Help needed with trig identities. (Or: My god, maths is a freaking nightmare) (1 Viewer)

[Heliotrope]

It might just be me, but I'm finding trig identities damn near impossible. It's really making me want to slit my wrists (not to be dramatic or anything). Is there a fool-proof method to working them out? If you help me, you will have my eternal gratitude.

I can do most of the basic ones, it's just the "Challenge" ones that are making me tear my hair out.

i) (1/sec a - tan a) minus (1/sec a + tan a) = 2 tan a

ii) (1 + cot a/1 + tan a) = cot a

iii) (cos a/1 + sin a) = sec a(1 - sin a)

Thank you thank you thank you.



Hey, can someone who does 2U Maths tell me IF I EVEN NEED TO KNOW THIS SHIT?

If not, I'm getting a new tutor.

 

[Zephyrio]

Umm. In the future, please place brackets around your terms because if I didn't correctly deduce where your terms went, I wouldn't have been able to figure them out.

Anyway, for the first one...

1 1
----------------- - ---------------------
sec a - tan a sec a + tan a

Okay, from here, you can use trig identities. If we multiplied the denominators of each term, we'd be heading for a 1, so cross-multiplication looks promising.

After cross multiplying, we arrive at:

sec a + tan a - sec a + tan a
------------------------------------
sec ^ 2 a - tan ^ 2 a

= 2tan a because we know that 1 + tan^2 a = sec^2 a

Mkay, for the second one, we can use our definitions of cot and tan and that will rescue us.

So,
1 + cot a
-----------
1 + tan a

1 + (cos a)/(sin a)
-------------------------
1 + (sin a )/(cos a)

sin a + cos a
----------------
sin a
--------------------------
cos a + sin a
-----------------
cos a

Then we do the wonky reciprocal thingies

(sin a + cos a) cos a
---------------- x -----------------
(sin a) cos a + sin a

= (cos a)/(sin a)
= cot a

voila!

I'm sorry for the wonky working out (the fraction signs!!)

Onto the third one now.

Well, all I've gotten so far is

cos^2 a = 1 - sin^2 a...

:S

I converted the sec a to 1 / (cos a).

Hope this all helps you in some way. And yeah, you need to know trig identities for 2U. And slitting your wrists is so emo!

 

[Le3sah]

Hey, can someone who does 2U Maths tell me IF I EVEN NEED TO KNOW THIS SHIT?

If not, I'm getting a new tutor.

You need to know it for 2u. If you use the 'Maths in focus' textbook, everything up to '3D trig' is 2u stuff.

It's okay, I hate trig identities too! I always end up with the damn question as my result! :mad1:

 

[Zephyrio]

Thank you, Without Wings

Yeah, trig identities used to annoy my brains out, but once I got the hang of it I started to like them more. I think that compound angles and trig equations are more annoying, because in trig identities, you always look to multiply things which will leave you with a trig identity where you can deduce its value, and using the definitions of certain values to your advantage.

I think the key is practice!

 

[Kujah]

Trig Identities can be annoying, but like Zephyrio said, practice is the key!

Oh btw, I <3 Schindler's List

 

[nelses]

And continuing with the solutions, for the third one just multiply it by (1-sina)/(1-sina), in other words one. Expand and yeah~

The key to trig identities for me is to look at the answer you want to obtain and the path to it becomes easy.

 

[watatank]

seems as though the first two have been done for you i'll tackle the third one

iii) (cos a/1 + sin a) = sec a(1 - sin a)

LHS = [ cos a / (1 + sin a) ]

look at the RHS. notice that it has a (1 - sin a) on it? you're tryna manipulate the LHS so it looks like the RHS. sorry can't explain further. but what you have to do is multiply by [ (1 - sin a) / (1 - sin a) ]. it still stays the same though.

LHS = [ cos a / (1 + sin a) ] * [ (1 - sin a) / (1 - sin a) ]

= [ cos a / (1 + sin a) ] * [ (1 - sin a) / (1 - sin a) ]

= [ cos a (1 - sin a) / (1 - sin²a) ]

= [ cos a (1 - sin a) / (cos²a) ]

= (1 - sin a) / cos a

reciprocal of cos is sec

.'. LHS = sec * (1 - sin a)

= RHS

hope it helps! :wave:

 

[z600]

its just converting things around, make sure you know your trig identities very very well.

 

[cccclaire]

My last maths test I had a trig identity.
I just couldn't get it out, I tried 4 different ways
=[
But my friend who got 99% in the last test couldn't get it either, so its not too bad.

my maths teacher says the trick is to try and change everything to sin's and cos's, but I don't think it always works because often i get these huge things with fractions over fractions and it just confuses me =/

 

[Heliotrope]

Sorry about the brackets! A bit late, but I fixed them.

Thanks for all your efforts. But, um, I'm still kind of lost.

After cross multiplying, we arrive at:

sec a + tan a - sec a + tan a
------------------------------------
sec ^ 2 a - tan ^ 2 a

= 2tan a because we know that 1 + tan^2 a = sec^2 a

Could you please explain cross-multiplication for me? What I did originally was I got the common denominator by multiplying the two denominators and merged the fractions, which got me... (sec a + tan a/sec^2 a - tan^2 a) - (sec a - tan a/sec^2 a - tan^2 a) = (sec a + tan a) - (sec a - tan a)/sec^2 a - tan^2 a. Is that right?

sin a + cos a
----------------
sin a
--------------------------
cos a + sin a
-----------------
cos a

You lost me here, sorry.

Watatank, I can follow everything, but I still don't quite get why you have to do the first thing. But I guess I know what to do now. Thank you!

 

[expertdkx]

WELL I GOT A FRIEND WHO DO 3U MATHS AND HE LIKE TRIGS
MSN HIM:
pokemon_master081@hotmail.com
HE SHOULD BE ABLE TO HELP

 

[watatank]

(sec a + tan a) - (sec a - tan a)/sec^2 a - tan^2 a. Is that right?

Watatank, I can follow everything, but I still don't quite get why you have to do the first thing. But I guess I know what to do now. Thank you!

that working out is correct. simple expansion now. the numerator goes to 2tana and the denominator goes to 1.

how did i kno that? know your trig identities.

cos^2x + sin^x = 1

divide whole equation by cos^2 and you've got

1 + tan^2 a = sec^2 a

your denominator is sec^2 a - tan^2 a. rearranging the above equation you will see that equals 1.

as for how i got my answer...as lottox said, look to get squares. if you multiply by (1 - sin a)/(1 - sin a) you make the denominator the difference of two squares. i also knew that cos i knew my answer needed to have (1 - sin a) in it so it was a good guess to try that. sorry i can't explain it any better

 

[watatank]

show: (1 + cot a/1 + tan a) = cot a

LHS = (1 + cot a/1 + tan a)

put in terms of sin and cos

LHS = [ ( 1 + {cosa/sina} ) / ( 1 + {sina/cosa} ) ]

put numerator and denominator respectively over a common denominator.

= [ ( sina + cosa)/sina ] / [ (cosa + sina) / cosa ]

now you're dividing one fraction by another. multiplying by the reciprocal (means you flip fraction over) has the same effect as dividing.

= [ ( sina + cosa)/sina ] * [ cos a /(cosa + sina) ]

( sina + cosa) in numerator and denom cancel each other out so youre left with

= cosa/sina

= cot a

= RHS

hope that helps! :wave:

 

[PrettyVacant]

RARGH LOL I HATE THOSE!!

Yeah. Okay.

Not really..

Okay




I love maths.