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C3 Trig Identities: Can you spot where I've gone wrong?

Given that a = 4secx, b=cosx and c=cotx

Show that c^2 = 16/a^2 - 16:

c^2 = cot^2x

= c0s^2x/1-cos^2x

secx = a/4 therefore cosx=4/a

=16/a^2/1 - 16/a^2

=16/a^2/a^2 - 16

So I've managed to get the denominator correct, but the not the numerator! Where have I gone wrong?
Reply 1
Avatar for jadys10
jadys10
14
Original post by creativebuzz
Given that a = 4secx, b=cosx and c=cotx

Show that c^2 = 16/a^2 - 16:

c^2 = cot^2x

= c0s^2x/1-cos^2x

secx = a/4 therefore cosx=4/a

=16/a^2/1 - 16/a^2

=16/a^2/a^2 - 16

So I've managed to get the denominator correct, but the not the numerator! Where have I gone wrong?


Can you retype this using the square in this πr²h so it's not ^2
I'm getting confused
Is it 16 divided by a squared minus 16 or 16 divided by a squared and then the fraction minus 16?
Original post by creativebuzz

=16/a^2/1 - 16/a^2


Your notation is slightly confusing. :smile:

I presume that by this, you mean:

16a21−16a2\dfrac{\frac{16}{a^2}}{1-\frac{16}{a^2}} ? :smile:

If so...

=16/a^2/a^2 - 16


How did you get to this line from the previous one? :smile:
Reply 3
Avatar for jadys10
jadys10
14
I've got as far to cot²x=cosx-4
(edited 9 years ago)
Original post by usycool1
Your notation is slightly confusing. :smile:

I presume that by this, you mean:

16a21−16a2\dfrac{\frac{16}{a^2}}{1-\frac{16}{a^2}} ? :smile:

If so...



How did you get to this line from the previous one? :smile:


Yup, I meant

because I squared cosx = 4/a to get 16/a²
Original post by creativebuzz
Yup, I meant

because I squared cosx = 4/a to get 16/a²


Yes, I get that and that's fine. However, how did you then go from:

16a21−16a2\dfrac{\frac{16}{a^2}}{1-\frac{16}{a^2}}

to:

16a2a2−16\dfrac{\frac{16}{a^2}}{a^2 - 16} ?

Spoiler

Original post by usycool1
Yes, I get that and that's fine. However, how did you then go from:

16a21−16a2\dfrac{\frac{16}{a^2}}{1-\frac{16}{a^2}}

to:

16a2a2−16\dfrac{\frac{16}{a^2}}{a^2 - 16} ?

Spoiler



I multiplied the denominator by a²!

Oh.. wait.. whatever I do to the top I do to the bottom. So that would have given the answer :tongue:
Original post by creativebuzz
I multiplied the denominator by a²!

Oh.. wait.. whatever I do to the top I do to the bottom. So that would have given the answer :tongue:


Indeed, good work. :awesome:
So I did RHS

16/[(4secx)2-16]
16/16sec2x-16
split them up
16/16 * 1/1+tan2x - 16/16 (from the trig rule 1+tan2x = sec2x)
1 - 1 = 0
so you're left with 1/1+tan2x
split that up
1/1 * 1/tan2x which is 1/tan2x = cot2x = c2

I think something may have gone a bit wrong in my method even though I obtained the right answer....
image.jpg233.6KB
Original post by usycool1
Indeed, good work. :awesome:


Hahah thanks for being so kind! I've given you a good rep :biggrin::smile:

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