The following is an attempted proof of the conjecture:
if tanθ > 0, then sinθ + cosθ > 1.
Suppose tanθ > 0, so in particular cosθ ≠ 0.
(I) Since tanθ = sinθcosθ, then sinθcosθ = tanθcos2θ > 0.
(II) It follows that 1 + 2sinθcosθ > 1.
(III) Therefore sin2θ + 2sinθcosθ + cos2θ > 1.
(IV) which factorises to give (sinθ + cosθ)2 > 1.
(V) Therefore sinθ + cosθ > 1.
Which one of the following is the case?
- AThe proof is correct.
- BThe proof is incorrect, and the first error occurs in line (I).
- CThe proof is incorrect, and the first error occurs in line (II).
- DThe proof is incorrect, and the first error occurs in line (III).
- EThe proof is incorrect, and the first error occurs in line (IV).
- FThe proof is incorrect, and the first error occurs in line (V).
Show the answer and worked solution
answer · F
- AThe proof is correct.
- BThe proof is incorrect, and the first error occurs in line (I).
- CThe proof is incorrect, and the first error occurs in line (II).
- DThe proof is incorrect, and the first error occurs in line (III).
- EThe proof is incorrect, and the first error occurs in line (IV).
- FThe proof is incorrect, and the first error occurs in line (V).
Lines (I) to (IV) are all sound: cos2θ > 0 so the product sinθcosθ is positive, doubling and adding 1 is legitimate, and sin2θ + cos2θ = 1 turns that into the perfect square. Line (V) is where it breaks: u2 > 1 gives u > 1 or u < −1, not u > 1 alone. The negative branch really occurs — in the third quadrant tanθ > 0 while sinθ and cosθ are both negative, so at θ = 5π4 the sum is −√2, and the conjecture itself is false.