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C3 Trig identities

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Can someone explain why you would have to use the cot identity and not solve in terms of cosec but cot instead ? In other words when putting all terms into either cot or cosec how do you choose which one to pick
Reply 1
Original post by ColeWorld98

Can someone explain why you would have to use the cot identity and not solve in terms of cosec but cot instead ? In other words when putting all terms into either cot or cosec how do you choose which one to pick

cotx=cosec2x7\cot x = \mathrm{cosec}^2 x - 7

The aim of these questions is to get them in terms of a single trig ratio. So here you correctly thought that you could either get it all in terms of cotx\cot x or cosecx\mathrm{cosec} x.

The relevant identity is

cot2x+1cosec2x\cot^2 x + 1 \equiv \mathrm{cosec}^2 x

Notice that the identity includes cot2x\cot^2 x and not cotx\cot x so you wouldn't be able to use the identity to change the cotx\cot x part of your equation. So your only choice is to change the cosec2x\mathrm{cosec}^2 x.

Does that make sense?
(edited 7 years ago)
Why not give it a whirl and see what happens? You can rearrange cosec^2(x) = 1+cot^2(x) to give cot(x) = (cosec^2(x) - 1)^0.5 and put that on the left hand side of the equation. You can then square both sides and arrive at a quadratic equation in cosec^2(x) which is perfectly solvable. (The equation will look like a quartic in cosec(x) but it can be handled as a quadratic in cosec^2(x)).

However, danger lurks when you square both sides of an equation as it throws up spurious solutions corresponding to (left hand side) = -(right hand side). Both sides will become positive on squaring.

Therefore it's best to go with the suggested working.

..and note what notnek said..
(edited 7 years ago)
Reply 3
Yes ! So for 1+cot^2x=cosec^2x - are you unable to subtract 1 and square root to get in terms of cot to then sub in ?
Reply 4
Original post by ColeWorld98
Yes ! So for 1+cot^2x=cosec^2x - are you unable to subtract 1 and square root to get in terms of cot to then sub in ?

If you did that you'd end up with an equation that includes a square root. These are never nice and you'd eventually get a quartic equation that may or may not be easy to solve. Plus you'll probably get extra solutions.

So that may lead to the correct solutions but it's 10 times harder. Always change the squared trig ratio for these types of questions.
Reply 5
So choose the term with the most available substation. So in this example sub in cosec^2x because it can be subbed in without changing the identity ?
Reply 6
Original post by ColeWorld98
So choose the term with the most available substation. So in this example sub in cosec^2x because it can be subbed in without changing the identity ?

Yes. Of course you can rearrange an identity and still use it but you never want to change it by square rooting it.
you wrote that cot α is cosec α - 1

this is not the case.
Original post by ColeWorld98
So choose the term with the most available substation. So in this example sub in cosec^2x because it can be subbed in without changing the identity ?


Curiosity about this kind of thing is good. To understand the issues here I suggest you start with the equation below and see where it leads. Then compare your findings with the suggested working.

cosec2x1=cosec2x7\sqrt {\mathrm{cosec}^2 x-1}=\mathrm{cosec}^2 x-7

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