Higher: translate sine graphs in degrees
Translate outputs and inputs, recover the signed shifts from labelled landmarks, and retain the correct graph domains.
Exam scope and assessment notes
Additional Higher only. Higher includes the shared Foundation content. The Graphs route has nine shared lessons, two Higher line-equation lessons, four Higher graph-family lessons, then these nine transformation lessons. H:3.3B requires all eight core transformation lessons: all four forms f(x)+a, f(x+a), af(x), f(ax) across linear, quadratic, degree sine and degree cosine graphs, with worked construction and every practice stage. H:3.3C requires equation recovery and analysis in all eight core lessons plus the ninth domain/equivalence/order synthesis lesson. This nine-lesson split is StudyForA’s teaching design, not a syllabus-prescribed lesson count. Map landmarks, write equations and draw the original and both target graphs on labelled axes; an automatic trace alone does not replace these constructions. Finite interval endpoints in these exercises are included and infinite bounds are not. Retain input preimages and exclusions; zero factors, even or periodic formulas do not restore invalid arguments or establish a unique history. The bounded interactive examples support exploration, not a general expression or composition parser. All nine Foundation and seven additional Higher Graphs source rows have teaching and staged-practice links. These links do not establish exhaustive subskill approval, mastery or examination readiness; specialist review and formal assessment validation remain pending. Written work is not automatically marked; explained answers support self-checking, not exam marks or proof of mastery. Specialist review and formal assessment validation remain unestablished. Pearson chapter-map and uncorrected-proof navigation remain separate evidence; final full-book and Teaching Hub alignment has not been verified.
Before you begin
- Evaluate the original sine rule and plot exact coordinate pairs.
- Read function notation and solve simple equations with signed fractions.
- Use interval domains and ranges and standard degree sine/cosine values.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What are the midline and amplitude of y=sin x−3?
What you will learn
- Apply f(x)+a and f(x+a) to sine graphs using exact point mappings.
- Recover unknown parameters and transformed equations from known forms and labelled correspondences.
- Draw the original and both targets, retaining exact domains and ranges and distinguishing amplitude from the full-rule period measured in degrees.
Original: f(x) = sin(x°). Target: y = f(x)+a, with a = 1. All angles and horizontal shifts use degrees.
The domain choice changes the original exercise deliberately. Every finite interval endpoint here is included; an infinite bound is never included.
Applied equation: y = sin(x°) + (1). Original domain: All real numbers; target domain: All real numbers. Original range: [-1, 1]; target range: [0, 2].
Point rule: (u,v) → (u, v + (1)).
The extending target formula has amplitude 1 and least positive period 360°. The all-real target has that period.
Blue dashed: original. Red solid: target. Filled points mark supplied landmarks or included endpoints; grey dotted joins show point correspondences, not additional curves. If traces coincide, the red trace can cover the blue. The window is x from -360 to 360, y from -5 to 5; it does not change either mathematical domain.
At target input x=0, use original argument 0. Exact target output = 1. The selected target point is ringed in white.
Show transformation values and mapped landmarks
| Target x | Original argument | Target output |
|---|---|---|
| -180 | -180 | 1 |
| -90 | -90 | 0 |
| 0 | 0 | 1 |
| 30 | 30 | 3/2 |
| 60 | 60 | √3/2 + (1) |
| 90 | 90 | 2 |
| 180 | 180 | 1 |
| Original (u,v) | Image (x,y) |
|---|---|
| (30, 1/2) | (30, 3/2) |
| (90, 1) | (90, 2) |
Exact fractions and standard-angle expressions may remain unsimplified. Other sine/cosine readings are marked approximate. The curve is a numerical illustration; your own drawn graph readings are approximate. Use the point correspondences, equation and domain together to check your construction.
Read a degree sine wave before moving it
For y=C sin x, take x in degrees: the full rule has period 360°, amplitude |C| and midline y=0. For the unit wave, useful exact points include (−90°,−1), (0°,0), (30°,1/2), (90°,1) and (180°,0). Multiplying by C changes their heights. Use these landmarks to draw a smooth wave rather than straight segments between a sparse table.
An output translation y=f(x)+a adds a to every height. It changes the midline to y=a and shifts the range, but the amplitude is still the distance from that midline to a peak. It does not add a to the amplitude or change the 360° period of the full sine rule.
