Melody

@@ End of Day Wrap :Thurs 17/7/14   Sydney, Australia   Time 3:40 am (Really Friday morning)   ♬        

Hi everyone,

Our great answers were today posted by Alan, DragonSlayer554, NinjaDevo, Kitty<3, Rosala, Stu, Zegroes and awesomeness.

I haven't got through everything today but most is done.  There are more for other people to answer if they are keen.

Here are some interesting posts forthe day.

We'll start with a riddle for the "rat pack" I am told that this is what CPhill calls the children of the forum!  Thanks awesomenes  

When things fall, they get faster and faster - they accelerate with the effect of gravity.

gravity on Earth (at sea level I think)  is 9.8m/sec²

I think that is 32 feet/sec²

A very good answer DragonSlayer,  thumbs up from me.

$$x^2*x^2=x^4$$

It is not infinity it is just a huge number.

It is already written in the best form.

Scientific notation is a perfectly acceptable presentation for both huge and tiny numbers.  

$$\\cot^{-1}(-\sqrt3)=tan^{-1}\left(\frac{-1}{\sqrt3}\right)\\\\
\mbox{ The first quadrant answer is } \pi/6\quad \mbox{ But the first quadrant in not correct.}\\\\$$

this is the range of the function.

so

answer is     $$-\pi/6$$

They are both fantastic answers.  Thank you very much.  

Actually I have  a 3rd answer.

(5a+5)(5a+5)    This is a perfect square so there is a nice short cut  

 $$\boxed{(a+b)^2=a^2+2ab+b^2}$$

so

$$(5a+5)^2\\\\
=(5a)^2+2*25a+5^2\\\\
=25a^2+50a+25$$

$$cosec^{-1}(5)= sin^{-1}(1/5)$$

$$\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}^{\!\!\mathtt{-1}}{\left({\frac{{\mathtt{1}}}{{\mathtt{5}}}}\right)} = {\mathtt{11.536\: \!959\: \!032\: \!815^{\circ}}}$$

On the web2 calculator on the home page you have to enter asin instead of sin^-1

Inverse sine and arc sine are the same thing.  

The range of cosec^-1x is              that is in radians of course 

so    

11.53 degrees is the only answer.

This is enough help DragonSlayer,  you can think about some of the written answers yourself.

A line Graph would have been just as good as my column graph.  Maybe better.   

Thanks Kitty,

Kitty has done it the usual traditional way (the way that Iwould normally do it) - I just thought I would take a look at a more basic method.

$$\\cot\;\theta=\frac{1}{tan\theta}=\frac{adj}{opp}\qquad \mbox{pos in 1st and 3rd quads}\\\\
sec\;\theta=\frac{1}{cos\theta}=\frac{hyp}{adj}\qquad \mbox{pos in 1st and 4th quads}\\\\
cosec\;\theta=\frac{1}{sin\theta}=\frac{hyp}{opp}\qquad \mbox{pos in 1st and 2nd quads}\\\\
\cot\theta\times sec\;\theta=\frac{adj}{opp}\times\frac{hyp}{adj}=\frac{hyp}{opp}=cosec\;\theta$$

Now, this obviously works in the first quadrant where everything is positive but what about in the other quadrants?

2nd quad - x - = +  true

3rd quad  + x - = -  true

4th quad - x + = -  true