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Help with C3 trig please?

Currently C3 trig is making me want to die because I've spent about an hour on about 3 questions. Here are the 3 I'm struggling with (hint, they all have sin2x in)
1) cos^2(x) - sin2(x) = sin^2(x)
2) 2sin2(x) = 3tan(x)
3) 5sin2(x) + 4sin(x) = 0

If anyone could help on any of them then I'd be really grateful.
I'll go one by one. For the first, assuming you mean cos⁡2(x)−sin⁡2(x)=sin⁡2(x)\cos^2(x) - \sin^2(x) = \sin^2(x), remember that tan⁡x=sin⁡xcos⁡x\displaystyle \tan x = \frac{\sin x}{\cos x} and so tan⁡2x=sin⁡2xcos⁡2x\displaystyle \tan^2 x = \frac{\sin^2 x}{\cos^2 x}. How does this help here?
Reply 2
Avatar for yusyus
yusyus
21
Original post by AxSirlotl
Currently C3 trig is making me want to die because I've spent about an hour on about 3 questions. Here are the 3 I'm struggling with (hint, they all have sin2x in)
1) cos^2(x) - sin2(x) = sin^2(x)
2) 2sin2(x) = 3tan(x)
3) 5sin2(x) + 4sin(x) = 0

If anyone could help on any of them then I'd be really grateful.


for 1.)
cos^2(x)-sin^2(x)=sin(2x)
cos^2(x)-sin^2(x)=cos(2x) so

cos(2x)=sin(2x)
divide by cos(2x) but remember you get cos(2x)=0 as a lost solution.
Reply 3
Avatar for AxSirlotl
AxSirlotl
OP
18
Original post by _gcx
I'll go one by one. For the first, assuming you mean cos⁡2(x)−sin⁡2(x)=sin⁡2(x)\cos^2(x) - \sin^2(x) = \sin^2(x), remember that tan⁡x=sin⁡xcos⁡x\displaystyle \tan x = \frac{\sin x}{\cos x} and so tan⁡2x=sin⁡2xcos⁡2x\displaystyle \tan^2 x = \frac{\sin^2 x}{\cos^2 x}. How does this help here?


Is there a guide on how to type maths equations like that because I've been looking but I can't find one.
What I mean is cos squared x - sin 2x = sin squared x (I hope that makes more sense.
Original post by AxSirlotl
Is there a guide on how to type maths equations like that because I've been looking but I can't find one.
What I mean is cos squared x - sin 2x = sin squared x (I hope that makes more sense.


There's a guide here. :smile: From cos⁡2(x)−sin⁡(2x)=sin⁡2(x)\cos^2(x) - \sin(2x) = \sin^2(x) you have cos⁡2(x)−sin⁡2(x)=sin⁡(2x)\cos^2(x) - \sin^2(x) = \sin(2x). Is there any identity that you've learnt that you can apply to the LHS here?
Original post by _gcx
I'll go one by one. For the first, assuming you mean cos⁡2(x)−sin⁡2(x)=sin⁡2(x)\cos^2(x) - \sin^2(x) = \sin^2(x)


But... but... that's not an identity otherwise cot⁡(x)≡2\cot(x) \equiv 2 :/
Original post by RDKGames
But... but... that's not an identity otherwise cot⁡(x)≡2\cot(x) \equiv 2 :/


Did I say it was an identity? :rofl:

Spoiler

(edited 6 years ago)
Reply 7
Avatar for Pangol
Pangol
15
Original post by RDKGames
But... but... that's not an identity otherwise cot⁡(x)≡2\cot(x) \equiv 2 :/


It's not clear from the question, but I think that these are equations to be solved, rather than identities. OP - can you clarify?
Original post by Pangol
It's not clear from the question, but I think that these are equations to be solved, rather than identities. OP - can you clarify?


I think it's pretty safe to assume that they're equations to be solved.
Reply 9
Avatar for AxSirlotl
AxSirlotl
OP
18
Original post by _gcx
There's a guide here. :smile: From cos⁡2(x)−sin⁡(2x)=sin⁡2(x)\cos^2(x) - \sin(2x) = \sin^2(x) you have cos⁡2(x)−sin⁡2(x)=sin⁡(2x)\cos^2(x) - \sin^2(x) = \sin(2x). Is there any identity that you've learnt that you can apply to the LHS here?


Thanks, should I get Sin2x=1 Sin2x = 1 ?
Reply 10
Avatar for AxSirlotl
AxSirlotl
OP
18
Original post by Pangol
It's not clear from the question, but I think that these are equations to be solved, rather than identities. OP - can you clarify?


They need to be solved
Original post by _gcx
Did I say it was an identity? :rofl:


Original post by Pangol
It's not clear from the question, but I think that these are equations to be solved, rather than identities. OP - can you clarify?


Eh yeah, they're equations to solve, I thought OP meant identities to prove without knowing he can use the equivalence relation in his post, then gcx just skipping the \equiv part by being lazy and use = instead

Most of these threads are identities so I just skipped to assume without reading the thread properly lol
Original post by AxSirlotl
Thanks, should I get Sin2x=1 Sin2x = 1 ?


Easy mistake to make, cos⁡2(x)+sin⁡2(x)≡1\cos^2(x) + \sin^2(x) \equiv 1, but cos⁡2(x)−sin⁡2(x)≢1\cos^2(x) - \sin^2(x) \not\equiv 1 . Think double angle formulas.
(edited 6 years ago)
Reply 13
Avatar for AxSirlotl
AxSirlotl
OP
18
Original post by _gcx
Easy mistake to make, cos⁡2(x)+sin⁡2(x)≡1\cos^2(x) + \sin^2(x) \equiv 1, but cos⁡2(x)−sin⁡2(x)≢1\cos^2(x) - \sin^2(x) \not\equiv 1 . Think double angle formulas.


Oh yeah I've just realised that it's the Cos2A Cos2A double angle formula.
Original post by AxSirlotl
Oh yeah I've just realised that it's the Cos2A Cos2A double angle formula.


So are you fine with saying we have cos⁡2x=sin⁡2x\cos 2x = \sin 2x? :smile: Where can we go from there?

Btw, you can click on LaTeX here to see what code was used to produce it :smile:
(edited 6 years ago)
Reply 15
Avatar for AxSirlotl
AxSirlotl
OP
18
Original post by _gcx
So are you fine with saying we have cos⁡2x=sin⁡2x\cos 2x = \sin 2x? :smile: Where can we go from there?

Btw, you can click on LaTeX here to see what code was used to produce it :smile:


I got to cos2x=sin2x cos2x = sin2x and then from there got to tan2(x)+2tan(x)−1=0 tan^2(x) + 2tan(x) -1 = 0 which is a quadratic but I don't know if I've made a mistake somewhere or if I've done the right thing.
Original post by AxSirlotl
I got to cos2x=sin2x cos2x = sin2x and then from there got to tan2(x)+2tan(x)−1=0 tan^2(x) + 2tan(x) -1 = 0 which is a quadratic but I don't know if I've made a mistake somewhere or if I've done the right thing.


Not quite sure what you did there. You can just divide both sides by cos⁡2x\cos 2x to get tan⁡2x=1\tan 2x = 1 :smile:
Reply 17
Avatar for AxSirlotl
AxSirlotl
OP
18
Original post by _gcx
Not quite sure what you did there. You can just divide both sides by cos⁡2x\cos 2x to get tan⁡2x=1\tan 2x = 1 :smile:


I only managed to get one of the correct solutions from what I did, thanks for the help, I have a habit of over-complicating things.

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