Question

When x+y = 120º , x>= 0º and y>=0º , the maximum of sin x + sin y is [??] and the minimum of that is [???] (Just to make sure that I did the right way, please)

Answer 1

So, I considerate the following values : 120,90,60,30,0
sin 120 = sqrt(3)/2
sin 90 = 1
sin 60 = sqrt(3)/2
sin 30 = 1/2
sin 0 = 0
And just combined the values to find the maximum (which is sin 60 + sin 60) and the minimum (which is sin 120 + sin 0)

Is this enough to guarantee both minimum and maximum?

Answer 2

You're considering only a few combinations

Answer 3

familiar with the identity
\[a\cos x+b\sin x = \sqrt{a^2+b^2}\sin(\star)\]

?

Answer 4

No... :(

Answer 5

How about calculus ?

Answer 6

I know the basics... Which part?

Answer 7

Familiar with finding the max/min of a function by setting the first derivative equal to 0 ?

Answer 8

Yes

Answer 9

You want to find the min/max of sin(x) + sin(120-x) over the interval [0, 120]

Answer 10

Let

f(x) = sin(x) + sin(120-x)

f'(x) = ?

Answer 11

f'(x) = cos(x) + cos(120-x)

Answer 12

Careful, derivative of sin(120-x) is not just cos(120-x)

Answer 13

Remember the chain rule ?

Answer 14

You must multiply by the derivative of `120-x`

Answer 15

What's the derivative of `120-x` with respect to `x` ?

Answer 16

it's -1 ?

Answer 17

Yes


f(x) = sin(x) + sin(120-x)

f'(x) = ?

Answer 18

f'(x) = cos(x) - cos(120-x)

Answer 19

Yep, set that equal to 0 and solve x over the given interval

Answer 20

cos(x) - cos(120-x) = 0

cos(x) = cos(120-x)

Answer 21

Recall that cos(x) = cos(-x)

Answer 22

cos(x) - cos(120-x) = 0

cos(x) = cos(120-x)

This is same as
cos(-x) = cos(120-x)


-x = 120-x
x = ?

Answer 23

x = 120

Answer 24

really ?

Answer 25

try again

Answer 26

x=60

Answer 27

Yes, that means x = 60 is a critical value.

Since [0, 120] is the range, other critical values are x=0, x=120.

Calculus guarantees us that the max/min values of f(x) can occur only at these critical values.

Answer 28

So it is sufficient to check the function at these 3 critical values :

x = 0, 60, 120

Answer 29

And then I just need to do sin(x) + sin(120-x) with these three values, right?

Answer 30

Yes

Answer 31

I didn't think about y being 120-x , that's interesting

Answer 32

Thank you!! :)

Answer 33

Np :) But I prefer solving this problem using the identity
\[a\cos x+b\sin x = \sqrt{a^2+b^2}\sin(\star)\]

This is my favorite identity from trig. You will see it a lot if you take differential equations...

Answer 34

Look it up when free
http://www.mash.dept.shef.ac.uk/Resources/web-rcostheta-alphaetc.pdf

Answer 35

I will, thanks