Time limit: 20 minutes
Total marks: 30
Instructions: Section A is multiple choice (1 mark each). Section B is matching (1 mark each pair). Section C is True/False — you MUST give a one-line justification for the mark.
Q1. Convert 135 ° 135° 135° 3 π 4 \dfrac{3\pi}{4} 4 3 π 2 π 3 \dfrac{2\pi}{3} 3 2 π 5 π 6 \dfrac{5\pi}{6} 6 5 π 3 π 2 \dfrac{3\pi}{2} 2 3 π [1]
Q2. In which quadrant are BOTH sin θ \sin\theta sin θ cos θ \cos\theta cos θ [1]
Q3. The reference angle for 210 ° 210° 210° 30 ° 30° 30° 60 ° 60° 60° 150 ° 150° 150° 210 ° 210° 210° [1]
Q4. The period of y = tan x y=\tan x y = tan x π / 2 \pi/2 π /2 π \pi π 2 π 2\pi 2 π 4 π 4\pi 4 π [1]
Q5. For y = 3 sin ( 2 x + π ) − 1 y = 3\sin(2x + \pi) - 1 y = 3 sin ( 2 x + π ) − 1 3 , π 3,\ \pi 3 , π 2 , π 2,\ \pi 2 , π 3 , 2 π 3,\ 2\pi 3 , 2 π 2 , 2 π 2,\ 2\pi 2 , 2 π [1]
Q6. Which is a correct Pythagorean identity?
(a) 1 + tan 2 θ = sec 2 θ 1 + \tan^2\theta = \sec^2\theta 1 + tan 2 θ = sec 2 θ 1 + cot 2 θ = sec 2 θ 1 + \cot^2\theta = \sec^2\theta 1 + cot 2 θ = sec 2 θ sin 2 θ − cos 2 θ = 1 \sin^2\theta - \cos^2\theta = 1 sin 2 θ − cos 2 θ = 1 1 + sec 2 θ = tan 2 θ 1 + \sec^2\theta = \tan^2\theta 1 + sec 2 θ = tan 2 θ [1]
Q7. The arc length of a sector of radius 6 6 6 π 3 \dfrac{\pi}{3} 3 π π \pi π 2 π 2\pi 2 π 3 π 3\pi 3 π 6 π 6\pi 6 π [1]
Q8. The range of y = arcsin x y=\arcsin x y = arcsin x [ 0 , π ] [0,\pi] [ 0 , π ] [ − π 2 , π 2 ] \left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right] [ − 2 π , 2 π ] ( − ∞ , ∞ ) (-\infty,\infty) ( − ∞ , ∞ ) [ 0 , 2 π ] [0,2\pi] [ 0 , 2 π ] [1]
Q9. cos 2 A \cos 2A cos 2 A cos 2 A − sin 2 A \cos^2A - \sin^2A cos 2 A − sin 2 A 1 − 2 sin 2 A 1 - 2\sin^2A 1 − 2 sin 2 A 2 cos 2 A − 1 2\cos^2A - 1 2 cos 2 A − 1 2 sin A cos A 2\sin A\cos A 2 sin A cos A [1]
Q10. The area of a triangle with sides a , b a,b a , b C C C 1 2 a b cos C \tfrac12 ab\cos C 2 1 ab cos C 1 2 a b sin C \tfrac12 ab\sin C 2 1 ab sin C a b sin C ab\sin C ab sin C 1 2 a b tan C \tfrac12 ab\tan C 2 1 ab tan C [1]
Q11. Match each identity/formula on the left to its correct expression on the right. [8]
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Left
1
sin ( A + B ) \sin(A+B) sin ( A + B )
2
cos ( A − B ) \cos(A-B) cos ( A − B )
3
tan 2 A \tan 2A tan 2 A
4
Law of cosines for c 2 c^2 c 2
5
csc θ \csc\theta csc θ
6
Co-function: cos θ \cos\theta cos θ
7
sin ( A 2 ) \sin\left(\dfrac{A}{2}\right) sin ( 2 A )
8
Sector area
Letter
Right
P
1 sin θ \dfrac{1}{\sin\theta} sin θ 1
Q
sin ( π 2 − θ ) \sin\left(\dfrac{\pi}{2}-\theta\right) sin ( 2 π − θ )
R
± 1 − cos A 2 \pm\sqrt{\dfrac{1-\cos A}{2}} ± 2 1 − cos A
S
sin A cos B + cos A sin B \sin A\cos B + \cos A\sin B sin A cos B + cos A sin B
T
a 2 + b 2 − 2 a b cos C a^2+b^2-2ab\cos C a 2 + b 2 − 2 ab cos C
U
cos A cos B + sin A sin B \cos A\cos B + \sin A\sin B cos A cos B + sin A sin B
V
2 tan A 1 − tan 2 A \dfrac{2\tan A}{1-\tan^2 A} 1 − tan 2 A 2 tan A
W
1 2 r 2 θ \tfrac12 r^2\theta 2 1 r 2 θ
Each: 1 mark correct T/F, 1 mark justification.
