Graphically find #x# for which #sinx=cosx#?

1 Answer
Shwetank Mauria
Sep 18, 2016

Please see below.

Explanation:

As #sin(x+pi/2)=cosx#, #sinx# function moves with a gap of #pi/2# with respect to #cosx#. Between #x=0# and #x=pi/2#, #cosx# falls from #1# to #0#and #sinx# rises from #0# to #1#.

Further #cos(-x)=cosx# and hence while #cosx# is symmetric around #y#-axis i.e. #x=0#, As #sinx# appears with a lag of #pi/2#, it is symmetric around #x=pi/2#.

Hence, #cosx=sinx# appears exactly at the midpoint between #0# and #pi/2# i.e. at #pi/4=0.7854#
graph{(y-sinx)(y-cosx)=0 [-1.365, 3.635, -1, 1.5]}

Related questions
See all questions in Solving Trigonometric Equations
Impact of this question
2196 views around the world
You can reuse this answer
Creative Commons License