Maximizing (sin x+cos x) on [0,2*pi]

[ritwik06]

Homework Statement

I was solving a problem involving sine and cosine ratios.
If I come to know the maximum possible value of (sin x+cos x), x belongs to [0,2*pi]. my problm would be solved.

Homework Equations

The Attempt at a Solution

I think x ill be in 1st quadrant as both sin and cos are +ve. If I am not wrong x=45?
but how do i mathematically prove
this?

 

[HallsofIvy]

Homework Statement

I was solving a problem involving sine and cosine ratios.
If I come to know the maximum possible value of (sin x+cos x), x belongs to [0,2*pi]. my problm would be solved.

Homework Equations

The Attempt at a Solution

I think x ill be in 1st quadrant as both sin and cos are +ve. If I am not wrong x=45?
but how do i mathematically prove
this?

One of the serious problems with showing no work at all is that we don't know what techniques you are familiar with. Do you know how to find the derivative of sin(x)+ cos(x)? Do you know what the derivative has to do with finding maximum values?

 

[ritwik06]

One of the serious problems with showing no work at all is that we don't know what techniques you are familiar with. Do you know how to find the derivative of sin(x)+ cos(x)? Do you know what the derivative has to do with finding maximum values?

No I have no idea. Sorry. But I had already written what I could think of

 

[pizzasky]

You may want to use the "R-formula", which states that $$a sin (x) \ + \ b cos(x) \ = \ R sin(x + \alpha)$$, where $$R \ = \ \sqrt{a^2+b^2}$$ and $$\alpha = tan^{-1} \frac{b}{a}$$.

 

[BrendanH]

Since you haven't learned about derivatives, I'll nudge you in another direction that could help.
It uses trigonometric identities.

Hint: sin x + cos x = sin x + sin (90-x)
&
sin x + sin y = 2 sin ((x+y)/2)cos((x-y)/2)

This reduces the problem to finding the max of just one function instead of the max of a sum of functions.

 

[BrendanH]

Alright, since the question was asked long ago, I shall go ahead and just give the answer, for my benefit as it passes the time (I've got to do SOMETHING at work, after all!)

$$\sin x + \cos x = \sin x + \sin (\pi/2 - x)$$
$$\sin x + \sin (\pi/2 - x) = 2 \sin ((x + \pi/2 - x)/2)\cos((2x-\pi/2)/2)$$
$$=2\cos(x-\pi/4)$$
This must be maximized, but is easy to do. We know that $$\cos x$$ has a maximum of 1.
$$\cos (x-\pi/4) = 1$$
$$x-\pi/4 = \arccos 1 *$$
$$x= \pi/4$$

*The only oversight here is not include that \arccos 1 = 2n\pi