Hi all.(adsbygoogle = window.adsbygoogle || []).push({});

Hey - i'm new here - so please excuse me if i have posted in the wrong thread etc

Now, for the details.

Firstly, I'm from New Zealand, 16yrs - at High School.

I'm doing NCEA (new system) level 3 Calculus (last year High School Calc)

I came here mainly to look around - but i have a problem, so i thought I might as well ask.

We're doing Trigonometry at the moment - and we're touching on Indentites - and proving them.

Me and my mate have been going through doing the worksheet we got given for the holidays - we've managed them all ok, except from one, so i thought i'd come hear for help (like i said earlier - this may be the wrong thread - o, and what is "college" and Grades K-12 - what would i fit into...)

I'm not to sure if this is standard stuff or not... but anyway, here goes...

We have to prove that:

cos²x cosec²x - sin²x sec²x = 4 cot2x cosec2x

Now, like I said - i tried it by myself - no luck, went to my classmate - we spent about 3 hrs straight, and 20-30 pages, and were completey puzzled.

If I do it in the simplistic way & change the Left side, I change:

cosec²x - which is 1/sin²x (correct?)

and sec²x to 1/cos²x (correct?)

mulitply cos²x with sin²x = cos²x/sin² x

and the other one = sin²x/cos²x

Subtract the two, and i get = 1 -- which is not true...

Another method we tried was...

as cosec²x = cot²x + 1

and

sec²x = tan²x + 1

you can subsitute them in, then change the cot²x to cos²x/sin²x (is that true? will they both be squared?)

and the tan²x to sin²x/cos²x

but even with this way we ended up in a jumble mess

We also tried starting by changing the right hand side (the 4 cot2x cosec2x) - even though we were talk not to use this method UNLESS YOU HAD TO, but we still got no where.

There were heaps of other methods we used - but I'd be here all night typing them up.

So if anyone has any ideas - i really only needed a pointer, or hint - i'll be able to sus out the rest! and I don't like 'cheating'

Thanks

Alex

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# Indentites - Calculus (newbie)

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