# Homework Help: Evaluate the maxium of a trig function

1. Sep 16, 2010

1. The problem statement, all variables and given/known data

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

2. Relevant equations

any trig identities.

3. 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?

2. Sep 16, 2010

### 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

3. Sep 16, 2010

4. Sep 16, 2010

### hunt_mat

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

Mat

5. Sep 16, 2010

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.

Last edited: Sep 16, 2010
6. Sep 16, 2010

### 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.

7. Sep 16, 2010

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

8. Sep 16, 2010

### 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.

9. Sep 16, 2010