For each statement, explain why the statement cannot be true

[bumblebee123]

question:



can anyone explain this to me? any help would be really appreciated!

 

[MarkFL]

Let's look at the first statement. Consider:

[MATH]-1\le\sin(x)\le1[/MATH]
[MATH]-1\le\cos(x)\le1[/MATH]
Even if the sine and cosine functions reached their maxima for the same value of \(x\) (which they don't), what would be the maximum value of their sum?

 

[pka]

question: Why FALSE??
II The equation \(\displaystyle \tan(x)+\cos(x)=2\) has a solution in \(\displaystyle \left[ { - \frac{\pi }{2},0} \right]\)
can anyone explain this to me? any help would be really appreciated!

If \(\displaystyle \alpha\in IV\) then \(\displaystyle 0\le\cos(\alpha)\le 1\\\tan(\alpha)\le 0\)

 

[bumblebee123]

Let's look at the first statement. Consider:

[MATH]-1\le\sin(x)\le1[/MATH]
[MATH]-1\le\cos(x)\le1[/MATH]
Even if the sine and cosine functions reached their maxima for the same value of \(x\) (which they don't), what would be the maximum value of their sum?

2?

 

[MarkFL]

2?

Yes, and so it would be impossible for the sum to be 3. Incidentally, it can be shown that the maximum value of the sum is \(\sqrt{2}\), but we needed only to show the maximum was below 3.

 

[bumblebee123]

for part 2

the Y values for tanx are infinite, and the maximum Y value for cosx is 1

so the sum of the maximum values would be way bigger than 2

 

[MarkFL]

For the 3rd statement, consider:

[MATH]-\infty<\tan(x)<0[/MATH]
[MATH]0<\sin(x)<1[/MATH]
And so, when you subtract the latter, from the former, what do you get?

 

[bumblebee123]

For the 3rd statement, consider:

[MATH]-\infty<\tan(x)<0[/MATH]
[MATH]0<\sin(x)<1[/MATH]
And so, when you subtract the latter, from the former, what do you get?

0 - 1 = -1

 

[MarkFL]

After thinking about this more completely, in order to maximize the expression, we want the maximum of the first term and the minimum of the second, so that we find:

[MATH]-\infty<\tan(x)-\sin(x)<0[/MATH]

 

[Steven G]

for part 2

the Y values for tanx are infinite, and the maximum Y value for cosx is 1

so the sum of the maximum values would be way bigger than 2

No No, the tanx values are NOT infinite. The tanx values go from -infinity to positive infinity and obtains every value in between. That is there is an x value such that tan(x) = 1.3, there is an x value such that tan(x) = -125.6, there is an x value such that tan(x) = 12,365,497,098,000,345,000,000.00125, ....

So no, tanx + cosx is not always much bigger that 2. In fact there are x values such that tanx + cosx < 10,000,000!

 

[bumblebee123]

Okay that makes more sense. How can I use this to prove tanx + cosx = 2 is wrong

 

[Deleted member 4993]

Okay that makes more sense. How can I use this to prove tanx + cosx = 2 is wrong

Domain of investigation is -90^o → 0^o

In that domain

tan(x) < 0

and

0 < cos(x) < 1

tan(x) + cos(x) < 1