Trigonometric functions · Teacher reference · answers included

← Senior 1 · short mathematical inquiries

Teacher reference: contains answers, conditions and proofs.

07 Trigonometric functions · g10-07-01 · 5–10 min · TEACHER · ANSWERS

What does one radian measure?

Does doubling a circle radius double its angle or its arc?

Predict: For a 90° turn, predict the arc lengths when r=1 and r=2.

Compare	Computed observation
Unit radius Turn (degrees) = 90; Radius r = 1	The radian ratio is independent of radius.
Double radius Turn (degrees) = 90; Radius r = 2	The radian ratio is independent of radius.
Zero turn Turn (degrees) = 0; Radius r = 2	The radian ratio is independent of radius.

Conclusion: A radian angle is the ratio s/r. A full turn has angle 2π, independently of radius; arc length is s=rθ when θ is in radians.

Conditions: Use r>0 and the magnitude of a turn for arc length; a zero turn has zero arc.

Justification: A full circumference 2πr gives s/r=2π. Proportional arcs give θ=πd/180 and s=rθ.

Ask: Which quotient remains unchanged when both radius and arc length double?

Watch for: Substituting 90 into s=rθ while θ is still measured in degrees.

Extension: Explain why the definition s/r cannot use r=0.

Exit: Find the arc of a 60° sector with radius 3.

θ=π/3 and s=π.

Before: Angles, coordinates and basic function graphs. · Timing: predict 1, compare 3, explain 2, exit 1 minute. Finite observations do not prove a universal claim.

07 Trigonometric functions · g10-07-02 · 5–10 min · TEACHER · ANSWERS

One circle, two coordinates

Which coordinate is sine, and which signs depend on the quadrant?

Predict: Compare sine and cosine at 30°,150°,210° and330°.

Compare	Computed observation
First quadrant Angle θ (degrees) = 30	The point is (cosθ,sinθ).
Second quadrant Angle θ (degrees) = 150	The point is (cosθ,sinθ).
Axis boundary Angle θ (degrees) = 270	The point is (cosθ,sinθ).

Conclusion: The unit-circle point is (cosθ,sinθ). A full turn returns to the same point, while opposite points negate both coordinates.

Conditions: Angles may be negative or exceed360°; these are directed turns, not new radii.

Justification: Coordinates on x²+y²=1 have x=cosθ,y=sinθ by the unit-circle definition. Rotation by360° preserves the point.

Ask: Can a negative sine value still be a valid circle coordinate?

Watch for: Treating sine and cosine as positive lengths in every quadrant.

Extension: Find all turns in[0°,720°] returning to(1,0).

Exit: Give the coordinates at270°.

(0,−1), so cos270°=0 and sin270°=−1.

Before: Angles, coordinates and basic function graphs. · Timing: predict 1, compare 3, explain 2, exit 1 minute. Finite observations do not prove a universal claim.

07 Trigonometric functions · g10-07-03 · 5–10 min · TEACHER · ANSWERS

Reflect an angle without recalculating

Why do θ and180°−θ share a sine but not always a cosine?

Predict: Predict which coordinate changes under reflection in the y-axis.

Compare	Computed observation
30° reflection Original angle θ (degrees) = 30	Reflection keeps y and reverses x.
Obtuse input Original angle θ (degrees) = 120	Reflection keeps y and reverses x.
Fixed point Original angle θ (degrees) = 90	Reflection keeps y and reverses x.

Conclusion: Reflection in the y-axis maps(cosθ,sinθ) to(−cosθ,sinθ), giving the supplementary-angle formulas.

Conditions: The identities hold for every real angle; matching plotted decimals is only a check.

Justification: The reflected ray makes angle π−θ and keeps y while reversing x. Therefore sin(π−θ)=sinθ and cos(π−θ)=−cosθ.

Ask: What happens when the point is already on the y-axis?

Watch for: Assuming equal sines imply equal angles.

Extension: Explain sin(θ+180°)=−sinθ using a half-turn.

Exit: Express cos150° in terms of cos30°.

cos150°=−cos30°=−√3/2.

Before: Angles, coordinates and basic function graphs. · Timing: predict 1, compare 3, explain 2, exit 1 minute. Finite observations do not prove a universal claim.

