⚡ Quick Memory Point
SOH — CAH — TOA
SOH = Sin = Opposite/Hypotenuse
CAH = Cos = Adjacent/Hypotenuse
TOA = Tan = Opposite/Adjacent
Question 01
Easy
In a right triangle, the side opposite to angle θ = 3 and the hypotenuse = 5. What is sin θ?
💡 Tip: Which side is "across" from the angle? That's opposite.
\( \sin\theta = \dfrac{\text{Opposite}}{\text{Hypotenuse}} \)
A\(\dfrac{4}{5}\)
B\(\dfrac{3}{5}\)
C\(\dfrac{3}{4}\)
D\(\dfrac{5}{3}\)
Question 02
Easy
Using the same triangle (opposite = 3, hypotenuse = 5), find the adjacent side, then calculate cos θ.
Example
By the Pythagorean theorem: \(a^2 + b^2 = c^2\), so adjacent = \(\sqrt{5^2 - 3^2} = \sqrt{16} = 4\).
A\(\dfrac{3}{5}\)
B\(\dfrac{5}{4}\)
C\(\dfrac{4}{5}\)
D\(\dfrac{4}{3}\)
Question 03
Easy
In a right triangle, opposite = 5 and adjacent = 12. What is tan θ?
\( \tan\theta = \dfrac{\text{Opposite}}{\text{Adjacent}} \)
A\(\dfrac{12}{13}\)
B\(\dfrac{12}{5}\)
C\(\dfrac{5}{13}\)
D\(\dfrac{5}{12}\)
⚡ Quick Memory Point
The "1–√2–√3" Ladder Rule
sin 30° = ½
sin 45° = √2/2
sin 60° = √3/2
cos = reverse order of sin !
Question 04
Easy
What is the value of cos 60°?
💡 Tip: cos goes DOWN the ladder → cos 30°=√3/2, cos 45°=√2/2, cos 60°=½
A\(\dfrac{\sqrt{3}}{2}\)
B\(\dfrac{\sqrt{2}}{2}\)
C\(\dfrac{1}{2}\)
D\(1\)
Question 05
Easy
Which of the following is NOT a valid special angle value?
⚠️ Common trap: Students often confuse sin 45° with sin 30°.
Asin 30° = \(\dfrac{1}{2}\)
Bcos 45° = \(\dfrac{\sqrt{2}}{2}\)
Ctan 60° = \(\dfrac{1}{\sqrt{3}}\)
Dsin 60° = \(\dfrac{\sqrt{3}}{2}\)
⚡ Quick Memory Point
The Holy Trinity of Trig Identities
sin²θ + cos²θ = 1
1 + tan²θ = sec²θ
1 + cot²θ = csc²θ
Question 06
Easy
If sin θ = 3/5, what is cos θ? (Assume θ is in the first quadrant.)
\( \sin^2\theta + \cos^2\theta = 1 \)
Example Strategy
Plug sin θ in: \(\left(\tfrac{3}{5}\right)^2 + \cos^2\theta = 1\) → \(\cos^2\theta = 1 - \tfrac{9}{25} = \tfrac{16}{25}\) → cos θ = ?
A\(\dfrac{3}{4}\)
B\(\dfrac{4}{5}\)
C\(\dfrac{5}{4}\)
D\(\dfrac{3}{5}\)
Question 07
Medium
If cos θ = −4/5 and θ is in Quadrant II, find sin θ.
⚠️ Common mistake: Forgetting that sin is POSITIVE in QII. Always check the quadrant sign!
A\(-\dfrac{3}{5}\)
B\(\dfrac{4}{5}\)
C\(\dfrac{3}{5}\)
D\(-\dfrac{4}{5}\)
⚡ Quick Memory Point
ASTC — "All Students Take Calculus"
QI: All positive
QII: Sin only positive
QIII: Tan only positive
QIV: Cos only positive
Question 08
Easy
In Quadrant III, which trig functions are positive?
💡 Tip: Use "All Students Take Calculus" — the 3rd word is "Take" = Tan.
Asin and cos
Bcos only
Ctan only
DAll three
Question 09
Medium
What is the exact value of sin(210°)?
Strategy — Reference Angle
210° is in QIII. Reference angle = 210° − 180° = 30°. In QIII, sin is negative. So sin(210°) = −sin(30°) = ?
A\(\dfrac{\sqrt{3}}{2}\)
B\(-\dfrac{\sqrt{3}}{2}\)
C\(\dfrac{1}{2}\)
D\(-\dfrac{1}{2}\)
Question 10
Medium
What is cos(−60°)?
