The diagram shows the graphs of y = sin 2x and y = cos 2x for −π2 < x < π2, each a full wave of amplitude 1 with the cosine curve peaking at x = 0 and the sine curve peaking at x = π4.
Which one of the following is not true?
- Acos 2x < sin 2x < tan x for some real number x with −π2 < x < π2
- Bcos 2x < tan x < sin 2x for some real number x with −π2 < x < π2
- Csin 2x < cos 2x < tan x for some real number x with −π2 < x < π2
- Dsin 2x < tan x < cos 2x for some real number x with −π2 < x < π2
- Etan x < sin 2x < cos 2x for some real number x with −π2 < x < π2
- Ftan x < cos 2x < sin 2x for some real number x with −π2 < x < π2
Show the answer and worked solution
answer · C
- Acos 2x < sin 2x < tan x for some real number x with −π2 < x < π2
- Bcos 2x < tan x < sin 2x for some real number x with −π2 < x < π2
- Csin 2x < cos 2x < tan x for some real number x with −π2 < x < π2
- Dsin 2x < tan x < cos 2x for some real number x with −π2 < x < π2
- Etan x < sin 2x < cos 2x for some real number x with −π2 < x < π2
- Ftan x < cos 2x < sin 2x for some real number x with −π2 < x < π2
The six options are the six possible orderings, so exactly one is impossible. Write everything in terms of t = tan x, which sweeps all of ℝ on this interval: sin 2x = 2t1+t2 and cos 2x = 1−t21+t2. Option C asks for cos 2x < tan x and sin 2x < cos 2x at once. The first needs t3 + t2 + t − 1 > 0, so t > 0.54 roughly; the second needs t2 + 2t − 1 < 0, so t < √2 − 1 ≈ 0.41. Those cannot both hold, and every other ordering does occur (try t = −3, −2, −12, 12, 0.8, 2).