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# Trigonometric identity + equation?! watch

1. (Original post by PLM98)
Hey Zacken, just a question. How do you jump from 2(1-cos2x) = 2sin2x tanx to 2sin2x tanx=tanx ?
The question gives you but you also know that

So: .

And if you end up with that then then 2sin2x=1 so sin2x=1/2 and you dont get any bracket with a tanx outside or anything.
You cancelled on both side, never do that!! It's like me saying so by cancelling x on both sides. That's not correct at all.

You should factorise instead, so you'd do: so x=0 or x=1. In this case, bring over everything to one side and factor out a tan x.
2. (Original post by drandy76)
i could post it if you like,then you could get around to doing it whenever you're free?
post it in a separate thread as it is not nice to hijack somebody else's thread ...
(Wednesday that I am off I can look at it)
3. (Original post by Zacken)
The question gives you but you also know that

So: .

You cancelled on both side, never do that!! It's like me saying so by cancelling x on both sides. That's not correct at all.

You should factorise instead, so you'd do: so x=0 or x=1. In this case, bring over everything to one side and factor out a tan x.
Ohhhh right, now I see it. Thanks!!
4. Hope this helps
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5. Did you put 4 as dy/dx then got a quadratic and solved for sin x? If so you should be right.
6. (Original post by PLM98)
Did you put 4 as dy/dx then got a quadratic and solved for sin x? If so you should be right.
are you talking to me?
7. (Original post by Maim56)
are you talking to me?
Yeah

EDIT: NOOO sorry got confused probably with a different thread HAHAHAHA I was answering to someone who had put a parametric equation, maybe it was in this forum and they deleted the post, I don't really know
8. (Original post by PLM98)
Yeah

EDIT: NOOO sorry got confused probably with a different thread HAHAHAHA
No you didn't. The person who you were replying to has deleted their post.
9. (Original post by Zacken)
No you didn't. The person who you were replying to has deleted their post.
Oh right (wrote that in my second edit because I thought that was more likely). I am not crazy then!!
10. (Original post by PLM98)
Yeah
Sorry I should have explained it

so your original equation is (1-cos2x)/sin2x

we know from trig identities that:

*Cos2x= 1 - 2sin^2 x(sin squared x)
*Sin2x=2sinxcosx

so using the identities...

1-cos2x becomes...

1-(1-2sin^2 x)(sin squared x)

and Sin2x becomes...

2sinxcosx

so now the equation becomes:

(1-1+2sin^x)/2sinxcosx

= sin^2 x/sinxcosx (there are two sin's on the numerator and one on the denominator so you can cancel one of the sin's on the numerator out to be left with one sin on the numerator and cos on the denominator)

=sinx/cosx
=tanx

i hope this helped haha sorry if my explanation isn't clear enough
11. (Original post by Maim56)
Sorry I should have explained it

so your original equation is (1-cos2x)/sin2x

we know from trig identities that:

*Cos2x= 1 - 2sin^2 x(sin squared x)
*Sin2x=2sinxcosx

so using the identities...

1-cos2x becomes...

1-(1-2sin^2 x)(sin squared x)

and Sin2x becomes...

2sinxcosx

so now the equation becomes:

(1-1+2sin^x)/2sinxcosx

= sin^2 x/sinxcosx (there are two sin's on the numerator and one on the denominator so you can cancel one of the sin's on the numerator out to be left with one sin on the numerator and cos on the denominator)

=sinx/cosx
=tanx

i hope this helped haha sorry if my explanation isn't clear enough
Nooo your explanation was clear don't worry, read the edit of the post you answered to, I was answering to someone who deleted their post, sorry!!

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