the list of recirpical function are as follows:
(cotangent)
(cosecant)
(secant)
and then you have the inverse functions where if we have
and y is given so (or depending on notation).
and for the recipricals
(or arccsc depending on notation)
(or arcsec depending on notation)
(or arccot depending on notation)
Hello jgv115Your working is correct, and that's a very good question!
The answer is that, strictly speaking, you never do need to use ; you can always use instead. And, indeed, 9 times of 10 you do so. However, there are situations where it's neater to have this reciprocal function available as well, to save writing all the while. I expect you'll meet these situations in due course, so I won't try to give you an example.
I'd would just add one thing, though. Has it occurred to you that you don't really need either? That's simply because . We have all we need in these two trig functions, so we don't need a third. However, I think you'd agree that it's extremely handy to have tangent when we need it - it's much easier than dividing sine by cosine all the while!
Grandad
Thanks Grandpa, I really appreciate your help.
I have drawn this diagram up:
By using pythagoras we can get
But the book I have here says you can also get these:
and
i'm so confused by this. I have no idea how to get these. Could anyone provide assistance?
Thanks in advanced
Hello jgv115Your book is quite correct (and, incidentally, it means that, theoretically, you hardly need cosine either, you can just about get by with sine, and !)
The other two equations are very easy to derive, if you remember that and therefore that . So start with:and divide both sides, first of all, by :Now go back and divide both sides by :
Grandad
Hello jgv115Yes, it does. I said I wouldn't show you a situation, but here's one. The equation:represents a 'conic section' called a hyperbola.
It is often easier to solve problems involving this curve by expressing this equation in 'parametric' form:which is possible because:(using the relationship )
If you're not familiar with using parametric equations, you'll just have to trust me on this one, and wait until you've met them in another part of your course.
Grandad
Thank you Grandad, once I reach that stage in my maths your name will be remembered
ok last thing before I head off (it's nearly 12 over here)
Prove that
ok on the LHS we have a we can change that into ,
Therefore:
On the RHS:
we have
we can change the into
So the fraction will be:
Put LHS and RHS side together and we have:
is this all good?
You went too far, stop where I put it in red. You weren't asked to show anything past where I put my edit
edit: seems I've misunderstood - when you're trying to prove an identity pick one side only and manipulate it to get the other. For example starting on the LHS:
As you said
As you then pointed out later so put the new expression into the original one
As we've obtained the RHS the identity is proven and there it stops
Hello jgv115The simple way is to use the 'double-angle' formula (which perhaps you haven't yet come across):which makes it rather simple, doesn't it?
Or, since is a special angle, we know that , and that . So perhaps you're expected to prove it that way. And you can, can't you?
Grandad
Hello jgv115Content: 5 stars!
Style: 1 star!
You have everything you need here, so what do I mean about style? It's OK to write down both sides of the equation, like this:when you're establishing what it is you have to prove. But when you begin the proof itself, just write down one side, and work your way around to the other; like this:
See the difference? Now you have a logical argument which works correctly from one end to the other, rather than starting with something that you don't know about (i.e. the thing you're asked to prove) and ending up with something that is self-evidently true .
That's the difference between good and bad style!
Grandad