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TMUA 2021 · Paper 2 · Question 5 of 20

TMUA 2021 Paper 2 Question 5

Proof and counterexample — Finding the first error · square roots of squares. Try it first; the answer and a full worked solution are below.

TMUA 2021 · Paper 2Proof and counterexampleFinding the first error · square roots of squares5 options
On which line is the first error in the following argument?
  1. Asin2 x + cos2 x = 1 for all values of x.
  2. BTherefore cos x = √1 − sin2 x for all values of x.
  3. CHence 1 + cos x = 1 + √1 − sin2 x for all values of x.
  4. DThus (1 + cos x)2 = (1 + √1 − sin2 x)2 for all values of x.
  5. ESubstituting x = π gives 0 = 4.
Show the answer and worked solution
answer · B
  1. Asin2 x + cos2 x = 1 for all values of x.
  2. BTherefore cos x = √1 − sin2 x for all values of x.
  3. CHence 1 + cos x = 1 + √1 − sin2 x for all values of x.
  4. DThus (1 + cos x)2 = (1 + √1 − sin2 x)2 for all values of x.
  5. ESubstituting x = π gives 0 = 4.
Here the options are the lines of the argument itself. Line A is the standard identity. Line B rearranges it to cos2 x = 1 − sin2 x and then takes a square root — but the square root symbol always returns the non-negative value, so this only holds when cos x ≥ 0. At x = π, cos x = −1 while √1 − sin2 x = 1. Everything after B is a valid consequence of B, so the first error is on line B, and the absurdity at line E is what exposes it.