Evaluate the maxium of a trig function

[AdrianZ]

Homework Statement

here is the function:
y = 8sin(x) + cos(2x). the question is to evaluate the point where the function is maximum.

Homework Equations

any trig identities.

The Attempt at a Solution

well, I tried to substitute cos(2x) with $$1 - 2sin^2(x)$$. then we will have the function $$y = -2sin^2(x) + 8sin(x) + 1$$. by setting sin(x) = u we will have a nice quadratic function in the form $$y = -2u^2 + 8u + 1$$. we know that the point where a quadratic function in the form $$ax^2+bx+c$$ is maximum is given by $${-b \over 2a}$$. so with respect to that formula, we get $${-b \over 2a} = - {8\over 2(-2)} = 2$$. now because we had set sin(x) = u the x we're looking for is given by solving the trig equation sin(x)=2 and we know that that equation has no solutions. so logically the answer to our question is there's no such point.

now let's solve it this way, we have the function f(x) = 8sin(x) + cos(2x), if we take the first derivative, we will have: f'(x) = 8cos(x) - 2sin(2x). x=pie/2 is a solution to this equation, that means the point x=pie/2 is an extremum and because the a coefficient in the previous equation was negative it might be a local or global maximum.

well, as you see the two solutions contradict each other. any idea?

 

[hunt_mat]

Why don't you just differentiate the trig function?
$$f(x)=-2sin^2(x) + 8sin(x) + 1$$
Differentiate to obtain:
$$\frac{df}{dx}=-4\sin x\cos x+8\cos x =\cos x(-4\sin x+8)=0$$
From there is shouldn't be too hard to work out the maximum/minimum.

Mat

 

[AdrianZ]

Why don't you just differentiate the trig function?
$$f(x)=-2sin^2(x) + 8sin(x) + 1$$
Differentiate to obtain:
$$\frac{df}{dx}=-4\sin x\cos x+8\cos x =\cos x(-4\sin x+8)=0$$
From there is shouldn't be too hard to work out the maximum/minimum.

Mat

please read the thread again. I want to know why the two solutions contradict each other. that's the aim of the thread.

 

[hunt_mat]

Because the first solution is wrong basically. The quadratic you have is effectively only defined from $$-1\leqslant u\leqslant 1$$

Mat

 

[AdrianZ]

Because the first solution is wrong basically. The quadratic you have is effectively only defined from $$-1\leqslant u\leqslant 1$$

Mat

well, that's obvious that $$-1\leqslant sin(x)\leqslant 1$$ but I don't see anything wrong with the quadratic equation itself. explain please. I'm a little bit lost why it's giving me a false answer.

 

[hunt_mat]

Your looking outside the range of applicability. Cinsidering the quadratic by itself is not enough, you must examine how u changes with respect to x also. The quadratic equation does not do this.

 

[AdrianZ]

Your looking outside the range of applicability. Cinsidering the quadratic by itself is not enough, you must examine how u changes with respect to x also. The quadratic equation does not do this.

yes, I understand that. but I don't know WHY this happens. would you give me an example of the same case in another maths problem? and tell me how to determine how u changes with respect to x?

thank you

 

[hunt_mat]

To determine how u changes with x, you simply differentiate. I can't think of an example off hand, you just have to be very careful about how you simplify your problems, that's all.

 

[AdrianZ]

To determine how u changes with x, you simply differentiate. I can't think of an example off hand, you just have to be very careful about how you simplify your problems, that's all.

well, would you determine it for this particular problem?

 

[hunt_mat]

I would (as you did) write the equation in terms of sin(x) and just differentiate. You will get your previous solution and the one which you need.