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1.5 Trigonometry

1.5 Trigonometry

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Question 128

A research team is modeling the vertical displacement, HHH, of a specialized underwater sensor. The displacement is given by the function

H(θ)=sin⁡2θsec⁡θ+cos⁡2θcsc⁡θ+sin⁡θ H(\theta) = \sin 2\theta \sec \theta + \cos 2\theta \csc \theta + \sin \theta H(θ)=sin2θsecθ+cos2θcscθ+sinθ

where θ\thetaθ is the tilt angle of the sensor.

a.

Show that the expression for H(θ)H(\theta)H(θ) can be written as

H(θ)=sin⁡θ+csc⁡θ H(\theta) = \sin \theta + \csc \theta H(θ)=sinθ+cscθ

where sin⁡θ≠0\sin \theta \neq 0sinθ=0 and cos⁡θ≠0\cos \theta \neq 0cosθ=0.

[4]
bi.

A technician attempts to find the tilt angles where the displacement is exactly 4.254.254.25 units by solving the equation

sin⁡2θsec⁡θ+cos⁡2θcsc⁡θ+sin⁡θ=4.25 \sin 2\theta \sec \theta + \cos 2\theta \csc \theta + \sin \theta = 4.25 sin2θsecθ+cos2θcscθ+sinθ=4.25

for 0∘≤θ≤360∘0^{\circ} \leq \theta \leq 360^{\circ}0∘≤θ≤360∘. They produce the following solution:

Step 1: sin⁡θ+csc⁡θ=4.25\sin \theta + \csc \theta = 4.25sinθ+cscθ=4.25

Step 2: sin⁡θ+1sin⁡θ=174\sin \theta + \frac{1}{\sin \theta} = \frac{17}{4}sinθ+sinθ1​=417​

Step 3: 4sin⁡2θ−17sin⁡θ+4=04\sin^{2} \theta - 17\sin \theta + 4 = 04sin2θ−17sinθ+4=0

Step 4: sin⁡θ=4\sin \theta = 4sinθ=4 or sin⁡θ=0.25\sin \theta = 0.25sinθ=0.25

Step 5: θ=14.5∘,165.5∘\theta = 14.5^{\circ}, 165.5^{\circ}θ=14.5∘,165.5∘

Explain why the technician should reject the value sin⁡θ=4\sin \theta = 4sinθ=4 in Step 4.

[1]
bii.

Determine if there are any other reasons, based on the original expression's domain, why solutions might need to be rejected, and state the final correct solutions for the technician's equation in the range 0∘≤θ≤360∘0^{\circ} \leq \theta \leq 360^{\circ}0∘≤θ≤360∘.

[3]
Markscheme

1.5 Trigonometry Questions

  1. A Level
  2. /Maths
  3. /1.5 Trigonometry

317 exam-style questions on OCR A Level Maths 1.5 Trigonometry, covering 1.5.1 Definitions for all arguments, 1.5.2 Sine and cosine rules, 1.5.3 Area of a triangle, 1.5.4 Radian measure (A-level only), 1.5.5 Small angle approximations (A-level only), 1.5.6 Graphs of basic trigonometric functions, 1.5.7 Exact values in radians (A-level only), 1.5.8 Reciprocal and inverse trigonometric ratios (A-level only), 1.5.9 Graphs of reciprocal and inverse functions (A-level only), 1.5.10 Trigonometric identities, 1.5.11 Further trigonometric identities (A-level only), 1.5.12 Double angle and compound angle formulae (A-level only), 1.5.13 Geometrical proofs of formulae (A-level only), 1.5.14 Harmonic form Rcos / Rsin (A-level only), 1.5.15 Trigonometric equations, 1.5.16 Proof involving trigonometric functions (A-level only), and 1.5.17 Trigonometric functions in context. Each one has a worked solution and a mark scheme showing where the marks go.

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