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

creativebuzz
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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?
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jadys10
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(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?
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usycool1
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(Original post by creativebuzz)
=16/a^2/1 - 16/a^2
Your notation is slightly confusing.

I presume that by this, you mean:

\dfrac{\frac{16}{a^2}}{1-\frac{16}{a^2}} ?

If so...

=16/a^2/a^2 - 16
How did you get to this line from the previous one?
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jadys10
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I've got as far to cot²x=cosx-4
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creativebuzz
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(Original post by usycool1)
Your notation is slightly confusing.

I presume that by this, you mean:

\dfrac{\frac{16}{a^2}}{1-\frac{16}{a^2}} ?

If so...



How did you get to this line from the previous one?
Yup, I meant

because I squared cosx = 4/a to get 16/a
²
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usycool1
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(Original post by creativebuzz)
Yup, I meant

because I squared cosx = 4/a to get 16/a
²
Yes, I get that and that's fine. However, how did you then go from:

\dfrac{\frac{16}{a^2}}{1-\frac{16}{a^2}}

to:

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

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I presume you got there because you multiplied by \dfrac{a^2}{a^2}.

If so, \dfrac{16}{a^2} \times a^2 = 16
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creativebuzz
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(Original post by usycool1)
Yes, I get that and that's fine. However, how did you then go from:

\dfrac{\frac{16}{a^2}}{1-\frac{16}{a^2}}

to:

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

Spoiler:
Show
I presume you got there because you multiplied by \dfrac{a^2}{a^2}.

If so, \dfrac{16}{a^2} \times a^2 = 16
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
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usycool1
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(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
Indeed, good work. :awesome:
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selenerrr__x
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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....
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langlitz
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creativebuzz
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(Original post by usycool1)
Indeed, good work. :awesome:
Hahah thanks for being so kind! I've given you a good rep
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