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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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marked as duplicate by lab bhattacharjee algebra-precalculus Aug 2 '17 at 17:51

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ 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 $y = \sin \theta$. $\endgroup$ – peterwhy Aug 2 '17 at 17:58
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It's $\sin(x-45^{\circ})>0$, which gives $0^{\circ}<x-45^{\circ}<180^{\circ}$

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