Question

Have to review trig. Can someone help me with:
cos(x)-sin(x)=1

Answer 1

cos(x)-sin(x)=1


(cos(x)-sin(x))^2 = 1^2


cos^2(x) - 2cos(x)*sin(x) + sin^2(x) = 1


(cos^2(x)+sin^2(x)) - 2cos(x)*sin(x) = 1


(1) - 2cos(x)*sin(x) = 1


1 - 2cos(x)*sin(x) = 1


I'll let you finish

Answer 2

Find all solutions.
I think I got it:
cosx-sinx=1
(cosx-sinx)^2=(1)^2
cos^2x-sinxcosx-sincosx+sin^2x=1
cos^2x-2sinxcosx+sin^2x=1
(sin2x=2sinxcosx;double angle?)
cos^2x-sin2x+sin^2x=1
(cos^2x => 1-sin^2x)
1-sin^2x-sin2x+sin^2x=1
-sin2x+1=1
sin2x=0
x=0,270

Answer 3

Haha. I wish I would have looked at your reply before typing that out. -_- Thanks though!

Answer 4

you are correct and yw, so it would be

x = 0 + 360n where n is any integer

or

x = 270 + 360n where n is any integer

Answer 5

My favorite: \(\cos(x) - \sin(x) = \sqrt{2}\cos\left(x + \dfrac{\pi}{4}\right)\)

This, of course, leads to the same solution.