TMUA 2018 Paper 2 Q13 — Worked Solution

P2·Trigonometry, Logic & Quantifiers·6.5Official · 2018 · Q13

proof-errorsquare-root-signcounterexample-q3

The following is an attempted proof of the conjecture:

if tan θ > 0, then sin θ + cos θ > 1.

Suppose

tan θ > 0

, so in particular

cos θ \neq = 0

(I)

Since

tan θ = \frac{sin θ}{cos θ}

, then

sin θ cos θ = tan θ cos^{2} θ > 0

(II)

It follows that

1 + 2 sin θ cos θ > 1

(III)

Therefore

sin^{2} θ + 2 sin θ cos θ + cos^{2} θ > 1

(IV)

which factorises to give

(sin θ + cos θ)^{2} > 1

(V)

Therefore

sin θ + cos θ > 1

Which one of the following is the case?

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