⚠️ Common trap: cos is an EVEN function → cos(−θ) = cos(θ). Many students apply the negative sign to the answer.
A\(-\dfrac{1}{2}\)
B\(-\dfrac{\sqrt{3}}{2}\)
C\(\dfrac{\sqrt{3}}{2}\)
D\(\dfrac{1}{2}\)
⚡ Quick Memory Point
y = A sin(Bx + C) + D
A = Amplitude (height)
Period = 2π/B
Phase shift = −C/B (left/right)
D = Vertical shift (up/down)
📈 Reference — One Full Period of y = sin(x)
Question 11
Easy
What is the amplitude of \( y = 3\sin(2x) \)?
💡 Tip: Amplitude = |A| — just the absolute value of the coefficient in front of sin.
A\(2\)
B\(\pi\)
C\(3\)
D\(6\)
Question 12
Medium
What is the period of \( y = \cos\!\left(\dfrac{x}{3}\right) \)?
\( \text{Period} = \dfrac{2\pi}{|B|} \)
⚠️ Common trap: B is the coefficient of x inside the function. Here B = 1/3, not 3.
A\(\dfrac{2\pi}{3}\)
B\(6\pi\)
C\(2\pi\)
D\(3\pi\)
Question 13
Medium
For \( y = 2\sin(3x - \pi) + 1 \), what is the phase shift?
\( \text{Phase Shift} = -\dfrac{C}{B} \)
Careful Reading
Rewrite as \(y = 2\sin(3(x - \pi/3)) + 1\). The shift is π/3 to the RIGHT. Negative means left, positive means right!
Aπ to the left
Bπ/3 to the left
Cπ/3 to the right
D3π to the right
Question 14
Medium
The graph of y = sin x is shifted 2 units up and the amplitude is tripled. Which equation matches?
A\(y = 3\sin(x+2)\)
B\(y = 3\sin(x) + 2\)
C\(y = \sin(3x) + 2\)
D\(y = \sin(x+3) + 2\)
📈 Reference — y = cos(x) vs y = sin(x)
⚡ Quick Memory Point
π = 180°, That's the Only Formula You Need
Degrees → Radians: × π/180
Radians → Degrees: × 180/π
Arc length: s = rθ (θ in radians)
Question 15
Easy
Convert 135° to radians.
\( \text{Radians} = \text{Degrees} \times \dfrac{\pi}{180} \)
A\(\dfrac{\pi}{4}\)
B\(\dfrac{3\pi}{4}\)
C\(\dfrac{2\pi}{3}\)
D\(\dfrac{5\pi}{6}\)
Question 16
Medium
A circle has radius 6. A central angle of π/3 radians subtends an arc. What is the arc length?
\( s = r\theta \)
A\(3\pi\)
B\(2\pi\)
C\(4\pi\)
D\(\pi\)
Question 17
Medium
Which radian measure is equivalent to 270°?
⚠️ Trap: 270° is 3/4 of a full circle (2π). Calculate carefully: 270 × π/180 = ?
A\(\pi\)
B\(\dfrac{3\pi}{2}\)
C\(2\pi\)
D\(\dfrac{4\pi}{3}\)
Question 18
Hard
The graph of a sinusoidal function has a maximum of 7 and a minimum of 1. What are the amplitude and vertical shift (D)?
\( A = \dfrac{\text{max} - \text{min}}{2}, \quad D = \dfrac{\text{max} + \text{min}}{2} \)
Key Insight
Amplitude = half the total "height" of the wave. Vertical shift = the "middle" (average of max and min). These two formulas are heavily tested!
AA = 7, D = 1
BA = 3, D = 4
CA = 6, D = 7
DA = 4, D = 3
Question 19
Hard
Which identity is used to simplify \(\dfrac{\sin^2\theta}{1 - \cos^2\theta}\)?
⚠️ Watch out: 1 − cos²θ = sin²θ. The expression simplifies to something beautifully simple.
AThe result is \(\tan^2\theta\)
BThe result is \(1\)
CThe result is \(\cos^2\theta\)
DThe result is \(0\)
Question 20
Hard
Which function has a period of π (not 2π)?
💡 Key insight: tan and cot naturally have a period of π. But sin/cos with B=2 also gives period π.
Think it through
For y = sin(Bx): Period = 2π/B. To get π, you need B = 2. But y = tan(x) already has period π by nature!
A\(y = \sin(x)\)
B\(y = \cos\!\left(\dfrac{x}{2}\right)\)
C\(y = \tan(x)\)
D\(y = \cos(x) + 1\)
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out of 20 correct