# For which values of $x$, $0$°$≤x<360$°, is $\sin x > \cos x$? [duplicate]

This question already has an answer here:

For which values of $x$ such that $0$°$≤x<360$°, we have: $\sin x > \cos x$?

The answer is "When $45$° $< x < 225°$, the $y$-coordinate of the point on the unit circle is greater than the $x$-coordinate", but I don't understand why this is the answer and how the book got it.

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• Draw a diagonal line $x=y$. The open half-plane towards the upper-left (the half-plane that contains $(-1, 1)$) is the set of $(x,y)$ pair where $y>x$. Then the idea in your book is to intersect with (the set of points on) a unit circle, where $x = \cos \theta$ and $y = \sin \theta$. – peterwhy Aug 2 '17 at 17:58

## 1 Answer

It's $\sin(x-45^{\circ})>0$, which gives $0^{\circ}<x-45^{\circ}<180^{\circ}$