Hi Guys,

I came across the following problem and have no idea how to solve it. Any ideas?

Let a and b be reals. Find the least possible value of

(2cos(a) + 5sin(b)-8)^2 + (2sin(a)+5cos(b)-15)^2

- May 25th 2013, 11:13 AMShadowKnight8702Finding the least possible value of a trigonometric expression
Hi Guys,

I came across the following problem and have no idea how to solve it. Any ideas?

Let a and b be reals. Find the least possible value of

(2cos(a) + 5sin(b)-8)^2 + (2sin(a)+5cos(b)-15)^2 - May 25th 2013, 10:35 PMibduttRe: Finding the least possible value of a trigonometric expression
What i think is when we open the brackets and simplify we get

318 + 2 sin ( a+b) -80 sin b -32 cos a -150 cosb -60 sin a

Only one expression is being added i.e., sin (a+b) so to make it zero we get a + b = 0 that gives a = -b

now the expression becomes 318 - 20 sinb - 182 cosb

Now if we suppose b to be 0 we get the value as 318 - 182 that i think should be the minimum value - May 27th 2013, 09:56 AMjohngRe: Finding the least possible value of a trigonometric expression
Hi,

Unfortunately, the previous answer is wrong. I'm almost 100% confident the minimum is 100. I considered the function:

Since both cos and sin are periodic with period 2pi, I can restrict the function f to values of x and y both in [0,2pi]. I used my software to investigate the function. On the boundary of this square, one can use one variable calculus to get the minimum is about 154.3. (Actually, I used software to get this.) I then tried to see where both partial derivatives are 0.

Here's the picture I got:

Attachment 28472

As the diagram indicates, there are x and y with f(x,y) "=" 100 (any calculator will verify this). Since the absolute minimum of f is not on the boundary, it must occur at a critical point. The critical points found are "nice" and the values at these critical points are all squares of integers!

So there undoubtedly is some geometric figure that gives rise to the function -- I thought maybe some cyclic quadrilateral, but couldn't follow through on the thought. I'm convinced the minimum is 100, but this is not an acceptable proof. - May 27th 2013, 10:13 AMShadowKnight8702Re: Finding the least possible value of a trigonometric expression
You are right that the answer is 100, but the problem is that there is very clearly a way to solve the problem without software, as it appeared in a mathematics competition called HMNT,where no calculators are allowed. The solution they gave with vectors seemed too weird. Your solution makes sense though.

- May 27th 2013, 07:35 PMjohngRe: Finding the least possible value of a trigonometric expression
Hi again,

I surely hope someone can shed some light on this problem. I changed the function of my previous post to:

Here, c1 and c2 are positive constants. Unless I'm missing something easy, I can't analytically find the critical points of the function. Experimentally, the critical points are independent of c1 and c2! (The actual partials of course change.) The only thing that changes is the value of the minimum. Example: for c1=10 and c2=13, the minimum is 36. Again, it looks like the minimum for integral c1 and c2 is the square of an integer!

Please, somebody, help.