All topic tests topics

P2: TrigonometryEdexcel International A Level Maths: Topic test

20 questions, 54 marks

Edexcel International A Level Maths

P2: Trigonometry topic test

Total 54 marks

Name

Class

Date

  1. 1
    tan⁡θ=−512\tan\theta=-\dfrac{5}{12} and 90∘<θ<180∘90^\circ<\theta<180^\circ.
    (a)
    Find the value of sin⁡θ\sin\theta.
    [1 mark]
    • A513\frac{5}{13}
    • B−513-\frac5{13}
    • C512\frac5{12}
    • D1213\frac{12}{13}
    (b)
    Find the value of cos⁡θ\cos\theta.
    [1 mark]
    • A1213\frac{12}{13}
    • B−1213-\frac{12}{13}
    • C−513-\frac{5}{13}
    • D−1312-\frac{13}{12}
    (c)
    Find the exact value of 1cos⁡2θ\dfrac{1}{\cos^2\theta}.
    [2 marks]

    Total for question 1: 4 marks

  2. 2
    f(θ)=sin⁡θtan⁡θ+sin⁡θtan⁡θ\mathrm{f}(\theta)=\dfrac{\sin\theta}{\tan\theta}+\sin\theta\tan\theta, for values of θ\theta for which f(θ)\mathrm{f}(\theta) is defined.
    (a)
    Which expression is equal to sin⁡θtan⁡θ\dfrac{\sin\theta}{\tan\theta}?
    [1 mark]
    • Atan⁡θ\tan\theta
    • Bsin⁡2θcos⁡θ\dfrac{\sin^2\theta}{\cos\theta}
    • Ccos⁡θ\cos\theta
    • Dsin⁡θcos⁡θ\sin\theta\cos\theta
    (b)
    Which expression is equal to f(θ)\mathrm{f}(\theta)?
    [1 mark]
    • A11
    • B1sin⁡θ\dfrac{1}{\sin\theta}
    • C2cos⁡θ\dfrac{2}{\cos\theta}
    • D1cos⁡θ\dfrac{1}{\cos\theta}
    (c)
    Solve f(θ)=2\mathrm{f}(\theta)=2 for 0∘≤θ<360∘0^\circ\le\theta<360^\circ.
    [2 marks]

    Total for question 2: 4 marks

  3. 3
    f(x)=sin⁡(2x+30∘)\mathrm{f}(x)=\sin\left(2x+30^\circ\right), for 0∘≤x<360∘0^\circ\le x<360^\circ.
    (a)
    Solve f(x)=0.5\mathrm{f}(x)=0.5.
    [3 marks]
    (b)
    Solve f(x)=−0.6\mathrm{f}(x)=-0.6, giving your answers to 11 decimal place.
    [4 marks]

    Total for question 3: 7 marks

  4. 4
    2cos⁡2x−sin⁡x=12\cos^2x-\sin x=1, for 0∘≤x<360∘0^\circ\le x<360^\circ.
    (a)
    Show that the equation can be written as 2sin⁡2x+sin⁡x−1=02\sin^2x+\sin x-1=0, and hence solve it for 0∘≤x<360∘0^\circ\le x<360^\circ.
    [6 marks]
    (b)
    Hence solve 2cos⁡2(x−20∘)−sin⁡(x−20∘)=12\cos^2(x-20^\circ)-\sin(x-20^\circ)=1 for 0∘≤x<360∘0^\circ\le x<360^\circ.
    [6 marks]

    Total for question 4: 12 marks

  5. 5
    g(θ)=cos⁡2θ1−sin⁡θ\mathrm{g}(\theta)=\dfrac{\cos^2\theta}{1-\sin\theta}, for sin⁡θ≠1\sin\theta\ne1.
    (a)
    Which expression is equal to g(θ)\mathrm{g}(\theta)?
    [1 mark]
    • A1−sin⁡θ1-\sin\theta
    • B1+sin⁡θ1+\sin\theta
    • Ccos⁡θ\cos\theta
    • D11−sin⁡θ\dfrac{1}{1-\sin\theta}
    (b)
    For 0∘≤θ≤90∘0^\circ\le\theta\le90^\circ, which value of θ\theta satisfies g(θ)=1.5\mathrm{g}(\theta)=1.5?
    [1 mark]
    • A30∘30^\circ
    • B60∘60^\circ
    • C48.6∘48.6^\circ
    • D41.8∘41.8^\circ
    (c)
    Solve g(θ)=0\mathrm{g}(\theta)=0 for 0∘≤θ<360∘0^\circ\le\theta<360^\circ.
    [2 marks]

    Total for question 5: 4 marks

  6. 6
    h(x)=cos⁡(x+π6)\mathrm{h}(x)=\cos\left(x+\dfrac{\pi}{6}\right), for −π≤x≤π-\pi\le x\le\pi, where xx is in radians.
    (a)
    Between which values must x+π6x+\frac\pi6 lie?
    [1 mark]
    • A−π≤x+π6≤π-\pi\le x+\frac\pi6\le\pi
    • B−7π6≤x+π6≤5π6-\frac{7\pi}{6}\le x+\frac\pi6\le\frac{5\pi}{6}
    • C−π6≤x+π6≤7π6-\frac{\pi}{6}\le x+\frac\pi6\le\frac{7\pi}{6}
    • D−5π6≤x+π6≤7π6-\frac{5\pi}{6}\le x+\frac\pi6\le\frac{7\pi}{6}
    (b)
    Which gives all the solutions of h(x)=12\mathrm{h}(x)=\frac12?
    [1 mark]
    • Ax=π6x=\frac\pi6 only
    • Bx=±π3x=\pm\frac\pi3
    • Cx=−π2x=-\frac\pi2 and x=π6x=\frac\pi6
    • Dx=−π6x=-\frac\pi6 and x=π2x=\frac\pi2
    (c)
    Solve h(x)=−32\mathrm{h}(x)=-\dfrac{\sqrt3}{2}.
    [2 marks]

    Total for question 6: 4 marks

  7. 7
    cos⁡θ1+sin⁡θ+1+sin⁡θcos⁡θ≡2cos⁡θ\dfrac{\cos\theta}{1+\sin\theta}+\dfrac{1+\sin\theta}{\cos\theta}\equiv\dfrac{2}{\cos\theta}.
    (a)
    Prove the identity.
    [3 marks]
    (b)
    Hence solve cos⁡θ1+sin⁡θ+1+sin⁡θcos⁡θ=5\dfrac{\cos\theta}{1+\sin\theta}+\dfrac{1+\sin\theta}{\cos\theta}=5 for 0∘≤θ<360∘0^\circ\le\theta<360^\circ, giving your answers to 11 decimal place.
    [4 marks]

    Total for question 7: 7 marks

  8. 8
    A passenger PP on a Ferris wheel has height h=12−10cos⁡(30t)∘h=12-10\cos(30t)^\circ metres after tt minutes, for 0≤t≤120\le t\le12. A second passenger QQ has height k=12−7sin⁡(30t)∘k=12-7\sin(30t)^\circ metres.
    (a)
    Find the times at which PP is at a height of 1717 metres.
    [6 marks]
    (b)
    Show that PP and QQ are at the same height when tan⁡(30t)∘=107\tan(30t)^\circ=\frac{10}{7}, and find the times in the first 1212 minutes when this happens, giving your answers to 22 decimal places.
    [6 marks]

    Total for question 8: 12 marks

End of questions

Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).