07 Trigonometric functions · g10-07-04 · 5–10 min · TEACHER · ANSWERS

A square loses a sign

Does sin²θ+cos²θ=1 determine the sign of cosine?

Predict: Compare30° and150° before computing their squared coordinates.

Compare	Computed observation
Positive cosine Angle θ (degrees) = 30	The square gives the magnitude; the quadrant supplies the sign.
Negative cosine Angle θ (degrees) = 150	The square gives the magnitude; the quadrant supplies the sign.
Zero cosine Angle θ (degrees) = 90	The square gives the magnitude; the quadrant supplies the sign.

Conclusion: The sum of squares equals1, but solving for cosine requires its quadrant sign. Opposite signs have the same square.

Conditions: The identity uses the same angle; square roots alone return a nonnegative value.

Justification: The unit-circle equation gives cos²θ+sin²θ=1. Thus |cosθ|=√(1−sin²θ), with the sign supplied by x-coordinate.

Ask: Which extra fact distinguishes the two possible cosine values?

Watch for: Always using cosθ=√(1−sin²θ).

Extension: Why does tanθ=sinθ/cosθ need an excluded case?

Exit: If sinθ=1/2 and θ lies in quadrantII, find cosθ.

cosθ=−√3/2.

Before: Angles, coordinates and basic function graphs. · Timing: predict 1, compare 3, explain 2, exit 1 minute. Finite observations do not prove a universal claim.

07 Trigonometric functions · g10-07-05 · 5–10 min · TEACHER · ANSWERS

Unroll the circle into a wave

How does a moving circle point produce a sine graph?

Predict: Predict the graph height after a quarter, half and full turn.

Compare	Computed observation
Quarter turn Turn (degrees) = 90	Green copies the circle y-coordinate; gold copies x.
Half turn Turn (degrees) = 180	Green copies the circle y-coordinate; gold copies x.
Full-turn boundary Turn (degrees) = 360	Green copies the circle y-coordinate; gold copies x.

Conclusion: The horizontal graph coordinate records angle; graph height copies the circle y-coordinate. Cosine similarly copies x.

Conditions: The graph axis uses radians even when the control reports degrees; a full turn is2π.

Justification: For every θ the point(θ,sinθ) is constructed by copying the unit-circle ordinate. Equal points after2π give periodicity.

Ask: Which quantity keeps increasing while the circle point returns?

Watch for: Reading the graph horizontal coordinate as the circle x-coordinate.

Extension: Locate matching sine values with different angle inputs.

Exit: What graph point corresponds to one full turn?

(2π,0); the angle is not0 although the height is0.

Before: Angles, coordinates and basic function graphs. · Timing: predict 1, compare 3, explain 2, exit 1 minute. Finite observations do not prove a universal claim.

07 Trigonometric functions · g10-07-06 · 5–10 min · TEACHER · ANSWERS

Where does tangent break?

Can a tangent graph pass through90°?

Predict: Predict tanθ at85°,90°,95°.

Compare	Computed observation
Before the pole Angle θ (degrees) = 85	The quotient is defined at this input.
At the pole Angle θ (degrees) = 90	The denominator is zero: there is no tangent value here.
After the pole Angle θ (degrees) = 95	The quotient is defined at this input.

Conclusion: Tangent is sinθ/cosθ. At90°+180°k its denominator is0, so these inputs are excluded, not assigned a very large value.

Conditions: θ≠90°+180°k for integer k. The basic period is180°.

Justification: Both sine and cosine reverse sign after180°, so their quotient is unchanged where defined. Cosθ=0 makes division invalid.

Ask: Is a large finite table value the same as undefined?

Watch for: Joining the two sides of a vertical asymptote with a straight line.

Extension: Compare tan30° and tan210° without calculating each separately.

Exit: State whether tan270° exists.

No: cos270°=0.

Before: Angles, coordinates and basic function graphs. · Timing: predict 1, compare 3, explain 2, exit 1 minute. Finite observations do not prove a universal claim.

07 Trigonometric functions · g10-07-07 · 5–10 min · TEACHER · ANSWERS

Can sine distribute over addition?

Is sin(α+β)=sinα+sinβ?

Predict: Test α=β=30° mentally before revealing the result.

