Vtyjkyuk

I am a little bit confused what $$|-sin t + cos t|$$ is.

I heard that is $1$ but I thought $sin t + cos t$ was $1$.

Is it just that the progression of $t$ is reversed and the size stays the same?

What if the absolute sign is removed? Does it still stay $1$?

----edit----
Well, they are vectors and the question is asking about the length of the vector. So the length is 1 no matter what the signs are.

Notice that $cos t - sin t$ can be written in the form of $c sin (t + alpha)$ where $c =sqrt1^2+(-1)^2=sqrt2$ by using R-formula. The magnitude is not a constant as shown from a graph. For the first graph, it takes value from $0$ to $sqrt2$.

Similarly for $sin t + cos t$, it is not a constant.

If you know the following special case of the Cauchy-Schwarz inequality, you get immediately the maximum value of your function:

• For two real vectors $x= beginpmatrixx_1 \ x_2 endpmatrix$, $y= beginpmatrixy_1 \ y_2 endpmatrix$ it holds:
  $$| x cdot y |^2 = | x_1 y_1 + x_2 y_2 |^2 leq (x_1^2+x_2^2)(y_1^2+y_2^2)$$


• In addition, you have equality if and only if one of the vectors is a multiple of the other one.

All together:
$$|-sin t + cos t|^2 = | -1cdotsin t + 1 cdot cos t |^2 leq ((-1)^2 + 1^2)(sin^2 t + cos^2 t)= 2$$
Equality is reached, if $-sin t = cos t$, which happens, for example, at $t =-fracpi4$. So,
$$boxed$$

Note:

$sin (t -π/4)=$

$ sin t cos π/4 -cos t sin π/4=$

$(1/2)√2(sin t -cos t)$.

Hence

$|sin t -cos t|= √2|sin(t-π/4)| le √2cdot 1 =√2$.

Used: $cos π/4 = sin π/4= (1/2)√2$,
and
$sin (a+b) = sin a cos b+ sin b cos a$.