Q12. sin ( 90 ° − θ ) = cos θ \sin(90° - \theta) = \cos\theta sin ( 90° − θ ) = cos θ θ \theta θ [2]
Q13. The graph of y = sec x y=\sec x y = sec x 1 1 1 [2]
Q14. π \pi π 180 ° 180° 180° [2]
Q15. tan θ = cos θ sin θ \tan\theta = \dfrac{\cos\theta}{\sin\theta} tan θ = sin θ cos θ [2]
Q16. The product-to-sum formula gives 2 sin A cos B = sin ( A + B ) + sin ( A − B ) 2\sin A\cos B = \sin(A+B) + \sin(A-B) 2 sin A cos B = sin ( A + B ) + sin ( A − B ) [2]
Q17. In the ambiguous case, the Law of Sines can produce two possible triangles. [2]
Answer key Mark scheme & solutions
Q1. (a) 135 ° × π 180 ° = 135 π 180 = 3 π 4 135° \times \dfrac{\pi}{180°} = \dfrac{135\pi}{180} = \dfrac{3\pi}{4} 135° × 180° π = 180 135 π = 4 3 π (Convert by ×π/180.) [1]
Q2. (c) III. By ASTC, only Tangent is positive in QIII, so sin and cos are both negative. [1]
Q3. (a) 30 ° 30° 30° 210 ° 210° 210° = 210 ° − 180 ° = 30 ° = 210° - 180° = 30° = 210° − 180° = 30° [1]
Q4. (b) π \pi π Tangent repeats every π \pi π 2 π 2\pi 2 π [1]
Q5. (a) 3 , π 3,\ \pi 3 , π Amplitude = ∣ A ∣ = 3 =|A|=3 = ∣ A ∣ = 3 = 2 π ∣ B ∣ = 2 π 2 = π =\dfrac{2\pi}{|B|}=\dfrac{2\pi}{2}=\pi = ∣ B ∣ 2 π = 2 2 π = π [1]
Q6. (a). 1 + tan 2 θ = sec 2 θ 1+\tan^2\theta=\sec^2\theta 1 + tan 2 θ = sec 2 θ sin 2 + cos 2 = 1 \sin^2+\cos^2=1 sin 2 + cos 2 = 1 cos 2 \cos^2 cos 2 [1]
Q7. (b) 2 π 2\pi 2 π s = r θ = 6 ⋅ π 3 = 2 π s = r\theta = 6\cdot\dfrac{\pi}{3}=2\pi s = r θ = 6 ⋅ 3 π = 2 π [1]
Q8. (b) [ − π 2 , π 2 ] \left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right] [ − 2 π , 2 π ] Standard range of arcsin. [1]
Q9. (d). 2 sin A cos A = sin 2 A 2\sin A\cos A = \sin 2A 2 sin A cos A = sin 2 A cos 2 A \cos 2A cos 2 A cos 2 A \cos 2A cos 2 A [1]
Q10. (b) 1 2 a b sin C \tfrac12 ab\sin C 2 1 ab sin C Standard triangle area with included angle. [1]
Q11. (1 mark each correct pair, 8 total)
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Answer
1 → S
sin ( A + B ) = sin A cos B + cos A sin B \sin(A+B)=\sin A\cos B+\cos A\sin B sin ( A + B ) = sin A cos B + cos A sin B
2 → U
cos ( A − B ) = cos A cos B + sin A sin B \cos(A-B)=\cos A\cos B+\sin A\sin B cos ( A − B ) = cos A cos B + sin A sin B
3 → V
tan 2 A = 2 tan A 1 − tan 2 A \tan 2A=\dfrac{2\tan A}{1-\tan^2A} tan 2 A = 1 − tan 2 A 2 tan A
4 → T
c 2 = a 2 + b 2 − 2 a b cos C c^2=a^2+b^2-2ab\cos C c 2 = a 2 + b 2 − 2 ab cos C
5 → P
csc θ = 1 / sin θ \csc\theta=1/\sin\theta csc θ = 1/ sin θ
6 → Q
cos θ = sin ( π / 2 − θ ) \cos\theta=\sin(\pi/2-\theta) cos θ = sin ( π /2 − θ )
7 → R
sin ( A / 2 ) = ± ( 1 − cos A ) / 2 \sin(A/2)=\pm\sqrt{(1-\cos A)/2} sin ( A /2 ) = ± ( 1 − cos A ) /2
8 → W
Sector area = 1 2 r 2 θ =\tfrac12 r^2\theta = 2 1 r 2 θ
Q12. TRUE. Justification: co-function identity — sin ( 90 ° − θ ) = cos θ \sin(90°-\theta)=\cos\theta sin ( 90° − θ ) = cos θ θ \theta θ [2]
Q13. FALSE. Justification: sec x \sec x sec x ∣ y ∣ ≥ 1 |y|\ge 1 ∣ y ∣ ≥ 1 [2]
Q14. TRUE. Justification: half a full revolution; 180 ° = π 180° = \pi 180° = π [2]
Q15. FALSE. Justification: quotient identity is tan θ = sin θ cos θ \tan\theta=\dfrac{\sin\theta}{\cos\theta} tan θ = cos θ sin θ cot θ \cot\theta cot θ [2]
Q16. TRUE. Justification: adding sin ( A + B ) \sin(A+B) sin ( A + B ) sin ( A − B ) \sin(A-B) sin ( A − B ) cos A sin B \cos A\sin B cos A sin B 2 sin A cos B 2\sin A\cos B 2 sin A cos B [2]
Q17. TRUE. Justification: SSA data (two sides + non-included angle) may give 0, 1, or 2 valid triangles — the ambiguous case. [2]
[ {"claim":"135 degrees = 3pi/4 radians","code":"result = simplify(Rational(135)*pi/180 - 3*pi/4) == 0"}, {"claim":"arc length r=6 theta=pi/3 equals 2pi","code":"result = simplify(6*(pi/3) - 2*pi) == 0"}, {"claim":"period of 3 sin(2x+pi)-1 is pi","code":"result = simplify(2*pi/2 - pi) == 0"}, {"claim":"2 sinA cosB = sin(A+B)+sin(A-B)","code":"A,B=symbols('A B'); result = simplify(sin(A+B)+sin(A-B) - 2*sin(A)*cos(B)) == 0"}, {"claim":"1+tan^2 = sec^2","code":"t=symbols('t'); result = simplify((1+tan(t)**2) - sec(t)**2) == 0"} ]