Plus-Minus Symbol In This Trig. Equation

AI Thread Summary
The discussion centers on a calculus problem involving a trigonometric equation where the placement of the plus-minus symbol in the solution is debated. One participant argues that the plus-minus should be inside the arccos function to yield two potential answers, while acknowledging that the solution correctly places it outside, which simplifies the process. The reasoning provided indicates that the secant function is always positive in the given domain, allowing the negative option to be disregarded. Additionally, the even nature of the secant and cosine functions is highlighted, which justifies the solution's approach. Ultimately, the conversation reflects on the nuances of applying the Mean Value Theorem and the implications of function properties in calculus.
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Hey guys,

The problem is #49 and it is a simple calculus problem, but the part that I am confused on is how the solution solves the trig. equation. In the solving, the solution brings out the plus-minus symbol and puts it outside the arccos, but I feel as if it should be inside the arccos.

I understand that putting the symbol outside solves the problem perfectly, but it seems like a shortcut. How I think it should be done is to put the plus-minus symbol inside the arcccos (it would give us two different answers) and choose the answer that fits into the domain of the problem which is (pi/4 and -pi/4). When I choose the answer that fits into the domain, I can make it positive or negative because the secant is squared.

Thanks!
 

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I suppose what isn't written in the solution is that the secant is always positive in the given domain. Hence, the solution skips checking the negative option. The plus-minus sign on the following line, however, is there for a different reason: the secant (and in turn cosine) is an even function i.e. f(x) = f(-x).
 
Admittedly, I haven't used the Mean Value Theorem in a while, but isn't the tangent at the point c supposed to equal the slope of the secant formed by the points at the ends of the interval? The slope of the secant on the function is 0. So there is only one place on the interval with a point that has a tangent line with a slope of 0.

Note the result of the solution in the jpg on the graph of the original equation.
 
thelema418 said:
Admittedly, I haven't used the Mean Value Theorem in a while, but isn't the tangent at the point c supposed to equal the slope of the secant formed by the points at the ends of the interval? The slope of the secant on the function is 0. So there is only one place on the interval with a point that has a tangent line with a slope of 0.

Note the result of the solution in the jpg on the graph of the original equation.
Note that the problem in question refers to the MVT for integrals i.e. the MVT you're suggesting is applied to the antiderivative.
 
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@da_nang, thanks! That's definitely what I missed when reading it!
 
da_nang said:
I suppose what isn't written in the solution is that the secant is always positive in the given domain. Hence, the solution skips checking the negative option. The plus-minus sign on the following line, however, is there for a different reason: the secant (and in turn cosine) is an even function i.e. f(x) = f(-x).

Just to reclarify, was I correct in my thinking and that the solution took a shortcut (the plus minus should have went inside)? A plus minus then should have been added AFTER solving and throwing out the solution outside the domain because the secant function is even.

Thanks!
 
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