# Math Help - Simple trig help!

1. ## Simple trig help!

I know what sine, cosine and tangent is.

I was looking through a few books and I see a lot of ones I'm not familiar with.

One of them is $\cot$ and $cosec$

what do they mean? are there anymore besides sin, cos and tan?

2. Hello jgv115
Originally Posted by jgv115
I know what sine, cosine and tangent is.

I was looking through a few books and I see a lot of ones I'm not familiar with.

One of them is $\cot$ and $cosec$

what do they mean? are there anymore besides sin, cos and tan?
$\cot x = \frac{1}{\tan x}=\frac{\text{Adjacent}}{\text{Opposite}}$

$\text{cosec } x = \frac{1}{\sin x}=\frac{\text{Hypotenuse}}{\text{Opposite}}$

And there's also $\sec x = \frac{1}{\cos x}=\frac{\text{Hypotenuse}}{\text{Adjacent}}$

3. the list of recirpical function are as follows:
(cotangent) $\cot (x) = \frac{1}{\tan(x)} = \frac{\cos(x)}{\sin(x)}$
(cosecant) $\csc (x) = \frac{1}{\sin(x)} = \frac{\mbox{hypothuse}}{\mbox{opposite side}}$
(secant) $\sec(x) = \frac{1}{\cos(x)} = \frac{\mbox{hypothuse}}{\mbox{adjacent side}}
$

and then you have the inverse functions where if we have

$y= \sin{x}$ and y is given so $x = \arcsin(y) = x$ (or $\sin^{-1}(x)$ depending on notation).

$y = \cos(x) \ \mbox{then} \ x = \arccos(y)$
$y = \tan(x) \ \mbox{then} \ x = \arctan(y)$

and for the recipricals

$y = \csc(x) \ \mbox{then} \ x = csc^{-1} (y)$ (or arccsc depending on notation)
$y = \sec (x) \ \mbox{then} \ x = sec^{-1} (y)$ (or arcsec depending on notation)
$y = \cot(x) \ \mbox{then} \ x = cot^{-1} (y)$ (or arccot depending on notation)

4. Ok, thanks for your help so far.

$\sec 60 = \frac {x}{4}$

$\sec 60$ can change into $\frac {1}{\cos 60}$

$x = \frac {4}{\cos60} = 8$

If we know that $\sec 60$ can change into $\frac {1}{\cos 60}$ how come we don't use $\frac {1}{\cos 60}$ instead?

5. Hello jgv115
Originally Posted by jgv115
Ok, thanks for your help so far.

$\sec 60 = \frac {x}{4}$

$\sec 60$ can change into $\frac {1}{\cos 60}$

$x = \frac {4}{\cos60} = 8$

If we know that $\sec 60$ can change into $\frac {1}{\cos 60}$ how come we don't use $\frac {1}{\cos 60}$ instead?
Your working is correct, and that's a very good question!

The answer is that, strictly speaking, you never do need to use $\sec x$; you can always use $\cos x$ 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 $\frac{1}{\cos x}$ 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 $\tan x$ either? That's simply because $\tan x = \frac{\sin x}{\cos x}$. 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!

6. Thanks Grandpa, I really appreciate your help.

I have drawn this diagram up:

By using pythagoras we can get

$\cos^2 \theta + \sin^2 \theta = 1$

But the book I have here says you can also get these:

$cot^2 \theta +1 = cosec^2 \theta$

and

$1+\tan^2\theta = \sec^2 \theta$

i'm so confused by this. I have no idea how to get these. Could anyone provide assistance?

7. Hello jgv115
Originally Posted by jgv115
Thanks Grandpa, I really appreciate your help.

I have drawn this diagram up:

By using pythagoras we can get

$\cos^2 \theta + \sin^2 \theta = 1$

But the book I have here says you can also get these:

$cot^2 \theta +1 = cosec^2 \theta$

and

$1+\tan^2\theta = \sec^2 \theta$

i'm so confused by this. I have no idea how to get these. Could anyone provide assistance?

Your book is quite correct (and, incidentally, it means that, theoretically, you hardly need cosine either, you can just about get by with sine, and $\cos^2\theta = 1 - \sin^2\theta$!)

The other two equations are very easy to derive, if you remember that $\tan \theta = \frac{\sin \theta}{\cos \theta}$ and therefore that $\cot\theta = \frac{\cos\theta}{\sin\theta}$. So start with:
$\cos^2\theta+\sin^2\theta = 1$
and divide both sides, first of all, by $\sin^2\theta$:
$\frac{\cos^2\theta}{\sin^2\theta}+\frac{\sin^2\the ta}{\sin^2\theta}=\frac{1}{\sin^2\theta}$

$\Rightarrow \cot^2\theta+1 = \csc^2\theta$
Now go back and divide both sides by $\cos^2\theta$:
$\frac{\cos^2\theta}{\cos^2\theta}+\frac{\sin^2\the ta}{\cos^2\theta}=\frac{1}{\cos^2\theta}$

$\Rightarrow 1 + \tan^2\theta = \sec^2\theta$

8. Ok, I understand

$5+5\tan^2\theta$

factorise:

$5(1+tan^2\theta)$

since we know that $1+tan^2\theta = \sec^2\theta$

$5\sec^2\theta$

But we only simplify to make the thing look easier to read? Does it actually make problems easier to solve?

