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This question already has an answer here:

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.

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marked as duplicate by Nosrati, Lord Shark the Unknown, Adrian Keister, Namaste, José Carlos Santos Sep 1 '18 at 17:28

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ Perhaps you misremembered $\sin^2 t+\cos^2t = 1$? $\endgroup$ – kimchi lover Sep 1 '18 at 2:07
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    $\begingroup$ Alt. hint: $\;|-\sin t + \cos t|^2 = \sin^2 t + \cos^2 t - 2 \sin t \cos t = 1 - \sin 2t\,$. $\endgroup$ – dxiv Sep 1 '18 at 2:37
  • $\begingroup$ @Nosrati Not sure I'd call this question a duplicate, since it is a particular case that allows for simpler answers that don't work in the general case, see my previous comment for example. $\endgroup$ – dxiv Sep 1 '18 at 2:38
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Notice that $\cos t - \sin t$ can be written in the form of $c \sin (t + \alpha)$ where $c =\sqrt{1^2+(-1)^2}=\sqrt{2}$ 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 $\sqrt{2}$.

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

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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 $x= \begin{pmatrix}x_1 \\ x_2 \end{pmatrix}$, $y= \begin{pmatrix}y_1 \\ y_2 \end{pmatrix}$ 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: $$|-\sin t + \cos t|^2 = | -1\cdot\sin t + 1 \cdot \cos t |^2 \leq ((-1)^2 + 1^2)(\sin^2 t + \cos^2 t)= 2$$ Equality is reached, if $-\sin t = \cos t$, which happens, for example, at $t =-\frac{\pi}{4}$. So, $$\boxed{|-\sin t + \cos t| \leq \sqrt{2}}$$

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Note:

$\sin (t -π/4)=$

$ \sin t \cos π/4 -\cos t \sin π/4=$

$(1/2)√2(\sin t -\cos t)$.

Hence

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

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

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