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Sec 4 A Math: Trigonometric Identities & R-Formula, practice questions & worked solutions

Trigonometric Identities & R-Formula practice questions selected from 2025 Singapore school A Math prelim papers, each with a full worked solution.

About this topic & key methods

These questions support Secondary 4 and O-Level Additional Mathematics revision.

Attempt each question on paper before opening its worked solution. Keep the source question number when checking against the original paper.

Key methods

  • Use reciprocal, quotient and Pythagorean identities.
  • Simplify or prove an identity by working on one side.
  • Solve an equation using a proved identity and the stated range.
  • Use addition and double-angle formulae to find exact values.
  • Express a sin θ + b cos θ in R-form and interpret its maximum or minimum.

For guided practice, see our Sec 4 A Maths tuition programme.

Questions & worked solutions

Unless the question specifies otherwise, give numerical answers to 3 significant figures and angles in degrees to 1 decimal place. Angles in radians are stated explicitly.

Q1Bedok South Secondary School2025Prelim P1 Q1●○○ Foundational4 marks

Trigonometric Identities

Prove that sin⁡2x−cos⁡2xtan⁡x=tan⁡x\sin 2x - \cos 2x \tan x = \tan x.

Show worked solution▾

LHS=sin⁡2x−cos⁡2xtan⁡x=2sin⁡xcos⁡x−(2cos⁡2x−1)sin⁡xcos⁡x=2sin⁡xcos⁡x−2sin⁡xcos⁡x+sin⁡xcos⁡x=tan⁡x=RHS\begin{aligned} \text{LHS} &= \sin 2x - \cos 2x \tan x\\ &= 2\sin x \cos x - (2\cos^2 x - 1)\frac{\sin x}{\cos x}\\ &= 2\sin x \cos x - 2\sin x \cos x + \frac{\sin x}{\cos x}\\ &= \tan x\\ &= \text{RHS} \end{aligned}

Q2Anglican High School2025Prelim P1 Q10●●○ Standard5 marks

Trigonometric Identities

Given that sin⁡A\sin A and cos⁡B\cos B are in the same quadrant such that sin⁡A=35\sin A = \dfrac{3}{5} and cos⁡B=−513\cos B = -\dfrac{5}{13}. Find, without using a calculator, the value of

  1. cos⁡(A−B)\cos(A - B),

  2. cos⁡A2\cos\dfrac{A}{2}.

Show worked solution▾

(a)

cos⁡(A−B)=cos⁡Acos⁡B+sin⁡Asin⁡B=(−45)(−513)+(35)(1213)=5665\begin{aligned} \cos(A - B) &= \cos A\cos B + \sin A\sin B\\ &= \left(-\frac{4}{5}\right)\left(-\frac{5}{13}\right) + \left(\frac{3}{5}\right)\left(\frac{12}{13}\right)\\ &= \frac{56}{65} \end{aligned}

(b)

cos⁡A=2cos⁡2A2−1cos⁡A+1=2cos⁡2A2−45+1=2cos⁡2A215=2cos⁡2A2cos⁡2A2=110cos⁡A2=±110\begin{aligned} \cos A &= 2\cos^2\frac{A}{2} - 1\\ \cos A + 1 &= 2\cos^2\frac{A}{2}\\ -\frac{4}{5} + 1 &= 2\cos^2\frac{A}{2}\\ \frac{1}{5} &= 2\cos^2\frac{A}{2}\\ \cos^2\frac{A}{2} &= \frac{1}{10}\\ \cos\frac{A}{2} &= \pm\sqrt{\frac{1}{10}} \end{aligned} 90∘<A<180∘45∘<A2<90∘cos⁡A2=1010(−1010 (NA))\begin{aligned} 90^\circ < A &< 180^\circ\\ 45^\circ < \frac{A}{2} &< 90^\circ\\ \cos\frac{A}{2} = \frac{\sqrt{10}}{10} &\left(-\frac{\sqrt{10}}{10}\ (\text{NA})\right) \end{aligned}

Q3Ahmad Ibrahim Secondary School2025Prelim P1 Q1●●○ Standard5 marks

Trigonometric Identities & Equations

It is given that PP, QQ and RR are the angles of a triangle.

