Trigonometry Help (I am Desperate)

Can anyone help me figure these out?

I need to prove both.

 

http://www.equationsheet.com/sheets/Equations-33.html

The section I am working on deals with double angle and half angle identities.

 

I will ask my Dad, next time I see him.

He is a math god in my family.

But, I assume you need this done quickly.

 

Preferably this weekend, definately before Tuesday.

Thanks

 

But, I can guess his answer.

Read the books, you can figure this out. Math is how the world turns.

I'll see what I can do......

 

LOL, thanks

As for reading the book, I've done that, many times.
--and another Trig text book too.

These are the only two that stumped me. I've fought with the first one for hours.

 

Do you use Math Cad?

 

No, sure don't.

I got this far on the first one.

This is the left side.

 

Right side:

Not sure if its a legal move to put the 2 in the denominator. Nor does it get me any closer to my answer if I did. Its got to be simple, and I am just missing it.

 

Wish I could help you out. Proofs are my downfall. I personally could use to do this, but, after 15 years you forget a lot of stuff.

But, once again, I'll bring it up with my Dad. Unfortunately, this won't happen until Thanksgiving timefram.

 

I'll let you know if I figure it out. Maybe I just need to sleep on it.

 

Sleeping is good. I tend to take breaks on certain problems. Thinking about other items that need to be done. Then when I work on those items, I tend to think on the original problem with a new angle.

In the end, I find that understanding the problem on multiple sides, may conclue my problem, in itself.

 

I'll print and give to my mum.

 

sh!t dude, I have done trig just about 2 weeks ago. I was getting all correct, those items are bloody hard... which method did you use? try doing subtraction of the LHS from the RHS etc... I find those easier.. will have another go now...

 

well i figured out why i cant do it... i have never done half angle identities... my syllabus only needs double angle formulae and then where you find alternatives of asinx +/- bcosx.... soz bud, that $hits hard... what syllabus you doing?

 

duuuuuuuuuuhhh

 

That is quite a hard couple of problems you have there adrynalyne, unless your familiar with complex numbers, and a couple of theorems then you wont be able to do them! Finally, being a maths student has paid off!
Ive done the second one for you, if you still cant do the first one let me know and ill post the solutiuon for that aswell

write the solutions out on paper as I have wrote powers as
2(to the power of 3) = 8 and it doesnt look like it makes much sense! let me know if there any typos

 

I just looked at what you needed to end up with and its a little different to my answer, I put it in its simplest form, silly me! im sure you can work out your answer from there though!

 

That went WAYYYY above my head, and skill level at this point.

I should be able to solve these problems with identities learned thus far, with basic identities, and double angle and half angle identities.

 

Its a customized book for the College I go to. However, I checked a book from when I took this same course like 7 years ago...and the section is identical. Pretty standard stuff, really.

Section 5.3-5.4 in any McKeague Trig text book should be the same.

 

Ok, im trying my best to remember double angle identities at the moment. Ive not done this for a couple of years so bare with me. Im not sure if it is possible to do that proof without what I did, but something I always found helpful when doing these was get every double identity that you have got in front of you and just play about with them.
cos(tothepower4)@ = cos(squared)@cos(squared)@ would be a good start.

 

ps working backwards from what you have to prove is sometimes helpful.
so start with the Right Hand Side and try to get the LHS

 

I've got one worked out with help from my friend's husband, who is a mathmetician.

Standy and I'll post it. I dunno why I never saw it before.

 

You have to use a little creativity on this one.

Recognize the second to last line is the half angle cosine formula flipped, and squared, making it a secant formula squared.

 

i always found division of LHS/RHS by the RHS/LHS to make 1 OR subtraction from one side on the other to make 0 was quite good. but seen as i couldn't do these i should shut up now...

 

Well, that wouldn't work. The idea is that I have to put both in the same form, left side and right side. If they both are the same, then they must be equal to each other, hence the proof.

 

umm, duh! if they are the same then subtraction would give zero.. e.g. 5-5 = 0.

if they are the same then division would give 1. e.g. 5/5 = 1..

you might wanna try it sometime. quite basic really!

 

While doing that is a "legal" move, it won't get me credit on a test. Thats why I said it won't work.

