For what value(s) of x, 0<x<pi/2, is sinx<cosx

If you have a graphing calculator and any hesitation at all about this, the best way to go is to graph. Make sure your calculator is in radians mode!

Between x=0 and x=\frac{\pi}{2}, the red y=\sin x curve is below the blue y=\cos x curve until x=\frac{\pi}{4}. After that point, the lines cross.

Without a calculator, you should still be able to reason this one out if you remember a few basics:

    \sin 0 = 0
    \cos 0 = 1
    \sin\frac{\pi}{2}=1 (or in degrees, \sin 90^\circ =1)
    \cos\frac{\pi}{2}=0 (or in degrees, \cos 90^\circ =0)
    \sin\frac{\pi}{4}=\cos\frac{\pi}{4} (or in degrees, \sin 45^\circ =\cos 45^\circ)

From the above, you know that the cosine curve is higher than the sine curve at x=0, and that the curves intersect at x=\frac{\pi}{4}. They don’t intersect again before x=\frac{\pi}{2}.

Leave a Reply