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

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.

## marked as duplicate by lab bhattacharjee algebra-precalculus StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Aug 2 '17 at 17:51

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