CALCULUS PROOF WITH --> show ABS(SIN X + COS X) <= SQRT OF 2 (FOR ALL X)?

[NourShaikh]

CAN SOMEONE HELP ME SHOW THAT:

ABS(SIN X + COS X) <= SQRT OF 2 (FOR ALL X)?

It appears to be an absolute max question so i did a first derivation. I know the range is [-1,1] and i know that f'(x) = cosx + sinx.
When I set f'(x)=0, I get cos x=- sin x or 1=-tan x....and then I get stumped...please assist.

 

[Deleted member 4993]

CAN SOMEONE HELP ME SHOW THAT:

ABS(SIN X + COS X) <= SQRT OF 2 (FOR ALL X)?

It appears to be an absolute max question so i did a first derivation. I know the range is [-1,1] and i know that f'(x) = cosx + sinx.
When I set f'(x)=0, I get cos x=- sin x or 1=-tan x....and then I get stumped...please assist.

sin(x) + cos(x) = √2 * sin (x + π/4)

-√2 ≤ sin(x) + cos(x) ≤ √2

|sin(x) + cos(x)| ≤ √2

 

[daon2]

You can do this the way you started as well but you'll need to break it into cases since f'(x) doesn't exist when sin(x)+cos(x)=0.

Case 1) sin(x)+cos(x) > 0. Then f'(x) = cos(x)-sin(x), which will have an extrema when tan(x)=1 (not -1). i.e. x=pi/4 + k*pi. Then show f''(x)<0 at these points.

Case 2) sin(x)+cos(x)=0

Case 3) sin(x)+cos(x) < 0

 

[NourShaikh]

I understand...what if it's abs(sin x - cos x), and the derivative of that is cos x + sin x? (I clearly wrote an addition sign instead of subtraction, very sorry)

When I set that = to zero, I get -1=tan x? Would that be 3pi/4 than?

Thanks for your help and sorry for the typo.

 

[daon2]

I understand...what if it's abs(sin x - cos x), and the derivative of that is cos x + sin x? (I clearly wrote an addition sign instead of subtraction, very sorry)

When I set that = to zero, I get -1=tan x? Would that be 3pi/4 than?

Thanks for your help and sorry for the typo.

3pi/4+k*pi for all integers k.