straight line continues beyond it.]

Change only the gradient of the line from $1/4$ to $1/2$. The shared origin stays fixed, but the other intersection moves left.

The height at an intersection is useful for checking; the equation asks for the input $x$. Always state the angle measure and the interval considered.

Quick Recap

• Sine and cosine repeat after $2\pi$; tangent repeats after $\pi$ and has vertical asymptotes.
• Cosine has even symmetry; sine and tangent have odd symmetry.
• Outside multipliers change heights; inside multipliers change periods by the reciprocal factor.
• Read equation solutions as intersection inputs. For $\sin x=x/2$ on $[0,2\pi]$, they are $0$ and approximately $1.9$ radians.