  1. Show that cos⁡P=−cos⁡(Q+R)\cos P = -\cos(Q + R).

  2. Given that Q=45∘Q = 45^\circ and R=60∘R = 60^\circ, find cos⁡P\cos P in the form 14(a−b)\dfrac{1}{4}\left(\sqrt{a} - \sqrt{b}\right), where aa and bb are integers.

Show worked solution▾

(a)

cos⁡P=cos⁡[180∘−(Q+R)]=cos⁡180∘cos⁡(Q+R)+sin⁡180∘sin⁡(Q+R)=−cos⁡(Q+R)+0=−cos⁡(Q+R)\begin{aligned} \cos P &= \cos[180^\circ - (Q + R)]\\ &= \cos 180^\circ \cos(Q + R) + \sin 180^\circ \sin(Q + R)\\ &= -\cos(Q + R) + 0\\ &= -\cos(Q + R) \end{aligned}

Alternative: supplementary angles

cos⁡P=−cos⁡(180∘−P)=−cos⁡(Q+R)\begin{aligned} \cos P &= -\cos(180^\circ - P)\\ &= -\cos(Q + R) \end{aligned}

(b)

cos⁡P=−cos⁡(45∘+60∘)=−cos⁡45∘cos⁡60∘+sin⁡45∘sin⁡60∘=−22(12)+22(32)=14(6−2)\begin{aligned} \cos P &= -\cos(45^\circ + 60^\circ)\\ &= -\cos 45^\circ \cos 60^\circ + \sin 45^\circ \sin 60^\circ\\ &= -\frac{\sqrt{2}}{2}\left(\frac{1}{2}\right) + \frac{\sqrt{2}}{2}\left(\frac{\sqrt{3}}{2}\right)\\ &= \frac{1}{4}\left(\sqrt{6} - \sqrt{2}\right) \end{aligned}

Q4Cedar Girls’ Secondary School2025Prelim P1 Q9●●○ Standard5 marks

Trigonometry (Compound Angle)

Cedar Girls’ Secondary School 2025 P1 Q9: question diagram

The diagram shows a triangle PQRPQR.

Point MM lies on QRQR such that QM=1QM = 1 unit and MR=3MR = 3 units.

Angle PRQ=90°PRQ = 90° and PRPR is xx units. Angles QPRQPR and MPRMPR are denoted by α\alpha and β\beta respectively as shown, where 0°<α<45°0° < \alpha < 45°.

Given tan⁡(α−β)=18\tan(\alpha - \beta) = \dfrac{1}{8}, find the value of xx.

Show worked solution▾

tan⁡α=4x,tan⁡β=3xtan⁡(α−β)=18tan⁡α−tan⁡β1+tan⁡αtan⁡β=184x−3x1+(4x) ⁣(3x)=181x=18 ⁣(1+12x2)8x=x2+12x28x2=x3+12xx3−8x2+12x=0x(x2−8x+12)=0x(x−2)(x−6)=0x=0 (rej)x=2 (rej, α<45∘, tan⁡α<1)x=6\begin{aligned} \tan\alpha &= \frac{4}{x},\qquad \tan\beta = \frac{3}{x}\\ \tan(\alpha-\beta) &= \frac{1}{8}\\ \frac{\tan\alpha-\tan\beta}{1+\tan\alpha\tan\beta} &= \frac{1}{8}\\ \frac{\frac{4}{x}-\frac{3}{x}}{1+\left(\frac{4}{x}\right)\!\left(\frac{3}{x}\right)} &= \frac{1}{8}\\ \frac{1}{x} &= \frac{1}{8}\!\left(1+\frac{12}{x^2}\right)\\ \frac{8}{x} &= \frac{x^2+12}{x^2}\\ 8x^2 &= x^3+12x\\ x^3-8x^2+12x &= 0\\ x(x^2-8x+12) &= 0\\ x(x-2)(x-6) &= 0\\ x &= 0\ (\text{rej})\\ x &= 2\ (\text{rej},\ \alpha<45^\circ,\ \tan\alpha<1)\\ x &= 6 \end{aligned}

