For what values of #x# is #sin x = cos x# ?

1 Answer
Jul 12, 2015

Answer:

From geometric definition of #sin# and #cos# and properties of right angled triangles, find:

#x = pi/4 + n pi# any #n in ZZ#

Explanation:

In Q1 this is basically the #sqrt(2)/2#, #sqrt(2)/2#, #1# right angled isosceles triangle with angles #pi/4#, #pi/4# and #pi/2#. It needs to be isosceles in order that #sin(x) = cos(x)#

So #x = pi/4# is a solution.

Also #sin(pi/4 + pi) = cos(pi/4 + pi) = -sqrt(2)/2# in Q3

There are no solutions in Q2 or Q4 since #sin# and #cos# have opposite signs in those quadrants.

So #x = pi/4 + n pi# captures all the solutions.