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

2 Answers
May 20, 2015

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.

May 22, 2015

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