Q5Dunman Secondary School2025Prelim P1 Q6●●○ Standard6 marks

Trigonometry (Compound Angles)

  1. Show that sin⁡105∘=6+24\sin 105^{\circ} = \dfrac{\sqrt{6}+\sqrt{2}}{4}.

  2. The diagram below shows triangle ABCABC, such that AB=4AB = 4 cm, angle ABC=105∘ABC = 105^{\circ} and the area of triangle ABCABC is 10+2310 + 2\sqrt{3} cm2^{2}.

    Find BCBC in the form 2 ⁣(a3+b)\sqrt{2}\!\left(a\sqrt{3}+b\right), where aa and bb are integers.

    Dunman Secondary School 2025 P1 Q6: question diagram

Show worked solution▾

(a)

sin⁡(105∘)=sin⁡(60∘+45∘)=sin⁡60∘cos⁡45∘+cos⁡60∘sin⁡45∘=32⋅12+12⋅12=3+122×22=6+24\begin{aligned} \sin(105^\circ) &= \sin(60^\circ+45^\circ)\\ &= \sin 60^\circ\cos 45^\circ+\cos 60^\circ\sin 45^\circ\\ &= \frac{\sqrt{3}}{2}\cdot\frac{1}{\sqrt{2}}+\frac{1}{2}\cdot\frac{1}{\sqrt{2}}\\ &= \frac{\sqrt{3}+1}{2\sqrt{2}}\times\frac{\sqrt{2}}{\sqrt{2}}\\ &= \frac{\sqrt{6}+\sqrt{2}}{4} \end{aligned}

(b)

12(4)(BC)sin⁡105∘=10+2312(4)(BC) ⁣(6+24)=10+23BC=2(10+23)6+2×6−26−2=4(5+3)(6−2)6−2=56+18−52−6=46+32−52=46−22=2(43−2)\begin{aligned} \frac{1}{2}(4)(BC)\sin 105^\circ &= 10+2\sqrt{3}\\ \frac{1}{2}(4)(BC)\!\left(\frac{\sqrt{6}+\sqrt{2}}{4}\right) &= 10+2\sqrt{3}\\ BC &= \frac{2(10+2\sqrt{3})}{\sqrt{6}+\sqrt{2}}\times\frac{\sqrt{6}-\sqrt{2}}{\sqrt{6}-\sqrt{2}}\\ &= \frac{4(5+\sqrt{3})(\sqrt{6}-\sqrt{2})}{6-2}\\ &= 5\sqrt{6}+\sqrt{18}-5\sqrt{2}-\sqrt{6}\\ &= 4\sqrt{6}+3\sqrt{2}-5\sqrt{2}\\ &= 4\sqrt{6}-2\sqrt{2}\\ &= \sqrt{2}(4\sqrt{3}-2) \end{aligned}

Q6Jurongville Secondary School2025Prelim P2 Q4●●○ Standard6 marks

Trigonometric Identities & Equations

  1. Prove that cot⁡(45∘−x)=cot⁡x+1cot⁡x−1\cot(45^\circ - x) = \dfrac{\cot x + 1}{\cot x - 1}.

  2. Hence express cot⁡15∘\cot 15^\circ in the form p+q3p + q\sqrt{3}, where pp and qq are integers.

Show worked solution▾

(a)

