Trigonometry: When x+y = (2 pi)/3, x,y >= 0, then max/min of sin(a)+sin(b)=?

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Shiroe05

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Hi, everybody
I have this problem :


(3) When \(\displaystyle x\, +\, y\, =\, \dfrac{2\pi}{3},\, x\, \geq\, 0,\, y\, \geq\, 0,\) the maximum of \(\displaystyle \sin(x)\, +\, \sin(y)\)

is \(\displaystyle \boxed{(1)\qquad\\ \qquad}\) and the minimum of that expression is \(\displaystyle \boxed{(2)\qquad\\ \qquad}\)


I found the maximum(√3) an minimum(√3/2) values using logic and a sketch of the trigonometric circle.
I want to know a way to find the answer using more math instead of the way I did, because maybe it was only luck to find it. Or even better a clue to see if I could find it. Thanks :D
 

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y=120-x,plug in you will get

sinx+sin(120-x)=sinx+sin120cosx-cos120sinx=sinx+(sqrt3)/2cosx+.5sinx=1.5sinx+(sqrt3)/2cosx=(sqrt3)((sqrt3)/2sinx+1/2cosx)=(sqrt3)(cos30sinx+sin30cosx)=(sqrt3)sin(x+30).

Therefore, min is (sqrt3)/2 when x=30. max is sqrt3 when x is 60
 
Generally speaking, maximization/minimization problems are in the domain of calculus. Have you learned any calculus techniques, such as derivatives, yet? If you have learned those techniques, that would be my go-to approach. If you've not learned these techniques, whatever strategy you used is most likely the best bet.
 
ksdhart2 said:
Generally speaking, maximization/minimization problems are in the domain of calculus. Have you learned any calculus techniques, such as derivatives, yet? If you have learned those techniques, that would be my go-to approach. If you've not learned these techniques, whatever strategy you used is most likely the best bet.
Click to expand...
I know a few things of calculus but when I tried to use it in this problem I don't get the right answer maybe i do something wrong , I only can get the maximum value.
 
yma16 said:
sinx+sin(120-x)=sinx+sin120cosx-cos120sinx=sinx+(sqrt3)/2cosx+.5sinx=1.5sinx+(sqrt3)/2cosx=(sqrt3)((sqrt3)/2sinx+1/2cosx)=(sqrt3)(cos30sinx+sin30cosx)=(sqrt3)sin(x+30).

Therefore, min is (sqrt3)/2 when x=30. max is sqrt3 when x is 60
Click to expand...
I can't understand the last step, how do you get those values?
 
i typed wrong.

The min is when x=0. The reason is sinx an increase function between 0 and 90 degree. So, the min occurs at the leftmost x. The max of sinx is sin90. If you choose x=60, then sin(x+30)=sin90. you can choose x=120, which gives you the same min.
 
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yma16 said:
The min is when x=0. The reason is sinx an increase function between 0 and 90 degree. So, the min occurs at the leftmost x. The max of sinx is sin90. If you choose x=60, then sin(x+30)=sin90. you can choose x=120, which gives you the same min.
Click to expand...
Ahhhh ok, thanks :)
 
Shiroe05 said:
Hi, everybody
I have this problem :


(3) When \(\displaystyle x\, +\, y\, =\, \dfrac{2\pi}{3},\, x\, \geq\, 0,\, y\, \geq\, 0,\) the maximum of \(\displaystyle \sin(x)\, +\, \sin(y)\)

is \(\displaystyle \boxed{(1)\qquad\\ \qquad}\) and the minimum of that expression is \(\displaystyle \boxed{(2)\qquad\\ \qquad}\)


I found the maximum(√3) an minimum(√3/2) values using logic and a sketch of the trigonometric circle.
I want to know a way to find the answer using more math instead of the way I did, because maybe it was only luck to find it. Or even better a clue to see if I could find it. Thanks :D
Click to expand...
If x + y = 2pi/3, then y = 2pi/3 - x.

Define f(x) = sin(x) + sin(2pi/3 - x).

