I am working on a problem, and one of the keys to the solution for it is to prove the following inequality:
$$|\cos(x+y)|\geq |\cos(x)|-|\sin(y)|$$ for $x,y\in \mathbb{R}.$
I have verified that this is true numerically, and I tried various things to show this, but none seem to work.
First, I tried to expand the LHS, $$|\cos(x+y)|=|\cos(x)\cos(y)-\sin(x)\sin(y)|\geq |\cos(x)\cos(y)|-|\sin(x)\sin(y)|,$$ but here I run into a problem, since $$|\cos(x)\cos(y)|-|\sin(x)\sin(y)|\not\geq |\cos(x)|-|\sin(y)|.$$
Next, I rearranged the terms in the initial inequality, so $$|\sin(y)|\geq |\cos(x+y)|-|\cos(x)|.$$ Then, I worked with the RHS,
$$|\cos(x+y)|-|\cos(x)|\leq |\cos(x+y)-\cos(x)|$$ with the intention of using sum to product formulas to prove this. However, again, we run into the problem that $$|\cos(x+y)-\cos(x)|\not\leq |\sin(y)|$$ everywhere.
I think it is possible to use the same ideas and prove by casework, but if possible, I would like something simpler than that.
A friend suggested that the inequality looked similar to the triangle inequality with some Law of Sines involved in this way:
$$\cos(x+y)=\sin\left(\frac{\pi}{2}-x-y\right)$$
$$\cos(x)=\sin\left(\frac{\pi}{2}+x\right)$$ and the angles add up to $\pi.$ Perhaps this provides some motivation for the solution, but I'm not able to see it yet.
Any help appreciated.
Best Answer
$\cos x=\cos ((x+y)-y)=\cos (x+y) \cos y +\sin (x+y) \sin y$ so $|\cos x|\leq |\cos (x+y)| +|\sin y|$. Take $|\sin y|$ to the left side.