Proving 1+cos⁡x2+sin⁡x&lt;43

If cos⁡(θ)=35 and sin⁡(θ)=45 (which is possible since 32+42=52), the inequality (with ≤ rather than &lt;) is equivalent to cos⁡(x−θ)≤1

We want to prove 1+cos⁡x2+sin⁡x&lt;43. By multiplying by 3(2+sin⁡x) (note that 2+sin⁡x&gt;0), we get that this is equivalent to 3(1+cos⁡x)≤4(2+sin⁡x) which is equivalent to prove that 3cos⁡x−4sin⁡x≤5 This inequality comes from Cauchy Schwartz. Note that (3cos⁡x−4sin⁡x)2≤(32+(−4)2)(cos2x+sin2x)=25

If cos⁡(θ)=35 and sin⁡(θ)=45 (which is possible since 32+42=52), the inequality (with ≤ rather than &lt;) is equivalent to cos⁡(x−θ)≤1

We want to prove 1+cos⁡x2+sin⁡x&lt;43. By multiplying by 3(2+sin⁡x) (note that 2+sin⁡x&gt;0), we get that this is equivalent to 3(1+cos⁡x)≤4(2+sin⁡x) which is equivalent to prove that 3cos⁡x−4sin⁡x≤5 This inequality comes from Cauchy Schwartz. Note that (3cos⁡x−4sin⁡x)2≤(32+(−4)2)(cos2x+sin2x)=25

"Proving 1+cos⁡x2+sin⁡x<43" solved and explained

High School
Calculus and Analysis
Precalculus
Trigonometry

Ahmed Stewart

Answered question

2022-01-25

Proving

\frac{1 + \cos x}{2 + \sin x} < \frac{4}{3}

Answer & Explanation

Johnny Cummings

Beginner2022-01-26Added 7 answers

\cos (θ) = \frac{3}{5}

and

\sin (θ) = \frac{4}{5}

(which is possible since

3^{2} + 4^{2} = 5^{2}

), the inequality (with

\leq

rather than <) is equivalent to

\cos (x - θ) \leq 1

utgyrnr0

Beginner2022-01-27Added 11 answers

We want to prove

\frac{1 + \cos x}{2 + \sin x} < \frac{4}{3}

. By multiplying by

3 (2 + \sin x)

(note that

2 + \sin x > 0

), we get that this is equivalent to

3 (1 + \cos x) \leq 4 (2 + \sin x)

which is equivalent to prove that

3 \cos x - 4 \sin x \leq 5

This inequality comes from Cauchy Schwartz. Note that

{(3 \cos x - 4 \sin x)}^{2} \leq (3^{2} + {(- 4)}^{2}) (\cos^{2} x + \sin^{2} x) = 25

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