cot⁡(45∘−x)=1tan⁡(45∘−x)=1+tan⁡45∘tan⁡xtan⁡45∘−tan⁡x=1+tan⁡x1−tan⁡x=1+1cot⁡x1−1cot⁡x=cot⁡x+1cot⁡x−1(proven)\begin{aligned} \cot(45^\circ - x) &= \frac{1}{\tan(45^\circ - x)}\\ &= \frac{1 + \tan 45^\circ \tan x}{\tan 45^\circ - \tan x}\\ &= \frac{1 + \tan x}{1 - \tan x}\\ &= \frac{1 + \dfrac{1}{\cot x}}{1 - \dfrac{1}{\cot x}}\\ &= \frac{\cot x + 1}{\cot x - 1} (\text{proven}) \end{aligned}

cot⁡(45∘−x)=cos⁡(45∘−x)sin⁡(45∘−x)=cos⁡45∘cos⁡x+sin⁡45∘sin⁡xsin⁡45∘cos⁡x−cos⁡45∘sin⁡x=cos⁡x+sin⁡xcos⁡x−sin⁡x=cos⁡xsin⁡x+1cos⁡xsin⁡x−1=cot⁡x+1cot⁡x−1(proven)\begin{aligned} \cot(45^\circ - x) &= \frac{\cos(45^\circ - x)}{\sin(45^\circ - x)}\\ &= \frac{\cos45^\circ\cos x + \sin45^\circ\sin x}{\sin45^\circ\cos x - \cos45^\circ\sin x}\\ &= \frac{\cos x + \sin x}{\cos x - \sin x}\\ &= \frac{\dfrac{\cos x}{\sin x}+1}{\dfrac{\cos x}{\sin x}-1}\\ &= \frac{\cot x + 1}{\cot x - 1} (\text{proven}) \end{aligned}

(b)

cot⁡15∘=cot⁡(45∘−30∘)=cot⁡30∘+1cot⁡30∘−1=1tan⁡30∘+11tan⁡30∘−1=3+13−1=3+13−1×3+13+1=3+23+13−1=4+232=2+3\begin{aligned} \cot 15^\circ &= \cot(45^\circ - 30^\circ)\\ &= \frac{\cot 30^\circ + 1}{\cot 30^\circ - 1}\\ &= \frac{\dfrac{1}{\tan 30^\circ} + 1}{\dfrac{1}{\tan 30^\circ} - 1}\\ &= \frac{\sqrt{3} + 1}{\sqrt{3} - 1}\\ &= \frac{\sqrt{3} + 1}{\sqrt{3} - 1} \times \frac{\sqrt{3} + 1}{\sqrt{3} + 1}\\ &= \frac{3 + 2\sqrt{3} + 1}{3 - 1}\\ &= \frac{4 + 2\sqrt{3}}{2}\\ &= 2 + \sqrt{3} \end{aligned}

Q7School of Science and Technology2025Prelim P2 Q6●●○ Standard6 marks

R-formula (Trigonometry)

The expression 3cos⁡θ+5sin⁡θ3\cos\theta + 5\sin\theta is defined for 0≤θ≤π20 \leq \theta \leq \dfrac{\pi}{2} radians.

  1. Express 3cos⁡θ+5sin⁡θ3\cos\theta + 5\sin\theta in the form Rcos⁡(θ−α)R\cos(\theta - \alpha), where R>0R > 0 and α\alpha is an acute angle in radian.

Hence,

  1. solve the equation 3cos⁡θ+5sin⁡θ=43\cos\theta + 5\sin\theta = 4,

  2. find the minimum value of 1(3cos⁡θ+5sin⁡θ)2+1\dfrac{1}{(3\cos\theta + 5\sin\theta)^2 + 1}.

