Pope
Lv 7
Pope asked in Science & MathematicsMathematics · 10 years ago

# Can you keep this proof on the students' level?

This is something that came from an exercise book. Some students got stuck on this advanced problem. I can do the proof, but for sake of the students, I would like to find a simpler and shorter way. They have strong algebra skills, and they are getting good with elementary trigonometry identities, but they have had no double angle formulas and no calculus at all.

This is what they have proved so far:

sinθcosθ = k

(sinθ + cosθ)² = 1 + 2k

(sinθ - cosθ)² = 1 - 2k

Now prove this inequality:

-1/2 ≤ k ≤ 1/2

Relevance
• Hemant
Lv 7
10 years ago

(1+2k) ≥ 0 ∴ 0 ≤ ( 1 + 2k ) ∴ -1 ≤ 2k ∴ (-1/2) ≤ k ........ (1)

(1-2k) ≥ 0 ∴ 1 ≥ 2k ∴ (1/2) ≥ k ∴ k ≤ (1/2) ................... (2)

Writing (1) and (2) together,

(-1/2) ≤ k ≤ (1/2) ............................. Q.E.D.

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• satszn
Lv 6
10 years ago

-1<=sin2A<=1 , -1<2sinAcosA<1 , -1/2<sinAcosA<1/2 , -1/2<k<1/2