Translate the degree input in the opposite direction
In y=f(x+a), solve x+a=u, obtaining the point mapping (u,v)→(u−a,v). A positive a moves the sine wave left by a degrees; a negative a moves it right. Heights, amplitude and full-rule period stay unchanged. Keep degree labels on the axis and inside any written shift.
On a restricted original interval, permit only x for which x+a lies in that interval. Subtract a from both endpoints and keep the corresponding range. A formula such as sin(x−90°)=−cos x can describe the same values, but rewriting it does not remove the interval inherited from the original function.
Draw waves and recover shifts from labels
Draw the original and both targets using mapped zeros, peaks and additional check points; choose a horizontal scale showing the stated shifts clearly. When full domains are allowed, include repeated landmarks separated by 360°. The exact landmarks are fixed by the rule; an intermediate reading from a hand-drawn wave is approximate.
For recovery, first use the given form and the labelled old-to-new point. An unchanged x and a changed height identify an output difference; an unchanged height with a supplied horizontal correspondence gives a=u−x. Periodicity means an unlabelled full trace alone cannot select a unique shift history. A finite restricted piece has no global repeating domain, even though its underlying sine formula has a period.
Sources for this lesson
Pause and explain
Which is a correct full-rule statement for y=sin(x+40°)?
Worked example
Start from f(x)=sin x, with domain all real x (in degrees) and range [-1, 1]. Original landmarks: (30, 1/2), (90, 1). A uses y=f(x)+a with a=1: a vertical translation up 1 output units. B uses y=f(x+a) with a=60: a horizontal translation left 60°. Draw the original and both targets yourself on labelled axes, choosing scales that show the landmarks and the valid portions. State each mapping, exact equation, domain and range; map the supplied points and verify one additional point. Use degree axes, and state amplitude, midline and the period of each underlying full rule.
Show the worked solution
- A: a vertical translation up 1 output units. Use (u,v) → (u,v+a) with a=1. The labelled points become (30, 1/2) → (30, 3/2); (90, 1) → (90, 2). B: a horizontal translation left 60°. Use (u,v) → (u−a,v) with a=60. The labelled points become (30, 1/2) → (-30, 1/2); (90, 1) → (30, 1).
- A: y=sin(x)+1; all angles in degrees. Domain all real x (in degrees); range [0, 2]. The input domain is unchanged; add the parameter to both range endpoints. Amplitude 1, midline y=1, underlying full-rule period 360°. B: y=sin(x+60); all angles in degrees. Domain all real x (in degrees); range [-1, 1]. The allowed input condition is x+(60) in the original domain; subtract that parameter from both finite endpoints. Amplitude 1, midline y=0, underlying full-rule period 360°.
- Check the extra original point (-90, -1): A gives (-90, 0) and B gives (-150, -1). Substitute each image into its target equation and keep it in the allowed domain. Draw the original and both target traces through the mapped points, using a straight line for linear portions and a smooth curve for the other families. Label A and B separately even if traces coincide. Exact coordinates come from these rules; readings between drawn points are approximate. Include consecutive repeated full-rule landmarks when choosing the view, or mark how the restricted endpoints prevent a complete repeat.
Answer A: y=sin(x)+1; all angles in degrees; D=all real x (in degrees), R=[0, 2]. Amplitude 1, underlying period 360°. B: y=sin(x+60); all angles in degrees; D=all real x (in degrees), R=[-1, 1]. Amplitude 1, underlying period 360°.
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Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Start from f(x)=sin x, with domain all real x (in degrees) and range [-1, 1]. Original landmarks: (0, 0), (90, 1). A uses y=f(x)+a with a=2: a vertical translation up 2 output units. B uses y=f(x+a) with a=-90: a horizontal translation right 90°. Draw the original and both targets yourself on labelled axes, choosing scales that show the landmarks and the valid portions. State each mapping, exact equation, domain and range; map the supplied points and verify one additional point. Use degree axes, and state amplitude, midline and the period of each underlying full rule. The identity sin(x−90°)=−cos x gives a second formula for B on the same full domain; it does not change its labelled horizontal-translation history.
Give me a hint
Track one coordinate at a time. For an input change set the new argument equal to the old input, then solve for the new horizontal coordinate. Apply the same rule to the domain endpoints.