Show worked solution▾

(a)

3cos⁡θ+5sin⁡θ=Rcos⁡(θ−α)3cos⁡θ+5sin⁡θ=32+52 cos⁡(θ−α),where α=tan⁡−1 ⁣(53)=34 cos⁡(θ−1.0303)(working value)=5.83cos⁡(θ−1.03) (3 s.f.)\begin{aligned} 3\cos\theta + 5\sin\theta &= R\cos(\theta - \alpha)\\ 3\cos\theta + 5\sin\theta &= \sqrt{3^2 + 5^2}\,\cos(\theta - \alpha), \text{where } \alpha = \tan^{-1}\!\left(\frac{5}{3}\right)\\ &= \sqrt{34}\,\cos(\theta - 1.0303) (\text{working value})\\ &= 5.83\cos(\theta - 1.03) \ (\text{3 s.f.}) \end{aligned}

(b)

3cos⁡θ+5sin⁡θ=434 cos⁡(θ−1.0303)=4cos⁡(θ−1.0303)=434\begin{aligned} 3\cos\theta + 5\sin\theta &= 4\\ \sqrt{34}\,\cos(\theta - 1.0303) &= 4\\ \cos(\theta - 1.0303) &= \frac{4}{\sqrt{34}} \end{aligned} Basic angle =cos⁡−1 ⁣(434)=0.81482= \cos^{-1}\!\left(\dfrac{4}{\sqrt{34}}\right) = 0.81482 θ−1.0303=0.81482, −0.81482θ=1.84512, 0.21548\begin{aligned} \theta - 1.0303 &= 0.81482,\ -0.81482\\ \theta &= 1.84512,\ 0.21548 \end{aligned} As θ\theta is acute, θ=0.215\theta = 0.215 radians (3 s.f.)

Alternatively, 0.2160.216 rad if the exact form is used.

(c)

−34≤3cos⁡θ+5sin⁡θ≤340≤(3cos⁡θ+5sin⁡θ)2≤341≤(3cos⁡θ+5sin⁡θ)2+1≤35135≤1(3cos⁡θ+5sin⁡θ)2+1≤1\begin{aligned} -\sqrt{34} \leq 3\cos\theta + 5\sin\theta &\leq \sqrt{34}\\ 0 \leq (3\cos\theta + 5\sin\theta)^2 &\leq 34\\ 1 \leq (3\cos\theta + 5\sin\theta)^2 + 1 &\leq 35\\ \frac{1}{35} \leq \frac{1}{(3\cos\theta + 5\sin\theta)^2 + 1} &\leq 1 \end{aligned} Minimum value =135= \dfrac{1}{35}

Q8Catholic High School2025Prelim P2 Q8●●○ Standard7 marks

Trigonometry (R-formula)

The diagram shows two rods ABAB and BCBC of length 2 m and 5 m respectively, joined together at BB such that angle ABC=90∘ABC = 90^{\circ}. Small rings are attached to AA and CC so that AA can move along the vertical pole OYOY and CC can move along the horizontal pole OXOX. The rod BCBC makes an angle θ\theta with OXOX.

  1. Show that OA=5sin⁡θ−2cos⁡θOA = 5\sin\theta - 2\cos\theta.

  2. Express OAOA in the form Rsin⁡(θ−α)R\sin(\theta - \alpha) where R>0R > 0 and 0∘<α<90∘0^{\circ} < \alpha < 90^{\circ}.

  3. Given that AA can be no higher than 3.5 m above OO, calculate the greatest possible value of θ\theta.

Catholic High School 2025 P2 Q8: question diagram

Show worked solution▾

(a)

BP5=sin⁡θ  ⇒  BP=5sin⁡θ,BQ2=cos⁡θ  ⇒  BQ=2cos⁡θ\frac{BP}{5}=\sin\theta \;\Rightarrow\; BP=5\sin\theta,\qquad \frac{BQ}{2}=\cos\theta \;\Rightarrow\; BQ=2\cos\theta OA=BP−BQ=5sin⁡θ−2cos⁡θOA = BP-BQ = 5\sin\theta-2\cos\theta

(b)

OA=5sin⁡θ−2cos⁡θ=R(sin⁡θcos⁡α−cos⁡θsin⁡α)OA = 5\sin\theta-2\cos\theta = R(\sin\theta\cos\alpha-\cos\theta\sin\alpha) Rcos⁡α=5,Rsin⁡α=2  ⇒  tan⁡α=25,α≈21.801∘R\cos\alpha=5, R\sin\alpha=2 \;\Rightarrow\; \tan\alpha=\frac{2}{5}, \alpha\approx 21.801^\circ R2=22+52=29  ⇒  R=29R^2=2^2+5^2=29 \;\Rightarrow\; R=\sqrt{29} OA=29 sin⁡(θ−21.8∘)OA = \sqrt{29}\,\sin(\theta-21.8^\circ)

