Trigonometric Identities

**AngusBurger** · June 3rd 2011, 09:28 AM

Hi guys.

So, twenty hours of my life have passed. I'm never going to get them back, but I can prevent another twenty hours elapsing and that cold, hard feeling of my head hitting the desk one more time. Here's why:

$2cosec2xcos2x = cotx-tanx$

Obviously. So I have to show that. My initial thinking was to convert all the terms into sin and cos, and then using a couple of trigonometric functions left me with:

$2 \frac{1}{2cosxsinx} {cos}^{2}x - {sin}^{2}x = cotx - tanx$

Because of the sum on the right hand side of the equation, I'm not entirely sure what I should do it. I've tried various different techniques but I've ultimately not been able to get an answer. I've come close to ${cos}^{2}x + {sin}^{2}x = 1$ , so that's what I've been aiming for.

Can anyone nudge me in the right direction? I've also tried converting all terms into tan variants but that has only succeeded in driving me into another brick wall.

Thanks

**skeeter** · June 3rd 2011, 09:42 AM

$2\csc(2x)\cos(2x) =$

$\frac{2\cos(2x)}{\sin(2x)} =$

$\frac{2(\cos^2{x}-\sin^2{x})}{2\sin{x}\cos{x}} =$

break it up into two fractions and finish it

**jgv115** · June 3rd 2011, 06:06 PM

I suggest altering the RHS if you have trouble going from sin/cos to tan

So..

$\cot(x) - \tan(x)$

$\frac {\cos(x)}{\sin(x)} - \frac {\sin(x)}{\cos(x)}$

Now take the common denominator...

**AngusBurger** · June 4th 2011, 05:01 AM

This is ridiculously more advanced than any trig I've done before and I'm afraid the scope of my notes just doesn't cover it.

I'm okay converting the right hand side of the equation, it's just that when I start multiplying and dividing across the two sides I don't know how to do it properly with the plus and minus signs floating around.

By breaking the left hand side of the equation up into two fractions, I get:

$2{cos}^{2}x....- 2{sin}^{2}x$

$sinx cosx...sinx cosx$

(Can't get the code working, but you can see what I mean)

Again I'm stumped. I've never had to do anything even remotely like this before. I can only hazard a guess that it ends up as cosx/sinx - sinx/cosx at some point.

**Quacky** · June 4th 2011, 05:18 AM

I disagree with your result.

Originally Posted by skeeter

$2\csc(2x)\cos(2x) =$

$\frac{2\cos(2x)}{\sin(2x)} =$

$\frac{2(\cos^2{x}-\sin^2{x})}{2\sin{x}\cos{x}} =$

break it up into two fractions and finish it

$=\frac{Cos^2(x)}{Sin(x)Cos(x)}-\frac{Sin^2(x)}{Sin(x)Cos(x)}$

Now just cancel like terms.

**AngusBurger** · June 4th 2011, 07:01 AM

Now that I have it all written down, it seems so simple, but it was utter anguish getting there. Of course, as I do more of these problems I'll probably get a better feel for what I should and should not be doing, but is there any tried and tested process that I should take into account when doing this?

For instance, my first thought was to express the equation in terms of sin and cosine, so as to standardise like-terms, and that did seem to work, but I suppose my knowledge of trigonometry ultimately prevented me from seeing this through to the end without some help. I also think I seemed a bit preoccupied with flinging terms across the formula rather than focussing on its actual composition.

Or is it all just horses for courses?

**Quacky** · June 4th 2011, 07:17 AM

Originally Posted by AngusBurger

Now that I have it all written down, it seems so simple, but it was utter anguish getting there. Of course, as I do more of these problems I'll probably get a better feel for what I should and should not be doing, but is there any tried and tested process that I should take into account when doing this?

For instance, my first thought was to express the equation in terms of sin and cosine, so as to standardise like-terms, and that did seem to work, but I suppose my knowledge of trigonometry ultimately prevented me from seeing this through to the end without some help. I also think I seemed a bit preoccupied with flinging terms across the formula rather than focussing on its actual composition.

Or is it all just horses for courses?

I dont think there is a set approach. Converting to sin and cos is usually a good start (not always - think of your common identities first, and check whether any of those will help), otherwise it really is just a case of combining and separating fractions or adapting your identities as appropriate. If everything looks like a dead end, then you really just have to consider how to get your stage to resemble the one that you are trying to reach. With practice, it will become far easier.

**jgv115** · June 4th 2011, 10:08 PM

Originally Posted by AngusBurger

Now that I have it all written down, it seems so simple, but it was utter anguish getting there. Of course, as I do more of these problems I'll probably get a better feel for what I should and should not be doing, but is there any tried and tested process that I should take into account when doing this?

For instance, my first thought was to express the equation in terms of sin and cosine, so as to standardise like-terms, and that did seem to work, but I suppose my knowledge of trigonometry ultimately prevented me from seeing this through to the end without some help. I also think I seemed a bit preoccupied with flinging terms across the formula rather than focussing on its actual composition.

Or is it all just horses for courses?

I would suggest writing up on one sheet of paper all the trigonometric identities that you require, if you use a textbook then just extract the identities from them. Then practice a lot, keep referring to your sheet of identities, the more you use these the more easier it will be for you to remember them. You will get to a certain stage where it almost becomes second nature. Then all this identity stuff will be easy!

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