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2 years ago

NEED HELP ASAP !! WILL MARK BRAINLEAST SHOW WORK TOO !!!

Mathematics

1 answer:

irinina [24]2 years ago

3 0

Step-by-step explanation:

$(1 + \cos(x) )(1 - \cos(x) )$

$1 - \cos {}^{2} (x)$

$= \sin {}^{2} (x)$

$\frac{1}{ \cot {}^{2} (x) } - \frac{1}{ \cos {}^{2} (x) }$

$\frac{1}{ \frac{ \cos {}^{2} (x) }{ \sin { }^{2} (x) } } - \frac{1}{ \cos {}^{2} (x) }$

$\frac{1}{ \cot {}^{2} (x) } - \frac{ \csc {}^{2} (x) {} }{ \cot {}^{2} (x) }$

$\frac{ - \cot {}^{2} (x) }{ \cot {}^{2} (x) }$

$- 1$

$\sec {}^{2} ( \frac{\pi}{2} - x ) )( \sin {}^{2} (x) - \sin {}^{4} (x))$

$( \csc {}^{2} (x) )( \sin {}^{2} (x) - \sin {}^{4} (x )$

$\csc {}^{2} (x) ( \frac{1}{ \csc {}^{2} (x) } - \frac{1}{ \csc {}^{4} (x) } )$

$1 - \frac{1}{ \csc {}^{2} (x) }$

$1 - \sin {}^{2} (x)$

$\cos {}^{2} (x)$

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Suppose x is a real number and epsilon > 0. Prove that (x - epsilon, x epsilon) is a neighborhood of each of its members; in

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See proof below

Step-by-step explanation:

We will use properties of inequalities during the proof.

Let $y\in (x-\epsilon,x+\epsilon)$ . then we have that $x-\epsilon$ . Hence, it makes sense to define the positive number delta as $\delta=\min\{x+\epsilon-y,y-(x-\epsilon)\}$ (the inequality guarantees that these numbers are positive).

Intuitively, delta is the shortest distance from y to the endpoints of the interval. Now, we claim that $(y-\delta,y+\delta)\subseteq (x-\epsilon,x+\epsilon)$ , and if we prove this, we are done. To prove it, let $z\in (y-\delta,y+\delta)$ , then $y-\delta$ . First, $\delta \leq y-(x-\epsilon)$ then $-\delta \geq -y+x-\epsilon$ hence $z>y-\delta \geq x-\epsilon$

On the other hand, $\delta \leq x+\epsilon-y$ then $z$ hence $z$ . Combining the inequalities, we have that $x-\epsilon$ , therefore $(y-\delta,y+\delta)\subseteq (x-\epsilon,x+\epsilon)$ as required.