Compare	Computed observation
Two30° turns α (degrees) = 30; β (degrees) = 30	The product identity agrees; the simple additive claim generally fails.
Unequal turns α (degrees) = 30; β (degrees) = 60	The product identity agrees; the simple additive claim generally fails.
A misleading special case α (degrees) = 30; β (degrees) = 0	The product identity agrees; the simple additive claim generally fails.

Conclusion: The sum-angle identity is sin(α+β)=sinαcosβ+cosαsinβ. Sine is not an additive function.

Conditions: All angles must use the same unit. A successful special case such as β=0 is not a general proof.

Justification: Rotate(cosα,sinα) by β. Its new y-coordinate is sinαcosβ+cosαsinβ, which is sin(α+β).

Ask: Which products account for the rotation of each original coordinate?

Watch for: Replacing sin(α+β) with sinα+sinβ.

Extension: Use β=−α to obtain a difference-angle relation.

Exit: Give an exact counterexample to the additive claim.

α=β=30°: sin60°=√3/2, but sin30°+sin30°=1.

Before: Angles, coordinates and basic function graphs. · Timing: predict 1, compare 3, explain 2, exit 1 minute. Finite observations do not prove a universal claim.

07 Trigonometric functions · g10-07-08 · 5–10 min · TEACHER · ANSWERS

Double an angle, not its sine

How does sin2θ relate to sinθ?

Predict: At θ=45°, compare sin2θ,2sinθ and2sinθcosθ.

Compare	Computed observation
45° Angle θ (degrees) = 45	The product identity agrees; the simple additive claim generally fails.
30° Angle θ (degrees) = 30	The product identity agrees; the simple additive claim generally fails.
Zero angle Angle θ (degrees) = 0	The product identity agrees; the simple additive claim generally fails.

Conclusion: Putting α=β=θ into the sum formula gives sin2θ=2sinθcosθ and cos2θ=cos²θ−sin²θ.

Conditions: The identities hold for every real θ; 2sinθ alone is generally different.

Justification: Substitute equal angles into the proved angle-addition identities; no finite slider sweep is needed for the universal result.

Ask: Which extra factor is lost in the incorrect formula?

Watch for: Doubling a function input by doubling its output.

Extension: Derive cos2θ=1−2sin²θ.

Exit: Find sin90° using θ=45° in the double-angle formula.

2(√2/2)(√2/2)=1.

Before: Angles, coordinates and basic function graphs. · Timing: predict 1, compare 3, explain 2, exit 1 minute. Finite observations do not prove a universal claim.

07 Trigonometric functions · g10-07-09 · 5–10 min · TEACHER · ANSWERS

Amplitude and a reflected wave

Does a negative coefficient mean a negative amplitude?

Predict: Compare the ranges of2sinx,−2sinx and0sinx.

Compare	Computed observation
Positive coefficient Coefficient A = 2	The grey reference is sinx; compare one parameter at a time.
Negative coefficient Coefficient A = -2	The grey reference is sinx; compare one parameter at a time.
No oscillation Coefficient A = 0	This is a constant: do not apply the nonconstant basic-period formula.

Conclusion: For A sinx, the range is[−|A|,|A|]. Negative A reflects the wave; its amplitude is |A|. A=0 gives a constant.

Conditions: For a nonconstant sine wave A≠0. The zero case has range{0} and no least positive period.

Justification: Since −1≤sinx≤1 and both bounds are attained, multiplication by A gives the stated range, reversing order when A<0.

Ask: Which quantity records reflection and which records size?

Watch for: Calling the amplitude−2 or treating A=0 as an ordinary oscillation.

Extension: Compare A sinx and |A|sinx when A<0.

Exit: State the amplitude and range of−3sinx.

Amplitude3; range[−3,3].

Before: Angles, coordinates and basic function graphs. · Timing: predict 1, compare 3, explain 2, exit 1 minute. Finite observations do not prove a universal claim.

07 Trigonometric functions · g10-07-10 · 5–10 min · TEACHER · ANSWERS

Frequency squeezes the horizontal axis

Why is the period2π/|ω| rather than2πω?

Predict: Compare sinx,sin2x andsin(−2x) on the same interval.