Compare my reasoning
- A: a vertical translation up 2 output units. Use (u,v) → (u,v+a) with a=2. The labelled points become (0, 0) → (0, 2); (90, 1) → (90, 3). B: a horizontal translation right 90°. Use (u,v) → (u−a,v) with a=-90. The labelled points become (0, 0) → (90, 0); (90, 1) → (180, 1).
- A: y=sin(x)+2; all angles in degrees. Domain all real x (in degrees); range [1, 3]. The input domain is unchanged; add the parameter to both range endpoints. Amplitude 1, midline y=2, underlying full-rule period 360°. B: y=sin(x−90); all angles in degrees. Domain all real x (in degrees); range [-1, 1]. The allowed input condition is x+(-90) in the original domain; subtract that parameter from both finite endpoints. Amplitude 1, midline y=0, underlying full-rule period 360°.
- Check the extra original point (-90, -1): A gives (-90, 1) and B gives (0, -1). Substitute each image into its target equation and keep it in the allowed domain. Draw the original and both target traces through the mapped points, using a straight line for linear portions and a smooth curve for the other families. Label A and B separately even if traces coincide. Exact coordinates come from these rules; readings between drawn points are approximate. The identity sin(x−90°)=−cos x gives a second formula for B on the same full domain; it does not change its labelled horizontal-translation history. Include consecutive repeated full-rule landmarks when choosing the view, or mark how the restricted endpoints prevent a complete repeat.
A: y=sin(x)+2; all angles in degrees; D=all real x (in degrees), R=[1, 3]. Amplitude 1, underlying period 360°. B: y=sin(x−90); all angles in degrees; D=all real x (in degrees), R=[-1, 1]. Amplitude 1, underlying period 360°.
Look for these in your work
- Draw and label the original plus both targets, with all supplied and one extra point mapped correctly.
- Give both exact equations and their actual domains and ranges, preserving restricted endpoints.
- Use degrees and separate amplitude from period; qualify periods of restricted portions.
Start from f(x)=2sin x, with domain all real x (in degrees) and range [-2, 2]. Original landmarks: (0, 0), (90, 2). A is known to have form y=f(x)+a; its labelled correspondences are (0, 0) → (0, -1); (90, 2) → (90, 1). B is known to have form y=f(x+a); its labelled correspondences are (0, 0) → (-30, 0); (90, 2) → (60, 2). Recover each unknown a from those correspondences and explain why the given form matters. Draw the original and both targets yourself on labelled axes, choosing scales that show the landmarks and the valid portions. State each mapping, exact equation, domain and range; map the supplied points and verify one additional point. Use degree axes, and state amplitude, midline and the period of each underlying full rule.
Give me a hint
Use the known form to decide which coordinate is preserved. Recover an output shift by a difference, an input shift from u=x+a, an output scale from y=av, or an input scale from u=ax; choose a nonzero coordinate for division.
Compare my reasoning
- A: Using (0, 0) → (0, -1), a=-1−(0)=-1. Use (u,v) → (u,v+a) with a=-1. The labelled points become (0, 0) → (0, -1); (90, 2) → (90, 1). B: Using (0, 0) → (-30, 0), u=x+a, so a=0−(-30)=30. Use (u,v) → (u−a,v) with a=30. The labelled points become (0, 0) → (-30, 0); (90, 2) → (60, 2).
- A: y=2sin(x)−1; all angles in degrees. Domain all real x (in degrees); range [-3, 1]. The input domain is unchanged; add the parameter to both range endpoints. Amplitude 2, midline y=-1, underlying full-rule period 360°. B: y=2sin(x+30); all angles in degrees. Domain all real x (in degrees); range [-2, 2]. The allowed input condition is x+(30) in the original domain; subtract that parameter from both finite endpoints. Amplitude 2, midline y=0, underlying full-rule period 360°.
- Check the extra original point (-90, -2): A gives (-90, -3) and B gives (-120, -2). Substitute each image into its target equation and keep it in the allowed domain. Draw the original and both target traces through the mapped points, using a straight line for linear portions and a smooth curve for the other families. Label A and B separately even if traces coincide. Exact coordinates come from these rules; readings between drawn points are approximate. Include consecutive repeated full-rule landmarks when choosing the view, or mark how the restricted endpoints prevent a complete repeat.
