(adsbygoogle = window.adsbygoogle || []).push({}); 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 [tex]1 - 2sin^2(x)[/tex]. then we will have the function [tex]y = -2sin^2(x) + 8sin(x) + 1[/tex]. by setting sin(x) = u we will have a nice quadratic function in the form [tex]y = -2u^2 + 8u + 1[/tex]. we know that the point where a quadratic function in the form [tex]ax^2+bx+c [/tex] is maximum is given by [tex]{-b \over 2a}[/tex]. so with respect to that formula, we get [tex]{-b \over 2a} = - {8\over 2(-2)} = 2[/tex]. 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?

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# Homework Help: Evaluate the maxium of a trig function

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