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TMUA 2021 · Paper 2 · Question 19 of 20

TMUA 2021 Paper 2 Question 19

Trigonometry — Surds and signs · counting solutions over a full turn. Try it first; the answer and a full worked solution are below.

TMUA 2021 · Paper 2TrigonometrySurds and signs · counting solutions over a full turn8 optionshard
The angle θ can take any of the values 1∘, 2∘, 3∘, …, 359∘, 360∘.

For how many of these values of θ is it true that sinθ√1+sinθ √1−sinθ + cosθ√1+cosθ √1−cosθ = 0 ?

  1. A0
  2. B1
  3. C2
  4. D4
  5. E93
  6. F182
  7. G271
  8. H360
Show the answer and worked solution
answer · F
  1. A0
  2. B1
  3. C2
  4. D4
  5. E93
  6. F182
  7. G271
  8. H360
Each pair of surds combines, since both radicands are non-negative: √1+sinθ√1−sinθ = √1−sin2θ = |cosθ|, and likewise the second pair gives |sinθ|. The equation becomes sinθ|cosθ| + cosθ|sinθ| = 0 — the modulus signs are the whole question. In the first quadrant the left side is 2sinθcosθ > 0, and in the third it is −2sinθcosθ < 0; in the second and fourth quadrants the two terms cancel exactly, and on the axes both terms are zero. That gives 89 + 89 values strictly inside the second and fourth quadrants, plus θ = 90∘,  180∘,  270∘,  360∘: 182 in all.