Question

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

Answer 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`.

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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`.

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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`

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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.

Answer 2

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