Graphing Transformations of Trigonometric Functions

A transformation of a trig function takes a basic graph, like \(y = \sin x\) or \(y = \cos x\), and stretches, compresses, flips, or slides it. Every transformation you'll ever see on a trig graph comes down to four ingredients: amplitude, period, phase shift, and vertical shift. Once you can read these four values out of an equation, you can sketch the graph without plotting dozens of points by hand.

This lesson builds directly on the parent graph of the sine graph: y = sin x. If you haven't looked at that shape yet, it's worth reviewing first, since every transformation below starts from that same basic wave.

Almost every transformed sine or cosine function can be written as:

\( y = A \sin\big(B(x - C)\big) + D \)   or   \( y = A \cos\big(B(x - C)\big) + D \)

Each letter controls exactly one visual change:

• \(A\) controls the amplitude (how tall the wave is)
• \(B\) controls the period (how wide one full cycle is)
• \(C\) controls the phase shift (how far the graph slides left or right)
• \(D\) controls the vertical shift (how far the graph slides up or down)

The same four parameters also apply to tangent and cotangent, and to the reciprocal functions, though their base shapes look different. If you're transforming those specifically, the -tan(x) graph lesson shows the parent shape you'd be stretching and shifting.

The amplitude of a sine or cosine graph is \(|A|\). It tells you how far the curve rises above and falls below its midline. The parent graph \(y = \sin x\) has amplitude 1, so it oscillates between \(-1\) and \(1\).

If \(|A| > 1\), the graph stretches vertically and the peaks get taller. If \(0 < |A| < 1\), the graph compresses vertically and the wave flattens. If \(A\) is negative, the graph also reflects over the midline, so a maximum becomes a minimum and vice versa.

The period of a transformed sine or cosine function is:

\( \)period\( = \dfrac{2\pi}{|B|} \)

The parent functions \(y = \sin x\) and \(y = \cos x\) have period \(2\pi\), since \(B = 1\). Making \(|B|\) larger squeezes the graph horizontally so it completes each cycle faster. Making \(|B|\) smaller (between 0 and 1) stretches the graph horizontally, so each cycle takes longer to complete.

The phase shift is the value \(C\) in \(B(x - C)\). It slides the entire graph horizontally without changing its shape. A positive \(C\) shifts the graph to the right, and a negative \(C\) shifts it to the left.

A common mistake is reading the shift straight off the equation before factoring \(B\) out. If the equation is written as \(y = \sin(2x - \pi)\), you need to factor first: \(\sin\big(2(x - \tfrac{\pi}{2})\big)\), which shows a phase shift of \(\tfrac{\pi}{2}\), not \(\pi\).

The vertical shift is the value \(D\). It moves the midline of the graph from \(y = 0\) to \(y = D\), sliding the whole curve up if \(D > 0\) or down if \(D < 0\). It does not affect the amplitude, period, or phase shift at all.

Start with the parent graph \(y = \sin x\), which has amplitude 1, period \(2\pi\), no phase shift, and midline \(y = 0\).

Now compare it to \(y = 3\sin\big(2(x - \tfrac{\pi}{4})\big) + 1\). Reading the parameters: \(A = 3\), \(B = 2\), \(C = \tfrac{\pi}{4}\), \(D = 1\). So the amplitude is 3, the period is \(\tfrac{2\pi}{2} = \pi\), the phase shift is \(\tfrac{\pi}{4}\) to the right, and the midline moves up to \(y = 1\).

Notice how the wave is taller (amplitude 3 instead of 1), completes each cycle in half the horizontal distance (period \(\pi\) instead of \(2\pi\)), starts its cycle a quarter turn later, and now oscillates around \(y = 1\) instead of \(y = 0\).

Identify each parameter for \( y = -2\cos\big(4x + \pi\big) - 3 \).

Step 1: Factor out \(B\) inside the parentheses so the equation matches the general form \(A\cos(B(x - C)) + D\).

\( 4x + \pi = 4\big(x + \tfrac{\pi}{4}\big) = 4\big(x - (-\tfrac{\pi}{4})\big) \)

Step 2: Read off the values: \(A = -2\), \(B = 4\), \(C = -\tfrac{\pi}{4}\), \(D = -3\).

Step 3: Compute amplitude and period.

Amplitude \(= |-2| = 2\). Period \(= \dfrac{2\pi}{4} = \dfrac{\pi}{2}\).

Step 4: State the shift and reflection.

Phase shift \(= -\tfrac{\pi}{4}\), meaning the graph shifts \(\tfrac{\pi}{4}\) to the left. Vertical shift \(= -3\), so the midline is \(y = -3\). Since \(A\) is negative, the graph is also reflected vertically compared to a plain cosine curve.

Sketch \( y = \tfrac{1}{2}\sin\big(x - \tfrac{\pi}{2}\big) + 2 \).

Step 1: Draw the midline. Since \(D = 2\), draw a horizontal dashed line at \(y = 2\). This replaces the x-axis as the graph's center.

Step 2: Mark the amplitude. Since \(A = \tfrac{1}{2}\), the graph reaches \(\tfrac{1}{2}\) unit above and below the midline, so it peaks near \(y = 2.5\) and dips near \(y = 1.5\).

Step 3: Find the period. Since \(B = 1\), the period stays \(2\pi\), the same as the parent sine graph.

Step 4: Apply the phase shift. Since \(C = \tfrac{\pi}{2}\), shift the entire starting point of the cycle \(\tfrac{\pi}{2}\) to the right before drawing the curve.

Step 5: Sketch the wave. Starting from the shifted point on the midline, draw one full sine-shaped cycle rising to the new maximum, back through the midline, down to the new minimum, and back to the midline, using the period from Step 3 to space out the key points.

Every transformation problem breaks down into the same short checklist: rewrite the equation so it matches \(A\sin(B(x-C)) + D\) or \(A\cos(B(x-C)) + D\), then read off amplitude, period, phase shift, and vertical shift in that order. Once you can do that for sine and cosine, the same logic extends to other trig graphs. If a problem gives you a graph instead of an equation and asks you to work backward, the lesson on writing trig equations from graphs walks through that reverse process directly.

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