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Pearson Edexcel 4MA1 IGCSE  ·  Topic G7.8  ·  Higher tier

G7.8 Trigonometric and Exponential Graphs

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01 — Explanation

What do the parent graphs of sin, cos, tan and a^x look like?

Over 0∘0^\circ to 360∘360^\circ:

GraphPeriodMax/minKey feature
y=sin⁡xy=\sin x360∘360^\circ(90°,1)(90°,1), (270°,−1)(270°,-1)passes through (0,0)(0,0)
y=cos⁡xy=\cos x360∘360^\circ(0°,1)(0°,1), (180°,−1)(180°,-1)passes through (0,1)(0,1)
y=tan⁡xy=\tan x180∘180^\circnone (unbounded)vertical asymptotes at x=90°,270°x=90°, 270°
y=axy=a^x——passes through (0,1)(0,1); horizontal asymptote y=0y=0

The asymptotes of tan⁡x\tan x exist because tan⁡x=sin⁡xcos⁡x\tan x=\dfrac{\sin x}{\cos x}, which blows up wherever cos⁡x=0\cos x=0.

02 — Explanation

How does y=af(x) transform the graph?

A vertical stretch, scale factor aa — every yy-value is multiplied by aa, so the amplitude becomes aa. The period is unchanged, and points on the xx-axis stay fixed.

03 — Explanation

How does y=f(x)+a transform the graph?

A vertical translation by aa — the midline moves from y=0y=0 to y=ay=a. For y=3sin⁡x+ky=3\sin x+k with maximum value 10: the base max is 3, so 3+k=10⇒k=73+k=10 \Rightarrow k=7, giving a new minimum of −3+7=4-3+7=4.

04 — Explanation

How does y=f(ax) transform the graph?

A horizontal stretch, scale factor 1a\tfrac1a — counter-intuitively, the graph gets squashed when a>1a>1. The new period is 360°a\dfrac{360°}{a}. For y=sin⁡(2x)y=\sin(2x): period =180°=180°, so it completes two full cycles in 360°360°.

05 — Explanation

How does y=f(x+a) transform the graph?

A horizontal translation, and the direction is the opposite of what the sign suggests: y=f(x+a)y=f(x+a) shifts left by aa; y=f(x−a)y=f(x-a) shifts right by aa.

Exam Tip: Rewrite sin⁡(x+30°)\sin(x+30°) as sin⁡(x−(−30°))\sin(x-(-30°)) to see the shift is −30°-30°, i.e. left — this avoids the classic "shifts the wrong way" error.

This explains why cos⁡x\cos x is just sin⁡x\sin x shifted 90°90° left: cos⁡x=sin⁡(x+90°)\cos x = \sin(x+90°).

06 — Explanation

How do I describe a combined transformation like y=3\sin(2x)?

State both transformations separately: vertical stretch scale factor 3 (parallel to yy-axis), and horizontal stretch scale factor 12\tfrac12 (parallel to xx-axis).

07 — Explanation

How do I find all solutions to \sin x=k in a given interval?

Find the calculator's principal value x1x_1, then use symmetry for the rest:

  • sin⁡x=k\sin x=k: second solution is 180°−x1180°-x_1.
  • cos⁡x=k\cos x=k: second solution is 360°−x1360°-x_1.
  • tan⁡x=k\tan x=k: solutions repeat every 180°180° (x1,x1+180°,…x_1, x_1+180°,\dots).

Example: sin⁡x=0.5\sin x=0.5 for 0°≤x≤360°0°\le x\le360°: x1=30°x_1=30°, so x=30°x=30° or 180°−30°=150°180°-30°=150°.

08 — Explanation

How do I solve an equation by rearranging it to match a graph I've already drawn?

Rearrange the target equation until one side exactly matches the equation of the curve you plotted, then the solution is where that curve meets the resulting straight line.

Worked exam-style question: Having plotted y=2(0.5)x−1y=2(0.5)^x-1, solve 2(0.5)x+2x−3.5=02(0.5)^x+2x-3.5=0.

  1. Isolate the exponential term: 2(0.5)x=−2x+3.52(0.5)^x=-2x+3.5.
  2. Subtract 1 from both sides to match the plotted curve: 2(0.5)x−1=−2x+2.52(0.5)^x-1=-2x+2.5.
  3. Draw y=−2x+2.5y=-2x+2.5 on the same axes; the xx-coordinate where it crosses the curve is the solution.

09 — Common mistakes

Common mistakes to avoid

  • Assuming y=sin⁡(2x)y=\sin(2x) stretches the period (it halves it, to 180°180°).
  • Shifting f(x+a)f(x+a) to the right instead of left.
  • Forgetting to draw asymptotes for tan⁡x\tan x or ax+ka^x+k graphs.
  • Stopping at the calculator's one solution for a trig equation instead of using symmetry to find the rest.