9. Hello jgv115
Originally Posted by jgv115
...

Does it actually make problems easier to solve?
Yes, it does. I said I wouldn't show you a situation, but here's one. The equation:
$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$
represents a 'conic section' called a hyperbola.

It is often easier to solve problems involving this curve by expressing this equation in 'parametric' form:
$x = a\sec\theta, y = b\tan\theta$
which is possible because:
$\frac{x^2}{a^2}-\frac{y^2}{b^2}= \sec^2\theta - \tan^2\theta = 1$
(using the relationship $1 + \tan^2\theta = \sec^2\theta$)

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.

10. 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 $\sin\theta\tan\theta = \frac{1-\cos^2\theta}{\cos\theta}$

ok on the LHS we have a $\tan\theta$ we can change that into $\frac {\sin\theta}{\cos\theta}$,

Therefore: $\sin\theta * \frac {\sin\theta}{\cos\theta} \rightarrow \frac {\sin^2\theta}{\cos\theta}$

On the RHS:

we have $\frac {1-\cos^2\theta}{\cos\theta}$

we can change the $1-\cos^2\theta$ into $\sin^2\theta$

So the fraction will be:

$\frac {\sin^2\theta}{\cos\theta}$

Put LHS and RHS side together and we have:

$\frac {\sin^2\theta}{\cos\theta} = \frac {\sin^2\theta}{\cos\theta}$

is this all good?

11. Originally Posted by jgv115
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 $\sin\theta\tan\theta = \frac{1-\cos^2\theta}{\cos\theta}$

ok on the LHS we have a $\tan\theta$ we can change that into $\frac {\sin\theta}{\cos\theta}$,

Therefore: $\sin\theta * \frac {\sin\theta}{\cos\theta} \rightarrow \frac {\sin^2\theta}{\cos\theta}$

On the RHS:

we have $\frac {1-\cos^2\theta}{\cos\theta}$

e^(i*pi) - Stop here - you've already proved the LHS = RHS

we can change the $1-\cos^2\theta$ into $\sin^2\theta$

So the fraction will be:

$\frac {\sin^2\theta}{\cos\theta}$

Put LHS and RHS side together and we have:

$\frac {\sin^2\theta}{\cos\theta} = \frac {\sin^2\theta}{\cos\theta}$

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 $sin(\theta)tan(\theta) =\frac {\sin^2\theta}{\cos\theta}$

As you then pointed out later $sin^2 \theta= 1-\cos^2\theta$ so put the new expression into the original one

$sin(\theta)tan(\theta) =\frac {1-cos^2 \theta}{\cos\theta}$

As we've obtained the RHS the identity is proven and there it stops

I've come across this question:

Show that:

$\frac {2\tan60}{1-\tan^260} = \tan120$

How would I go about starting that?
Please don't write the answer or at least put it in spoiler tags.

13. Hello jgv115
Originally Posted by jgv115

I've come across this question:

Show that:

$\frac {2\tan60}{1-\tan^260} = \tan120$

How would I go about starting that?
Please don't write the answer or at least put it in spoiler tags.
The simple way is to use the 'double-angle' formula (which perhaps you haven't yet come across):
$\tan2A =\frac{2\tan A}{1-\tan^2A}$
which makes it rather simple, doesn't it?

Or, since $60^o$ is a special angle, we know that $\tan60^o = \sqrt3$, and that $\tan(180-A) = -\tan A$. So perhaps you're expected to prove it that way. And you can, can't you?

14. haha to be honest Grandad I have no idea what to do...

but this is my crack (it's probably wrong)

$\frac {2\tan60}{1-\tan^260} = \tan120$

$\frac {2 * \sqrt3}{1-\sqrt{3}*\sqrt{3}} = -\sqrt3$

$\frac {2 * \sqrt3}{1-3} = -\sqrt3$

$\frac {2 * \sqrt3}{-2} = -\sqrt3$

We can cross the 2 and -2 out.

$-\sqrt3 = -\sqrt3$

HAHA!

15. Hello jgv115
Originally Posted by jgv115
haha to be honest Grandad I have no idea what to do...

but this is my crack (it's probably wrong)

$\frac {2\tan60}{1-\tan^260} = \tan120$

$\frac {2 * \sqrt3}{1-\sqrt{3}*\sqrt{3}} = -\sqrt3$

$\frac {2 * \sqrt3}{1-3} = -\sqrt3$

$\frac {2 * \sqrt3}{-2} = -\sqrt3$

We can cross the 2 and -2 out.

$-\sqrt3 = -\sqrt3$

HAHA!
Content: 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:
$\frac {2\tan60}{1-\tan^260} = \tan120$
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:
$\frac {2\tan60}{1-\tan^260}=\frac{2\sqrt3}{1-(\sqrt3)^2}$
$= \frac {2\sqrt3}{1-3}$
$= \frac {2\sqrt3}{-2}$
$= -\sqrt3$
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 $(-\sqrt3 = -\sqrt3)$.

That's the difference between good and bad style!