A: y=2sin(x)−1; all angles in degrees; D=all real x (in degrees), R=[-3, 1]. Amplitude 2, underlying period 360°. B: y=2sin(x+30); all angles in degrees; D=all real x (in degrees), R=[-2, 2]. Amplitude 2, underlying period 360°.
Look for these in your work
- Draw and label the original plus both targets, with all supplied and one extra point mapped correctly.
- Give both exact equations and their actual domains and ranges, preserving restricted endpoints.
- Recover each unknown parameter from labelled correspondences and the stated form.
Start from f(x)=sin x, with domain [-90°, 90°] and range [-1, 1]. Original landmarks: (-30, -1/2), (90, 1). A uses y=f(x)+a with a=1/2: a vertical translation up 1/2 output units. B uses y=f(x+a) with a=-180: a horizontal translation right 180°. Draw the original and both targets yourself on labelled axes, choosing scales that show the landmarks and the valid portions. State each mapping, exact equation, domain and range; map the supplied points and verify one additional point. Use degree axes, and state amplitude, midline and the period of each underlying full rule. Include the finite endpoints. Explain why extending a drawing or rewriting a formula does not restore an excluded original input.
Give me a hint
Track one coordinate at a time. For an input change set the new argument equal to the old input, then solve for the new horizontal coordinate. Apply the same rule to the domain endpoints.
Compare my reasoning
- A: a vertical translation up 1/2 output units. Use (u,v) → (u,v+a) with a=1/2. The labelled points become (-30, -1/2) → (-30, 0); (90, 1) → (90, 3/2). B: a horizontal translation right 180°. Use (u,v) → (u−a,v) with a=-180. The labelled points become (-30, -1/2) → (150, -1/2); (90, 1) → (270, 1).
- A: y=sin(x)+1/2; all angles in degrees. Domain [-90°, 90°]; range [-1/2, 3/2]. The input domain is unchanged; add the parameter to both range endpoints. Amplitude 1, midline y=1/2, underlying full-rule period 360°. This finite restricted graph has no global repetition domain; the stated period belongs to its full extension. B: y=sin(x−180); all angles in degrees. Domain [90°, 270°]; range [-1, 1]. The allowed input condition is x+(-180) in the original domain; subtract that parameter from both finite endpoints. Amplitude 1, midline y=0, underlying full-rule period 360°. This finite restricted graph has no global repetition domain; the stated period belongs to its full extension.
- Check the extra original point (-90, -1): A gives (-90, -1/2) and B gives (90, -1). Substitute each image into its target equation and keep it in the allowed domain. Draw the original and both target traces through the mapped points, using a straight line for linear portions and a smooth curve for the other families. Label A and B separately even if traces coincide. Exact coordinates come from these rules; readings between drawn points are approximate. The endpoint images for A are (-90, -1) → (-90, -1/2); (90, 1) → (90, 3/2). The endpoint images for B are (-90, -1) → (90, -1); (90, 1) → (270, 1). Include consecutive repeated full-rule landmarks when choosing the view, or mark how the restricted endpoints prevent a complete repeat.
A: y=sin(x)+1/2; all angles in degrees; D=[-90°, 90°], R=[-1/2, 3/2]. Amplitude 1, underlying period 360°. B: y=sin(x−180); all angles in degrees; D=[90°, 270°], R=[-1, 1]. Amplitude 1, underlying period 360°.
Look for these in your work
- Draw and label the original plus both targets, with all supplied and one extra point mapped correctly.
- Give both exact equations and their actual domains and ranges, preserving restricted endpoints.
- Use degrees and separate amplitude from period; qualify periods of restricted portions.
Common mistakes
- Moving f(x+a) in the direction of the written sign instead of solving x+a=u.
- Changing a stated domain after expanding or rewriting the target formula.
- Inferring a unique history from an unlabelled trace or confusing the midline with amplitude.
What can you explain now?
Does translating sin x up by 4 make its amplitude 5?
Compare with the explanation
No. Its amplitude remains 1 and its midline is y=4.
The range becomes [3,5], so the distance from the midline to either extreme is 1.
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Make it stick
Explain the input and output mappings, recover a parameter from a labelled point, and state how a restricted domain survives the change. State amplitude and period separately, using degrees.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.