Compare	Computed observation
Unit frequency Frequency ω = 1	The grey reference is sinx; compare one parameter at a time.
Double frequency Frequency ω = 2	The grey reference is sinx; compare one parameter at a time.
Zero frequency Frequency ω = 0	This is a constant: do not apply the nonconstant basic-period formula.

Conclusion: The angle must change by2π to repeat. A nonzero ω therefore gives basic period2π/|ω|; negative ω reverses direction.

Conditions: This basic-period formula requires nonzero frequency and a nonzero sine coefficient. ω=0 gives a constant.

Justification: Solve |ω|T=2π for the least positive T. The sine function has least positive angular period2π.

Ask: How much must x change to add one full turn inside the sine?

Watch for: Dividing by ω without an absolute value or dividing by0.

Extension: Explain why a constant has periods but no least positive period.

Exit: Find the basic period ofsin(3x).

2π/3.

Before: Angles, coordinates and basic function graphs. · Timing: predict 1, compare 3, explain 2, exit 1 minute. Finite observations do not prove a universal claim.

07 Trigonometric functions · g10-07-11 · 5–10 min · TEACHER · ANSWERS

Phase is divided by frequency

Does adding φ shift every sine graph by−φ?

Predict: Compare sin(x+90°) andsin(2x+90°), using x in degrees for this comparison.

Compare	Computed observation
Frequency1 Frequency ω = 1; Phase φ (degrees) = 90	The shift is −φ/ω, not just −φ.
Frequency2 Frequency ω = 2; Phase φ (degrees) = 90	The shift is −φ/ω, not just −φ.
Zero frequency Frequency ω = 0; Phase φ (degrees) = 90	This is a constant: do not apply the nonconstant basic-period formula.

Conclusion: Factoring ωx+φ=ω(x+φ/ω) shows horizontal shift−φ/ω. Phase and horizontal displacement are different quantities.

Conditions: ω≠0 is required for a horizontal-shift interpretation; convert φ to radians before plotting against radian x.

Justification: Use algebraic factorisation inside the sine, then apply the definition of horizontal translation to x−h.

Ask: Where is the input at which the inner angle becomes0?

Watch for: Shifting by−φ even when the frequency is not1.

Extension: Predict the shift if ω is negative.

Exit: Find the shift ofsin(2x+π/2).

−π/4, meaning π/4 to the left.

Before: Angles, coordinates and basic function graphs. · Timing: predict 1, compare 3, explain 2, exit 1 minute. Finite observations do not prove a universal claim.

07 Trigonometric functions · g10-07-12 · 5–10 min · TEACHER · ANSWERS

A wheel needs a starting position

Can amplitude and period alone determine a height model?

Predict: A wheel has radius5, centre height6 and period40s. Predict heights at0,10 and30s when starting on the right and moving upwards.

Compare	Computed observation
Upward quarter turn Radius R (m) = 5; Centre height H (m) = 6; Period T (s) = 40; Time t (s) = 10; Starting phase (degrees) = 0	Check starting position, direction and units before trusting the model.
Three-quarter turn Radius R (m) = 5; Centre height H (m) = 6; Period T (s) = 40; Time t (s) = 30; Starting phase (degrees) = 0	Check starting position, direction and units before trusting the model.
Invalid zero period Radius R (m) = 5; Centre height H (m) = 6; Period T (s) = 0; Time t (s) = 10; Starting phase (degrees) = 0	Need R>0,T>0 and H≥R for an above-ground wheel.

Conclusion: With an upward start at centre height, h(t)=H+Rsin(2πt/T). Changing the starting phase changes the height at the same time.

Conditions: Use T>0,R>0 and consistent time units. A physical wheel above ground additionally needs H≥R.

Justification: Uniform rotation gives angle2πt/T+φ; the circle y-coordinate is Rsin of that angle, then add centre height H.

Ask: What must the formula predict at t=0 before using it later?

Watch for: Using cosine automatically without checking the starting point.

Extension: Modify the model for a start at the lowest point.

Exit: For the stated wheel, find h(10) andh(30).

11 and1 metres.

Before: Angles, coordinates and basic function graphs. · Timing: predict 1, compare 3, explain 2, exit 1 minute. Finite observations do not prove a universal claim.