Transformations of Trigonometric Functions

Every transformation of a trig function is a modification to one of four parameters in the general form. Recognising which parameter does what is the master key to this entire topic.

Starting from the base curves $y = \sin x$, $y = \cos x$, $y = \tan x$:

• Vertical reflection: $y = -\sin x$ flips the graph over the $x$-axis (peaks become troughs).
• Horizontal reflection: $y = \sin(-x) = -\sin x$ also flips vertically for sine (since sine is odd). For cosine, $y = \cos(-x) = \cos x$ — no change (cosine is even).
• Vertical translation: $y = \sin x + d$ shifts the graph up or down by $d$ units.
• Horizontal translation (phase shift): $y = \sin(x + c)$ shifts left by $c$ units (for $c > 0$).
• Vertical dilation: $y = a \cdot \sin x$ changes the amplitude to $|a|$.
• Horizontal dilation: $y = \sin(bx)$ changes the period to $\dfrac{2\pi}{|b|}$.

$y = a \cdot f(bx + c) + d$: $a$ = amplitude/reflection, $b$ = period change, $c$ = phase shift, $d$ = vertical shift.; Vertical reflection: $a < 0$ flips the graph. Amplitude = $|a|$ (always positive).

Pause — copy the general form $y = a \cdot f(bx+c)+d$ with the role of each parameter: $a$ = amplitude and reflection, $b$ = period change (period = $2\pi/|b|$), $c/b$ = phase shift, $d$ = vertical shift — into your book.

We just saw the four parameters of $y = a \cdot f(bx+c)+d$: $a$, $b$, $c$, and $d$ each change the shape or position. That raises a question: for $a$ and $b$ specifically — what is the exact formula for amplitude and period, and why is amplitude always positive even when $a < 0$? This card answers it → amplitude $= |a|$ (max $= d + |a|$, min $= d - |a|$); period $= 2\pi/|b|$ for sin/cos, $\pi/|b|$ for tan.

Amplitude and period are the two most commonly tested features of a transformed trig graph. Amplitude tells you how tall the wave is; period tells you how long one cycle takes.

Definitions and formulas:

• Amplitude $= |a|$: The distance from the midline to either a peak or a trough. Maximum $= d + |a|$; minimum $= d - |a|$.
• Period of sin/cos $= \dfrac{2\pi}{|b|}$: How far along the $x$-axis before the pattern repeats.
• Period of tan $= \dfrac{\pi}{|b|}$: Tan has a shorter natural period of $\pi$, not $2\pi$.
• Midline $= y = d$: The equilibrium line. Even if $d = 0$, the midline is the $x$-axis.

Amplitude $= |a|$. Max $= d + |a|$, min $= d - |a|$. Never negative.; Period (sin/cos) $= 2\pi / |b|$. Period (tan) $= \pi / |b|$.

Pause — copy amplitude $= |a|$, max $= d + |a|$, min $= d - |a|$, period $= 2\pi/|b|$ (sin/cos) and $\pi/|b|$ (tan) — noting amplitude is always positive, even when $a < 0$ — into your book.

We just saw that amplitude $= |a|$ and period $= 2\pi/|b|$ describe the shape of the wave. That raises a question: the parameters $c$ and $d$ shift the entire graph horizontally and vertically — but why is the phase shift $-c/b$, not simply $c$, and why must you factor out $b$ first? This card answers it → rewrite as $y = a\sin\!\left(b\!\left(x + c/b\right)\!\right) + d$; the phase shift is $-c/b$ (positive = right, negative = left).

Phase shift moves the entire graph left or right along the $x$-axis. Vertical shift moves it up or down. Both leave the shape (amplitude and period) unchanged.

Reading phase shift carefully:

• Phase shift $= -\dfrac{c}{b}$: In $y = a \cdot f(bx + c) + d$, the phase shift is $-c/b$.
• Positive result → shift to the right. Negative result → shift to the left.
• In $y = \sin(x + c)$ (where $b = 1$): phase shift $= -c$. So $y = \sin(x + \pi/4)$ shifts left by $\pi/4$.
• Vertical shift $= d$: Positive $d$ raises the midline; negative $d$ lowers it.

Phase shift $= -c/b$. Positive → right; negative → left.; To read phase shift from $y = a\sin(bx + c) + d$: factor out $b$ first: $y = a\sin\!\left(b\!\left(x + c/b\right)\!\right) + d$. Phase shift $= -c/b$.

Pause — copy the phase-shift rule: factor out $b$ first from $y = a\sin(bx+c)+d$, giving phase shift $= -c/b$ (positive = right, negative = left); vertical shift $= d$ (midline $y = d$) — into your book.

We just saw all four parameters: amplitude, period, phase shift $-c/b$, and vertical shift $d$. That raises a question: with all these values in hand, what is the minimum number of points you need to plot to produce an accurate sketch — and where exactly should those points go? This card answers it → the five-key-point method: start at the phase shift, then step through $T/4$, $T/2$, $3T/4$, and $T$ to capture one complete cycle.

To sketch a transformed trig graph, you don't need to plot dozens of points. The five-key-point method reduces any transformed sine or cosine to exactly five strategic points that define one complete cycle.

Five-key-point method — step by step:

• Step 1: Identify $a$, $b$, $c$, $d$ from the equation.
• Step 2: Calculate amplitude, period, phase shift and midline.
• Step 3: Divide one period into 4 equal quarters to find the $x$-values of the 5 key points.
• Step 4: For each key $x$, calculate the $y$-value using the pattern: start → max/min → midline → min/max → midline (for sin); or max → midline → min → midline → max (for cos).
• Step 5: Apply any vertical shift $d$ to all $y$-values. Plot and join with a smooth curve.

Five-key-point method: phase shift → phase shift + T/4 → phase shift + T/2 → phase shift + 3T/4 → phase shift + T.; For reflected graphs ($a < 0$): flip all $y$-values. A sine that would peak first now troughs first.

Pause — copy the five-key-point $x$-values: phase shift; $+T/4$; $+T/2$; $+3T/4$; $+T$ — and the rule that when $a < 0$, flip all $y$-values so a sine starts with a trough — into your book.

We just saw the five-key-point method for sketching one complete cycle of a transformed trig graph. That raises a question: if you need to find all $x$-values where a transformed trig function equals a constant $k$, how do you use the sketch — and the substitution $u = bx + c$ — to find every solution in the domain? This card answers it → isolate the trig ratio, substitute $u = bx + c$, adjust the domain for $u$, find all $u$-solutions using symmetry, then back-substitute for $x$.

A trig equation within a domain is asking: for what $x$-values does the transformed graph intersect a horizontal line? Finding all solutions requires understanding the period and domain together.

Graphical approach to solving trig equations:

• Set up: Rewrite as $f(\text{transformed}) = k$. Sketch the transformed graph over the given domain.
• Draw $y = k$: The solutions are the $x$-coordinates of all intersection points within the domain.
• Count solutions: Use the period — each full period of the graph can produce at most 2 solutions for sin/cos (one on the way up, one on the way down).
• Domain restriction: Only include solutions that fall within the given interval. Clearly justify why no others exist.

To solve $a\sin(bx + c) + d = k$: isolate the trig function, substitute $u = bx + c$, adjust the domain for $u$, find all $u$-solutions, back-substitute for $x$.; For each full period in the domain, sin/cos can produce 2 solutions...

Pause — copy the solving method: isolate the trig ratio → substitute $u = bx + c$ → adjust domain for $u$ → find all $u$-solutions using reference angle and symmetry → back-substitute to find $x$ — into your book.