(c)

29 sin⁡(θ−21.801∘)=3.5sin⁡(θ−21.801∘)=3.529θ−21.801∘=sin⁡−13.529≈40.536∘greatest θ=62.3∘\begin{aligned} \sqrt{29}\,\sin(\theta-21.801^\circ) &= 3.5\\ \sin(\theta-21.801^\circ) &= \frac{3.5}{\sqrt{29}}\\ \theta-21.801^\circ &= \sin^{-1}\frac{3.5}{\sqrt{29}} \approx 40.536^\circ\\ \text{greatest } \theta &= 62.3^\circ \end{aligned}

Q9Tanjong Katong Secondary School2025Prelim P1 Q7●●○ Standard7 marks

Trigonometry (Factor Formulae)

  1. Find the value of kk such that sin⁡(A+B)+sin⁡(A−B)=ksin⁡Acos⁡B\sin(A + B) + \sin(A - B) = k\sin A\cos B.

  2. By letting P=A+BP = A + B and Q=A−BQ = A - B, show that sin⁡P+sin⁡Q=2sin⁡ ⁣(P+Q2)cos⁡ ⁣(P−Q2).\sin P + \sin Q = 2\sin\!\left(\frac{P + Q}{2}\right)\cos\!\left(\frac{P - Q}{2}\right).

  3. Find the exact value of sin⁡75∘+sin⁡15∘\sin 75^\circ + \sin 15^\circ.

Show worked solution▾

(a)

sin⁡(A+B)+sin⁡(A−B)=sin⁡Acos⁡B+cos⁡Asin⁡B+sin⁡Acos⁡B−cos⁡Asin⁡B=2sin⁡Acos⁡Bk=2\begin{aligned} \sin(A + B) + \sin(A - B) &= \sin A\cos B + \cos A\sin B + \sin A\cos B - \cos A\sin B\\ &= 2\sin A\cos B\\ k &= 2 \end{aligned}

(b)

P+Q=2AA=P+Q2—– (1)P−Q=2BB=P−Q2—– (2)\begin{aligned} P + Q &= 2A\\ A &= \frac{P + Q}{2} \text{----- (1)}\\ P - Q &= 2B\\ B &= \frac{P - Q}{2} \text{----- (2)} \end{aligned} sin⁡(A+B)+sin⁡(A−B)=2sin⁡Acos⁡Bsin⁡P+sin⁡Q=2sin⁡ ⁣(P+Q2)cos⁡ ⁣(P−Q2)\begin{aligned} \sin(A + B) + \sin(A - B) &= 2\sin A\cos B\\ \sin P + \sin Q &= 2\sin\!\left(\frac{P + Q}{2}\right)\cos\!\left(\frac{P - Q}{2}\right) \end{aligned}

(c)

sin⁡75∘+sin⁡15∘=2sin⁡ ⁣(90∘2)cos⁡ ⁣(60∘2)=2(22)(32)=62\begin{aligned} \sin 75^\circ + \sin 15^\circ &= 2\sin\!\left(\frac{90^\circ}{2}\right)\cos\!\left(\frac{60^\circ}{2}\right)\\ &= 2\left(\frac{\sqrt{2}}{2}\right)\left(\frac{\sqrt{3}}{2}\right)\\ &= \frac{\sqrt{6}}{2} \end{aligned}

Q10Ngee Ann Secondary School2025Prelim P1 Q6●●● Challenging8 marks

Trigonometry (Compound & Half Angles)

  1. It is given that sin⁡A=−15\sin A = -\dfrac{1}{\sqrt{5}} and tan⁡B=−34\tan B = -\dfrac{3}{4}, where AA and BB are in the same quadrant. Without using a calculator, find the exact value of

    1. tan⁡(−A)\tan(-A),

    2. sin⁡(A+B)\sin(A + B),

    3. cos⁡B2\cos\dfrac{B}{2}.

