Pedja

New member
Hi. Do you know what is the value of [imath]\cos (\pi)[/imath] and [imath]\cos(2\pi)[/imath]?

cos pi = -1
cos 2pi = 1

darongkun

New member
But cos (pi^2) not equal ( cos pi )^2

Otis

Elite Member
But cos (pi^2) not equal ( cos pi )^2
That's true, and your exercise uses the latter. (It's called 'function notation'.)

$$\displaystyle \frac{1}{\cos(m\pi)^2} = \frac{1}{\cos(m\pi) \cdot \cos(m\pi)}$$

Evaluate factors [imath]\frac{1}{\cos(m\pi)^2}[/imath] for m=1, m=2, m=3 and m=4.

$$\displaystyle \frac{1}{\cos(\pi)^2}$$

$$\displaystyle \frac{1}{\cos(\pi)^2} \times \frac{1}{\cos(2\pi)^2}$$

$$\displaystyle \frac{1}{\cos(\pi)^2} \times \frac{1}{\cos(2\pi)^2} \times \frac{1}{\cos(3\pi)^2}$$

$$\displaystyle \frac{1}{\cos(\pi)^2} \times \frac{1}{\cos(2\pi)^2} \times \frac{1}{\cos(3\pi)^2} \times \frac{1}{\cos(4\pi)^2}$$

How does the aggregate product change, as m increases? Using that pattern, extrapolate to m=1000.

darongkun

New member
This is my understanding.

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darongkun

New member
Thank you so much.

Dr.Peterson

Elite Member
This is my understanding.
Is that just your uninformed opinion which you are setting against those of more experienced people, or were you told this?

The way I read it, [imath]\cos(\pi x)[/imath] is function notation, and must be evaluated before applying the exponent. So [imath]\cos(\pi x)^2[/imath] means [imath][\cos(\pi x)]^2[/imath]. It would have been better if they had written the latter explicitly, but I think anyone with experience will take it this way.

In addition, if you interpret it this way, the problem becomes easily solvable! Otherwise it is not. Sometimes what you need to do is to try one interpretation (especially if someone with knowledge tells you to) and see what happens, then try another to see how it compares. Have you at least tried doing what you've been told? If not, why did you bother to ask here?