How do you solve # cos(theta) - sin(theta) = 1#?

Whenever #cos(theta) = 1#,

we get #sin(theta) = +-sqrt(1-cos^2theta) = 0#

and #cos(theta)-sin(theta) = 1 - 0 = 1#.

#cos(theta) = 1# for #theta = 2npi# for all #n in ZZ#.

Whenever #sin(theta) = -1#,

we get #cos(theta) = +-sqrt(1-sin^2theta) = 0#

and #cos(theta)-sin(theta) = 0 - (-1) = 1#.

#sin(theta) = -1# for #theta = -pi/2+2npi# for all #n in ZZ#.

Putting two cases together, we have solutions when:

#theta = 2npi# for all #n in ZZ#

and when

#theta = -pi/2 + 2npi# for all #n in ZZ#

To make sure that these are the only solutions:

Starting with #cos(theta)-sin(theta)=1#, first add #sin(theta)# to both sides:

#cos(theta)=sin(theta)+1#

Then square both sides:

#cos^2(theta)=sin^2(theta)+2sin(theta)+1#

Then use #cos^2(theta)=1-sin^2(theta)# to get:

#1-sin^2(theta)=sin^2(theta)+2sin(theta)+1#

Add #sin^2(theta)-1# to both sides to get:

#0=2sin^2(theta)+2sin(theta)=2sin(theta)(sin(theta)+1)#

So either #sin(theta) = 0# or #sin(theta) = -1#

We have already accounted for #sin(theta) = -1# in our solutions.

What about #sin(theta) = 0#?

If this is so, then

#cos^2(theta) = 1 - sin^2(theta) = 1 - 0 = 1#

So #cos(theta) = +-sqrt(1) = +-1#

Only the case #cos(theta) = 1# satisfies #cos(theta)-sin(theta) = 1# and we have accounted for that case already too.

So we have found all the solutions.

There is another method that has been popular to solve this type of trig equation.

cos x - sin x = 1 .

Call a the arc whose #tan a = 1 = tan (pi/4)#

#cos x - sin a/cos a(sinx) = 1#

#cos x.cos a - sina.sin x = cos pi/4#

#cos (x + pi/4) = cos (pi/4)#

a. #(x + pi/4) = pi/4 -> x = 0 #

b.# (x + pi/4) = - pi/4 -> x = -pi/4 - pi/4 = -pi/2#

Check:
a. #x = 0 -> cos 0 - sin 0 = 1 - 0 = 1# . OK
b.# x = -pi/2 -> cos (-pi/2) - sin (-pi/2) = 0 - (-1) = 1 # OK