I'm not seeing how it would help with these proofs either. To me, it looks like it is adding extra work.

 

that is the way all 3 of my text books teach it.. they give 4 ways to prove an identity at advanced level. the first is to start with LHS. second is to start with RHS. Third is division of LHS by RHS and vice versa and ensure the result is equal to 1.. fourth is to subtract LHS from RHS and vice versa and make sure the result = 0..

they are legit proofs that i have been taught. straight out of at least 2 text books. not sure on the 3rd one...if they are not actually stated in the books then my teacher who got her degree or whatever in the UK taught them to me.. all the teachers at my school know of those methods... if you don't normally use em then maybe just stick to the ones you know... cheers

 

Don show him my solution then!

You found the solution to the cos@ one yet? I'll try and dig up my double angle identity notes if you havent!

 

Yeah, sure did!

Thanks for the help everyone.

Here is how the other one worked out. You have to recognize, the right hand side is a perfect square.

 

They might teach it differently in the UK. Here, at least with the books I am using, they don't have you make one side equal to zero, except in Algebra.

 

I didn't recognize most of what you did hehe...did you use calculus to prove it?

I'm sure there are numerous ways to do these.

 

Yeah I used calculus (btw making a trig equation = 0 isnt going to help prove anything they dont teach you to do that in England thats fo sure!). I could never remember any of the double identities for my test so I just use the calculus method.

If you're interested you need to know about complex numbers (i.e i =squarre root of -1), de moivres theorem and binomial theorem (which is for series expansion).

Are you doing a maths course or something?

 

I'm building up my math for engineering. Next semester is Calc 1, then summer (if I get brave) is Calc 2, then next winter, Calc 3.

There is more after that, such as Differential Equations, and such.

 

Yep im on differential equatons, I have renamed it to difficult equations!
Also do vector calculus which i find harder still! I keep telling myself only one more year to go!

You have any more problems then post and ill be here to baffle you with more irellevant material!

 

LOL, thanks

 

well thats what i got taught.

think about it... say you have 5=5... to prove that 5=5 you can rearrange to have 5-5=0 and then subtract (if they were not equal you would have a result of 1 or 2305026 depending on what numbers you got given). if they are equal then you will get 0...

then showing that 5/5 =1 (which is actually 1=1 if you do the division) means that they are both equal. therefore proving in the end that 5=5! thats how i was taught it and i have been getting good grades for those trig items... you guys just stick to what you been taught and i will mine...

we started doing differentiation, like dy/dx of e^x and all those items. damn hard..
we will be doing square roots of negatives quite soon, like end of november i think...

 

Yeah, but 5=5...its already proven, both sides are the same. Why throw in the extra step.

 

it was just an example. for your trig identity, on one side you have [(sec^2)A/2] = 2secA/{secA+1}.. you have to prove that they are equal. if you do the whole process of division/subtraction you too will find that they are already the same...

 

Can you show me?

I'm not seeing how its gonna work out any better.

 

k let me just do one in word or whatever.. i will do em all, like all the ways to show you how to do it... k chill

 

here is subtraction...

http://img446.imageshack.us/img446/9282/math10nc.jpg

 

i don't feel much like doing the division... you understand the subtraction though right? and then you already got the LHS and RHS methods squared away, so....

 

Thats damn hard? dy/dx of e^x = e^x!!!!!!!
e^x is the only function that remains the same after differentiation
(and integration!)

 

That would work fine, except there is an error in your arithmetic.

On line 4, you dropped a power.

On line 5, -2(1-tanA) multiplied out is -2+2tanA, not -2+tan2A. I assume you multipled it out and dropped the two.

tan2A= tan(A+A)
2tanA is 2 times tanA, which is correct.

--I think. I'm still learning this stuff. I'm certain of the typo on line 4 though.

 

no, thats the first thing we learnt and the easiest to type quickly. we have been doing lately stuff like - differentiation of products and quotients and their rules...

 

here ya go, fixed it, my working on paper hadn't dropped anything... just a typo..

http://img415.imageshack.us/img415/4999/math14sx.jpg

 

Thats important stuff, make sure you remember the chain rule and the product rule!

 

To each their own I guess. Just seems like extra work