Melody

This question made me think 

$$\\cos\;\frac{\pi}{2}=0,\qquad cos\;\frac{-\pi}{2}=0,\qquad and \qquad f(0)=0\\\\$$

Now since g(x)=f(x)*cosx it follows that g(x)=0 when x  = -pi/2, 0 or pi/2

Hence g(x) must be at least one turning point between  -pi/2 and 0  and at least one between 0 and pi/2

If you are not sure about this, plot the 3 points and think about what the graph will have to look like.

NOW  when g(x) turns the gradient of the tangent must be 0 SO g'(x)=0 

HENCE

1 and 3 must both be correct. 

Thanks TOK002 and anon. 

I like it when people provide questions and answers in physics that are basic enough and clear enough for me to learn from.  

This is a good question and has made me think.  

If the line was at 90 degrees to the tangent of the earth (that means vertical) then the curvature of the Earth would make no difference.

If the line were not perpendicular I still do not think it would make any difference.  A line is not a solid, it has no substance and hence it not affected by gravity. So in theory at least the line could just go in any direction and not be effected by the the planet on which it originates.

Others may have a different view.  I will include this question in my next wrap.     

1 radian =57.3 DEGREES 

then 3 radians =  57.3*3  degrees

$${\mathtt{180}}{\mathtt{\,-\,}}{\mathtt{42}}{\mathtt{\,-\,}}{\mathtt{80.4}} = {\frac{{\mathtt{288}}}{{\mathtt{5}}}} = {\mathtt{57.6}}$$

You are right it does :))

This is because the calc is giving you the option of a fraction answer (in this case an improper fraction)  and a decimal answer.   That is all :))

It is not really a part of the working - just an alternate answer.   

Heureka, could you (or someone else) explain the phi function to me please?

I thought phi was an irrational number like pi.     

Yes well Nauseated, you are correct about one thing, troll droppings are the only private offerings that us moderator 'Gods' are ever given. 

Doesn't really pay to be a 'God' does it.  

I always draw a quick pic for these.

let 

$$\\\theta=sec^{-1}u\qquad so \\
u=sec\;\theta\\
sec\;\theta = \frac{1}{cos\theta}=\frac{hyp}{adj}=\frac{u}{1}$$

Now I can read the positive answer straight off the triangle. 

$$cot(sec^{-1}(u))=cot\;\theta = \frac{adj}{opp}=\frac{1}{\sqrt{u^2-1}}$$

NOW u is positive so sec(theta) and cos(theta) are positive. So theta is in the 1st or 4th quadrant.

So cot(theta) can be positive or negative so

$$cot(sec^{-1}(u))=\pm\;\frac{1}{\sqrt{u^2-1}}$$

Well Chris, you have given lots of choices there  LOL    :))

$$\\24=4*6=4\div \frac{1}{6}\\\\
and\\\\
24=3*8=(4-1)*8\\$$