  2. Given that sin⁡A=q\sin A = q, where AA is an obtuse angle, express tan⁡2A\tan^2 A in terms of qq.

Show worked solution▾

(a)

(i) tan⁡(−A)=−tan⁡A=12\begin{aligned} \tan(-A) &= -\tan A = \frac{1}{2} \end{aligned}

(ii) sin⁡(A+B)=sin⁡Acos⁡B+cos⁡Asin⁡B=(−15) ⁣(45)+(25) ⁣(−35)=−1055=−25=−255\begin{aligned} \sin(A+B) &= \sin A\cos B+\cos A\sin B\\ &= \left(-\frac{1}{\sqrt{5}}\right)\!\left(\frac{4}{5}\right)+\left(\frac{2}{\sqrt{5}}\right)\!\left(-\frac{3}{5}\right)\\ &= -\frac{10}{5\sqrt{5}}\\ &= -\frac{2}{\sqrt{5}}\\ &= -\frac{2\sqrt{5}}{5} \end{aligned}

(iii) cos⁡B=2cos⁡2 ⁣(B2)−1=45cos⁡2 ⁣(B2)=0.9cos⁡ ⁣(B2)=±310\begin{aligned} \cos B &= 2\cos^2\!\left(\frac{B}{2}\right)-1=\frac{4}{5}\\ \cos^2\!\left(\frac{B}{2}\right) &= 0.9\\ \cos\!\left(\frac{B}{2}\right) &= \pm\frac{3}{\sqrt{10}} \end{aligned} Since 135∘≤B2≤180∘135^\circ\le\dfrac{B}{2}\le 180^\circ, cos⁡ ⁣(B2)=−31010\cos\!\left(\frac{B}{2}\right) = -\frac{3\sqrt{10}}{10}

(b)

tan⁡A=−q1−q2tan⁡2A=(−q1−q2)2=q2(1+q)(1−q)orq21−q2\begin{aligned} \tan A &= -\frac{q}{\sqrt{1-q^2}}\\ \tan^2 A &= \left(-\frac{q}{\sqrt{1-q^2}}\right)^2\\ &= \frac{q^2}{(1+q)(1-q)} \text{or} \frac{q^2}{1-q^2} \end{aligned}

Q11Bedok View Secondary School2025Prelim P2 Q8●●● Challenging9 marks

Trigonometry (Exact Values)

Do not use a calculator in this question.

  1. Find the exact value of cosec⁡A\cosec A when cos⁡2A=13\cos 2A = \dfrac{1}{3} and AA is acute.

  2. Hence, find the exact value of cosec⁡A−2sin⁡105∘\cosec A - \sqrt{2}\sin 105^\circ.

Show worked solution▾

(a)

cos⁡2A=131−2sin⁡2A=13sin⁡2A=13sin⁡A=13 or −13 (rej)cosec⁡A=1sin⁡A=1÷13=3\begin{aligned} \cos 2A &= \frac{1}{3}\\ 1 - 2\sin^2 A &= \frac{1}{3}\\ \sin^2 A &= \frac{1}{3}\\ \sin A &= \frac{1}{\sqrt{3}} \ \text{or} \ -\frac{1}{\sqrt{3}} \ (\text{rej})\\ \cosec A &= \frac{1}{\sin A}\\ &= 1 \div \frac{1}{\sqrt{3}}\\ &= \sqrt{3} \end{aligned}

