Question: > 
> For Cosine:
> 
> 
> 

For a triangle with angles$\angle ABC$ there exist lengths $a,b,c$ opposite to the angles.
By using the trigonometric function for cosines, I multiplied the length of $AC\times\cos\angle A$ for example:

> 
> 5+5+4=14 where $\cos\angle B=0.4,\cos\angle A=0.4,\cos\angle C=0.68$ and there are $3$ $\theta$,
> 
> 
> $AC\times\cos\angle A=2$ or $BC\times\cos\angle B=2$ which is the place of the altitude and $2+2=4$,
> 
> 
> $AB\times\cos\angle B=1.6$
> $BC\times\cos\angle C=3.4$
> $1.6+3.4=5$
> Next step is all I have to use is the Pythagorean theorem.
> Can anyone explain if this is true or false?
> 
> 
> 

for Sine:

> 
> $AC\times\sin\angle B=$
> 
> 
> $BC\times\sin\angle A=$
> 
> 
> 

And to find Sine I subtract 1 from all three $\cos ^2 A,\cos^2 B,\cos^2 C$

And in general the sum of all angles is:

> 
> $\sin^2 A+\sin^2 B+\sin^2 C+\cos^2 A+\cos^2 B+\cos^2 C=3$
> 
> 
>

Answer: 
A $\sigma$-algebra is a Boolean algebra. A Boolean algebra contains the complements of its members. $ B(R)$ contains all open sets so it contains their complements, which is all closed sets. Like $[a,b].$ It will also contain the intersection of any closed set with any open set, like $(a,b]=(-\infty,b]\cap (a,\infty).$

$B(R)$ can be generated by just the bounded open intervals because any open set in $R$ is the union of countably many bounded open intervals.
There is a more general stronger result that is no harder to prove. Suppose $\tau$ is a topology on a space $X.$ Then by definition, $\mathscr B(X)=\sigma(\tau).$ 

Claim: if $X$ is a second countable topological space, then $\mathscr B(X)$ is generated by any countable base for $X.$ 

To see this, let $\mathscr E$ be a countable base for the topology $\tau.$ Since $\mathscr E \subseteq \tau,\ \sigma (\mathscr E) \subseteq \sigma (\tau)=\mathscr B(X).$ For the reverse inclusion, we simply note that if $U\in \tau$ then $U$ is a countable union of elements of $\mathscr E$ so $U\in \sigma (\mathscr E),$ which means that $\sigma (\tau)\subseteq \sigma (\mathscr E).$

So in the case $X=\mathbb R$ we have that $\mathscr B(\mathbb R)$ is generated by the collection of open intervals with rational endpoints. Furthermore, since the closed and half-closed intervals are countable unions of open intervals, $\mathscr B(\mathbb R)$ is also generated by the collection of closed and half-closed intervals.