21Hint: 1= cos(x+y-(x+y)) = .....≤....≤....
Prove that for all real numbers x,y, |cos(x)|+ |cos(y)| + |cos(x+y)|\geq 1
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5 Answers
341WLOG |\cos x| \ge |\cos y| so that it follows that |\sin x| \le |\sin y|
Now,
|\cos x| + |\cos y| + |\cos(x+y)| = |\cos x| + |\cos y| + |\cos x \cos y - \sin x \sin y|
\ge |\cos x| + |\cos y| - |\cos x| \cos y| + |\sin x| |\sin y|
= |\cos y|(1-|\cos x|) + |\cos x| + |\sin x| |\sin y|
\ge |\cos y|(1-|\cos x|) + \cos^2 x + \sin^2 x \ge 1
21oh sorry its perfectly correct ...i was flipping inequalities in mind
|a|+|b|≥|a-b|≥|a|-|b| is correct i'm fine
deleted my post to keep things clean
21Somehow ordering did not come in my mind.
I did this . Do tell me if it is wrong.
1= |cos^2(x+y) + sin^2(x+y)| \leq |cos(x+y)| + |sin(x+y)| \leq |cos(x+y)|+ |sin x cos y+ sin y cos x |\leq |cos(x+y)|+ |sin x||cos y| + |cos x||sin y| \leq |cos(x+y)| + |cos x| + |cos y|