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I'm trying to help a friend of mine solve this problem. It's been like 10 years since I took calculus please help:

Find min/max values of $y = \sin x + \cos x$ on $[0, 2\pi)$

Thanks.

I've got this much:

  • f'(x) = \cos x - \sin x.
  • $0 = \cos x - \sin x$.
  • $\sin x = \cos x$.
  • $\tan x = 1$.

Thanks

Evaulate at end points and at the critical point:

  • $f(0) = 1$
  • $f(2\pi) = 1$
  • $\sin x = \cos x$ at $\pi/4$

Hence, $f(x)$ has a minimum at $\pi/4$ and $f(x)$ has a maximum at $x = 0$ on $[0,2\pi)$ (Extreme Value Theorem)

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    $\sin x=\cos x$ at another point in your interval. Also at $\pi/4$ the value of $\sin x + \cos x$ exceeds 1, so the value at 0 is not the maximum.2011-01-16

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No, you don't get $\tan x = 0$; you would get, at best, $\tan x = 1$ (remember that $\tan x = \frac{\sin x}{\cos x}$. If $\sin x$ and $\cos x$ are equal, then the quotient is equal to $1$, not to $0$).

So, you remembered that you want to find the critical points (points where the derivative is zero). That's good. What you also need to remember is that if you have a continuous function on a closed interval, then the maximum and the minimum will each be achieved at either a critical point or an endpoint.

Here, you might as well work over $[0,2\pi]$ (the value at $2\pi$ is the same as the value at $0$). So the maximum and the minimum of $f(x)$ will be achieved either at $x=0$, or at a point where f'(x)=0, that is, a point where $\sin(x) = \cos(x)$ in the interval (better to work with these, since this equality does not depend on $\cos(x)\neq 0$, whereas $\tan(x)=1$ does; of course, it does not really matter here because if $\cos(x)=0$, then $\sin(x)\neq \cos(x)$).

So, the question is: for what points $x$, $0\leq x\leq 2\pi$, do you have $\sin(x)=\cos(x)$? There are two such points; once you have them, simply evaluate the original function at these points, and at $x=0$ (where it has the same value as at $x=2\pi$, which is why we were able to add $2\pi$ to the interval for simplicity). The largest value you get is the maximum, the smallest value you get is the minimum.

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    I suppose I should try doing this in the morning instead at midnight hahaha2011-01-16
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An other thing you might try is to simplify the original function using trigonometric formulas.

Recall the addition formula for the sine function: $\sin(A+B)=\sin(A)\cdot\cos(B)+\sin(B)\cdot\cos(A).$ Using this we see $f(A)\cdot C=\sin(A)\cdot C+\cos(A)\cdot C=\sin(A+B)$, provided $\cos B =\sin B=C$. Next, we have $1=\sin^2 B+\cos^2 B=2C^2$ hence we get $f(A)=\sqrt{2}\sin(A+B) \quad \text{ or }\quad f(A)=-\sqrt{2}\sin(A+B)$ which shows that the maximal possible value of $f$ is $\sqrt{2}$ and the minimal possible value is $-\sqrt{2}$. (Looking at the "circle-definition" of sine and cosine it is easy to see where these values are attained -- on the line $y=x$).

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    @subt13: Ok, thanks - it is a standard method that works for expressions of the form $A\cdot\sin x+B\cdot\cos x$.2011-01-16
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Let $f(x)=\sin(x)+\cos(x)$

Thus

$f(x)^2 = (\sin(x)+\cos(x))^2 = \sin(x)^2+2\sin(x)\cos(x)+\cos^2(x) = 1+\sin(2x)\Rightarrow$

$0\le f(x)^2 \le 2 \Rightarrow -\sqrt{2}\le f(x)\le\sqrt{2}$

How $f(\frac{\pi}{4})=\sqrt{2}$ and $f(\frac{5\pi}{4})=-\sqrt{2}$, then the maximum value is $\sqrt{2}$ and the minimum is $-\sqrt{2}$

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    This is a nice alternative too.2018-02-23