(b)

sin⁡105∘=sin⁡(45∘+60∘)=sin⁡45∘cos⁡60∘+cos⁡45∘sin⁡60∘=12⋅12+12⋅32=1+32⋅2\begin{aligned} \sin 105^\circ &= \sin(45^\circ + 60^\circ)\\ &= \sin 45^\circ \cos 60^\circ + \cos 45^\circ \sin 60^\circ\\ &= \frac{1}{\sqrt{2}}\cdot\frac{1}{2} + \frac{1}{\sqrt{2}}\cdot\frac{\sqrt{3}}{2}\\ &= \frac{1 + \sqrt{3}}{\sqrt{2}\cdot 2} \end{aligned} cosec⁡A−2sin⁡105∘=3−2⋅1+32⋅2=3−1+32=3−12\begin{aligned} \cosec A - \sqrt{2}\sin 105^\circ &= \sqrt{3} - \sqrt{2}\cdot\frac{1 + \sqrt{3}}{\sqrt{2}\cdot 2}\\ &= \sqrt{3} - \frac{1 + \sqrt{3}}{2}\\ &= \frac{\sqrt{3} - 1}{2} \end{aligned}

Q12Methodist Girls’ School2025Prelim P1 Q11●●● Challenging9 marks

R-formula

An architect is designing a building, ABCDABCD, using the model shown below. To allow more natural light into the building, a 12 m glass roof, BCBC, is installed at an angle of θ\theta to the vertical. Another glass roof, ABAB, of 7 m, is installed such that it is perpendicular to BCBC. The vertical height of the brick wall, ADAD, is given as hh m.

Methodist Girls’ School 2025 P1 Q11: question diagram

  1. Show that h=12cos⁡θ−7sin⁡θh = 12\cos\theta - 7\sin\theta.

  2. Express hh in the form Rcos⁡(θ+α)R\cos(\theta + \alpha), where R>0R > 0 and 0<α<π20 < \alpha < \dfrac{\pi}{2}.

  3. Find the value of θ\theta for which h=8h = 8 m.

  4. The building owner makes a request to have the brick wall to have a height of at least 14 m. Is the architect able to meet the building owner’s request? Explain your answer.

Show worked solution▾

(a)

sin⁡θ=BX7BX=7sin⁡θcos⁡θ=BY12BY=12cos⁡θh=BY−BX=12cos⁡θ−7sin⁡θ\begin{aligned} \sin\theta &= \frac{BX}{7}\\ BX &= 7\sin\theta\\ \cos\theta &= \frac{BY}{12}\\ BY &= 12\cos\theta\\ h = BY - BX &= 12\cos\theta - 7\sin\theta \end{aligned}

(b)

R=122+72=193α=tan⁡−1(712)=0.52807h=193cos⁡(θ+0.528)\begin{aligned} R &= \sqrt{12^2 + 7^2} = \sqrt{193}\\ \alpha &= \tan^{-1}\left(\frac{7}{12}\right) = 0.52807\\ h &= \sqrt{193}\cos(\theta + 0.528) \end{aligned}

(c)

193cos⁡(θ+0.52807)=8cos⁡(θ+0.52807)=8193Basic angle=0.95714θ+0.52807=0.95714θ=0.429 (3sf)\begin{aligned} \sqrt{193}\cos(\theta + 0.52807) &= 8\\ \cos(\theta + 0.52807) &= \frac{8}{\sqrt{193}}\\ \text{Basic angle} &= 0.95714\\ \theta + 0.52807 &= 0.95714\\ \theta &= 0.429 \text{ (3sf)} \end{aligned}

(d)

193cos⁡(θ+0.52807)≥14cos⁡(θ+0.52807)≥14193cos⁡(θ+0.52807)≥1.0077\begin{aligned} \sqrt{193}\cos(\theta + 0.52807) &\geq 14\\ \cos(\theta + 0.52807) &\geq \frac{14}{\sqrt{193}}\\ \cos(\theta + 0.52807) &\geq 1.0077 \end{aligned} Since −1≤cos⁡(θ+0.52807)≤1-1 \leq \cos(\theta + 0.52807) \leq 1, Not possible.

Alternate Solution

Since max h=193=13.9<14h = \sqrt{193} = 13.9 < 14, it is not possible.

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