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Trigonometry · eight-question check

Read the idea, work independently, then explain what changed.

TOPIC 01

Trigonometry

Try two tasks per skill independently, then revisit the linked skill.

Try it. Leave your reasoning visible.

Use one hint at a time. A correction explains what changed, not just the final answer.

01 / Foundation#Your turn

Which sign pair is correct in quadrant III?

(sin⁡θ,cos⁡θ)(\sin\theta,\cos\theta)
  • Angle lies strictly in quadrant III.
Skills and prerequisite lessons

Scope: JM01 / JM02

  1. (+,+)
  2. (+,−)
  3. (−,+)
  4. (−,−)

Working and explanation

BUILD THE REASONING

Hint 1
Think y then x.
Hint 2
Both coordinates are negative.
Worked solution
  1. Normalisation divides by positive r.

    r>0r>0
  2. Both signs remain negative.

    (sin⁡θ,cos⁡θ)=(−,−)(\sin\theta,\cos\theta)=(-,-)

D: (−,−).

Checks and common pitfalls: Sine and cosine are signed coordinates, not positive lengths.

Think first. Reveal a hint when the class is ready.

02 / Foundation#Your turn

Evaluate the radian input.

sin⁡(π/6)\sin(\pi/6)
  • Input in radians.
Skills and prerequisite lessons

Scope: JM01 / JM02

  1. 1/2
  2. √3/2
  3. π/6
  4. 1

Working and explanation

BUILD THE REASONING

Hint 1
Convert π radians to 180°.
Hint 2
π/6 is 30°.
Worked solution
  1. Match angle units.

    π/6=30∘\pi/6=30^\circ
  2. Read the vertical coordinate.

    sin⁡(π/6)=1/2\sin(\pi/6)=1/2

A: 1/2.

Checks and common pitfalls: The value is not the angle measure itself.

Think first. Reveal a hint when the class is ready.

03 / Foundation#Your turn

Which value is sin(30°+30°)?

sin⁡60∘\sin60^\circ
  • Degrees.
Skills and prerequisite lessons

Scope: JM01 / JM02

  1. 1
  2. √3/2
  3. 1/2
  4. √3

Working and explanation

BUILD THE REASONING

Hint 1
Combine the angle first.
Hint 2
Use the 60° sine value.
Worked solution
  1. The combined angle is 60°.

    30∘+30∘=60∘30^\circ+30^\circ=60^\circ
  2. Read its unit-circle height.

    sin⁡60∘=3/2\sin60^\circ=\sqrt3/2

B: √3/2.

Checks and common pitfalls: Adding the two sine values incorrectly gives 1.

Think first. Reveal a hint when the class is ready.

04 / Foundation#Your turn

Is the original expression defined at x=0?

(1−cos⁡2x)/sin⁡x(1-\cos^2x)/\sin x
  • Radians.
Skills and prerequisite lessons

Scope: JM01 / JM02

  1. Defined, value 0
  2. Defined, value 1
  3. Undefined
  4. Infinite real value

Working and explanation

BUILD THE REASONING

Hint 1
Check the original denominator.
Hint 2
sin 0=0.
Worked solution
  1. Substitute into the unaltered quotient.

    (1−1)/0=0/0(1-1)/0=0/0
  2. Simplification cannot restore this input.

    x=0 excludedx=0\text{ excluded}

C: undefined.

Checks and common pitfalls: The simplified sine formula has a larger domain.

Think first. Reveal a hint when the class is ready.

05 / Foundation#Your turn

What is the horizontal shift?

y=sin⁡(2x+π/3)y=\sin(2x+\pi/3)
  • Radians.
Skills and prerequisite lessons

Scope: JM01 / JM02

  1. Left π/3
  2. Right π/6
  3. Left π/6
  4. Up π/3

Working and explanation

BUILD THE REASONING

Hint 1
Factor 2.
Hint 2
The displacement has the opposite sign inside x−h.
Worked solution
  1. Factor the full angle.

    2(x+π/6)2(x+\pi/6)
  2. Read the leftward displacement.

    h=−π/6h=-\pi/6

C: left π/6.

Checks and common pitfalls: Both division by frequency and direction matter.

Think first. Reveal a hint when the class is ready.

06 / Foundation#Your turn

For a constant function, what is the least positive period?

y=4y=4
  • All inputs are real unless a smaller domain is stated.
Skills and prerequisite lessons

Scope: JM01 / JM02

  1. 2π
  2. 0
  3. 4
  4. No least positive period

Working and explanation

BUILD THE REASONING

Hint 1
Every positive shift works.
Hint 2
For any proposed positive period, half of it also works.
Worked solution
  1. Compare shifted outputs.

    f(x+T)=4=f(x)f(x+T)=4=f(x)
  2. There is no smallest positive choice.

    T/2<T(T>0)T/2<T\quad(T>0)

D: none exists.

Checks and common pitfalls: Zero is not a positive period, and choosing 2π is not least.

Think first. Reveal a hint when the class is ready.

07 / Foundation#Your turn

Choose the complete solution set.

cos⁡x=1/2,0≤x<360∘\cos x=1/2,\quad0\le x<360^\circ
  • Degrees; interval half-open.
Skills and prerequisite lessons

Scope: JM01 / JM02

  1. {60°}
  2. {60°,300°}
  3. {60°,120°}
  4. {60°,240°}

Working and explanation

BUILD THE REASONING

Hint 1
Use both points with the same horizontal coordinate.
Hint 2
Cosine is positive in quadrants I and IV.
Worked solution
  1. Find the reference angle.

    60∘60^\circ
  2. Reflect across the x-axis.

    360∘−60∘=300∘360^\circ-60^\circ=300^\circ

B: {60°,300°}.

Checks and common pitfalls: The inverse principal value alone omits the reflected solution.

Think first. Reveal a hint when the class is ready.

08 / Foundation#Your turn

When 0≤x<360°, what interval must be searched for 2x?

u=2xu=2x
  • Degrees.
Skills and prerequisite lessons

Scope: JM01 / JM02

  1. [0,180°)
  2. [0,360°)
  3. [0,720°)
  4. [0,720°]

Working and explanation

BUILD THE REASONING

Hint 1
Multiply both endpoint bounds by positive 2.
Hint 2
Keep the endpoint openness.
Worked solution
  1. Transform the lower and upper bounds.

    0≤2x<720∘0\le2x<720^\circ
  2. The upper endpoint remains excluded.

    u∈[0,720∘)u\in[0,720^\circ)

C: [0,720°).

Checks and common pitfalls: Keeping [0,360°) would miss one phase cycle.

Think first. Reveal a hint when the class is ready.

Focus on one question

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