Now using calculus solve f '(x) = 0.

If you can continue from there then go for it, if not please post back with what you have so far.
 
yma16 said:
The min is when x=0. The reason is sinx an increase function between 0 and 90 degree. So, the min occurs at the leftmost x. The max of sinx is sin90. If you choose x=60, then sin(x+30)=sin90. you can choose x=120, which gives you the same min.
Click to expand...
I agree that sin x has a min at x = 0 (if 0<= x <= 90) but the problem is that sin (0) + sin (2pi/3) does not have to be the min value. Suppose (in theory) that sin (2pi/3) was the max value. Yes, it still is possible that sin (0) + sin (2pi/3) is the min value (subject to the constraint that x + y = 2pi/3). But it is also possible that if we increase x to say pi/100 and decrease 2pi/3 to (2pi/3 - pi/100), that sin (pi/100) + sin (2pi/3 - pi/100) < sin (0) + sin (2pi/3). The only way to be absolutely sure about this is to define some function to be sin (x) + sin (2pi/3 - x) and find the min point on the given interval using calculus.
 
Jomo said:
If x + y = 2pi/3, then y = 2pi/3 - x.

Define f(x) = sin(x) + sin(2pi/3 - x).

Now using calculus solve f '(x) = 0.

If you can continue from there then go for it, if not please post back with what you have so far.
Click to expand...
hyhy-1.jpg
i did it like this but i only can get one value, maybe i make a mistake or something
 
Shiroe05 said:
View attachment 9660
i did it like this but i only can get one value, maybe i make a mistake or something
Click to expand...
What you did was fine. But you need to recall that the cosine function is periodic. That is, cos(x) = cos (x +2pi) = cos(x + 4pi) = ... = cos(x - 2pi) = cos (x - 4pi) = .... = cos (2pi/3 - x)
This should give you enough x values to keep you busy.
See what you can do with this.
 
Jomo said:
What you did was fine. But you need to recall that the cosine function is periodic. That is, cos(x) = cos (x +2pi) = cos(x + 4pi) = ... = cos(x - 2pi) = cos (x - 4pi) = .... = cos (2pi/3 - x)
This should give you enough x values to keep you busy.
See what you can do with this.
Click to expand...
the values i get, dont fit with the equations x+y=2pi/3 and x ≥ 0 and y ≥ 0
 
Well I tried this to get many answers but the answers doesn't fit with the problem, for instance:
Cos(x)= Cos(2pi+2pi/3-x), x= 8pi/3-x, 2x=8pi/3, x= 4pi/3.

When I put this value in x+y= 2pi/3, y = -2pi/3 but y can't be a negative number because need to be more o equal to zero.

When cos(x)=Cos(-2pi+2pi/3-x), x = -2pi/3.

I tried with +4pi and -4pi but again the values of x doesn't fit.( +4pi, x=7pi/3 and -4pi, x= -5pi/3).

I know that when x = 0 and y = 2pi/3, I get the minimum value of the function, but when I plug in 0 or 2pi/3 in the derivative of sin (x) + sin (y) it doesn't seem like a minimum point
So maybe I get the derivate wrong or something?
The derivative I get is cos(x) -cos(2pi/3-x)
 
I changed it to (sqrt3)sin(x+30)

Jomo said:
I agree that sin x has a min at x = 0 (if 0<= x <= 90) but the problem is that sin (0) + sin (2pi/3) does not have to be the min value. Suppose (in theory) that sin (2pi/3) was the max value. Yes, it still is possible that sin (0) + sin (2pi/3) is the min value (subject to the constraint that x + y = 2pi/3). But it is also possible that if we increase x to say pi/100 and decrease 2pi/3 to (2pi/3 - pi/100), that sin (pi/100) + sin (2pi/3 - pi/100) < sin (0) + sin (2pi/3). The only way to be absolutely sure about this is to define some function to be sin (x) + sin (2pi/3 - x) and find the min point on the given interval using calculus.
Click to expand...

if you work with (sqrt3)sin(x+30), you do not need to use calculus.
 
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