^{2}.

^{2}= π

^{2}/4 > 1 = sin(x) + cos(x).

^{2}. It is continuous between 0 and π/2.

^{2}.

Noelle W.

asked • 10/10/14True or False and explain: there is at least one point x between 0 and π/2 where cos(x)+sin(x)=x^2.

Should I try plugging in the two numbers given into the equation or do I need to simplify the equation first? I would greatly appreciate help solving this! Thanks so much!!

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True.

At x = 0, cos(x) + sin(x) = 1 > 0 = x^{2}.

At x = π/2, x^{2} = π^{2}/4 > 1 = sin(x) + cos(x).

Form a the function f(x) = sin(x) + cos(x) - x^{2}. It is continuous between 0 and π/2.

Note that f(0) > 0 and f(π/4) < 0. By the intermediate value theorem, there is some x between 0 and π/4 that will give you f(x) = 0. This means that for some x, sin(x) + cos(x) = x^{2}.

Pierce O. answered • 10/10/14

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Hi Noelle,

You are looking at the equality cos(x)+sin(x)=x^{2}. Let's simplify this a bit and look at cos(x)+sin(x)-x^{2}=0. Take the left hand side of the equation to be f:

f(x)=cos(x)+sin(x)-x^{2}

We are concerned about what is happening to this function in the interval [0,Π/2]. Well, we know

f(0)=cos(0)+sin(0)-0^{2}

=1

and

f(Π/2)=cos(Π/2)+sin(Π/2)-(Π/2)^{2}

=1-Π^{2}/4

= (4-Π^{2})/4

Notice that f(0) is positive, and f(Π/2) is negavive.

The intermediate value theorem states that if f is a continuous function over the interval [a,b], and if u is a number such that f(a)>u>f(b), then there exists some c∈[a,b] such that f(c)=u.

Taking a step back and putting this to use, we are using f(x)=cos(x)+sin(x)-x^{2} over the interval [0,Π/2], and we are saying that 0 is a number such that

f(0)=1>0>(4-π^{2})/4=f(Π/2)

or just

f(0)>0>f(Π/2)

Hence, by the intermediate value theorem, we know that there exists some real number c∈[0,Π/2] such that f(c)=0.

But if f(c)=0, then

f(c)=cos(c)+sin(c)-c^{2}=0

or

cos(c)+sin(c)=c^{2}

So the statement is true, there exists at least one point c between 0 and Π/2 such that cos(c)+sin(c)=c^{2}, by the intermediate value theorem.

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