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C2: trig id

Hi everyone,

I've got a two part question that I'm struggling with. The first part, which I'm assuming is relevant to the second part is:

Show 3sin⁡θ=2cos⁡2=0 3 \sin \theta = 2 \cos^2 = 0 . I did this fine using the id tanx=sinxcosx tanx = \frac{sinx}{cosx}

I'm stuck on the second part though, and I'd be really appreciative if anyone could help. It is:

Hence, use an appropriate identity to show that 2sin⁡2θ+3sin⁡θ−2=0 2 \sin^2 \theta + 3 \sin \theta - 2 = 0 . To be honest, I'm not sure where to start!
Original post by marcsaccount
Hi everyone,

I've got a two part question that I'm struggling with. The first part, which I'm assuming is relevant to the second part is:

Show 3sin⁡θ=2cos⁡2=0 3 \sin \theta = 2 \cos^2 = 0 . I did this fine using the id tanx=sinxcosx tanx = \frac{sinx}{cosx}

I'm stuck on the second part though, and I'd be really appreciative if anyone could help. It is:

Hence, use an appropriate identity to show that 2sin⁡2θ+3sin⁡θ−2=0 2 \sin^2 \theta + 3 \sin \theta - 2 = 0 . To be honest, I'm not sure where to start!


sin⁡2x+cos⁡2x=1\sin^2 x + \cos^2 x =1
Reply 2
Avatar for marcsaccount
marcsaccount
OP
5
Original post by Mr M
sin⁡2x+cos⁡2x=1\sin^2 x + \cos^2 x =1



Hi MrM, I thought about using that ID initially but that just gave me 2(1−cos⁡2θ)+3sin⁡θ−2 2(1- \cos^2 \theta ) + 3 \sin \theta - 2 and I wasn't sure how to progress from there.
Original post by marcsaccount
Hi MrM, I thought about using that ID initially but that just gave me 2(1−cos⁡2θ)+3sin⁡θ−2 2(1- \cos^2 \theta ) + 3 \sin \theta - 2 and I wasn't sure how to progress from there.


You've made a mistake there. Look again. You need to substitute into the first identity.

Obviously your version will simplify to the first if you expand the bracket but it is a peculiar way to do it.
(edited 10 years ago)
Original post by marcsaccount
Hi MrM, I thought about using that ID initially but that just gave me 2(1−cos⁡2θ)+3sin⁡θ−2 2(1- \cos^2 \theta ) + 3 \sin \theta - 2 and I wasn't sure how to progress from there.

3sinx= 2cos^2x
Reply 5
Avatar for marcsaccount
marcsaccount
OP
5
Original post by Mr M
You've made a mistake there. Look again. You need to substitute into the first identity.

Obviously your version will simplify to the first if you expand the bracket but it is a peculiar way to do it.

The first identity being tanx=(sanx)/(cosx)?

I'm really confused!
Original post by marcsaccount
The first identity being tanx=(sanx)/(cosx)?

I'm really confused!


3sin⁡x−2cos⁡2x=03 \sin x - 2 \cos^2 x = 0

(your version is a bit mangled)

3sin⁡x−2(1−sin⁡2x)=03 \sin x - 2 (1 - \sin^2 x) = 0

Expand the bracket.
Reply 7
Avatar for tomtjl
tomtjl
18
Original post by marcsaccount
Hi MrM, I thought about using that ID initially but that just gave me 2(1−cos⁡2θ)+3sin⁡θ−2 2(1- \cos^2 \theta ) + 3 \sin \theta - 2 and I wasn't sure how to progress from there.


Multiply out the brackets and you get what you got in part a).

You just proved this is equal to 0, so it's done.
Reply 8
Avatar for marcsaccount
marcsaccount
OP
5
Original post by Mr M
3sin⁡x−2cos⁡2x=03 \sin x - 2 \cos^2 x = 0

(your version is a bit mangled)

3sin⁡x−2(1−sin⁡2x)=03 \sin x - 2 (1 - \sin^2 x) = 0

Expand the bracket.


3sin⁡x−2(1−sin⁡2x)=0[br]2sin⁡2x+3sin⁡x−23 \sin x - 2 (1 - \sin^2 x) = 0[br] 2 \sin^2 x + 3 \sin x - 2

Should I factorise it? I'm just not sure what the aim is?!
Reply 9
Avatar for marcsaccount
marcsaccount
OP
5
Actually factorising is the next part so I shouldn't do that now. I still don't know what I'm supposed to do with it to prove it equals zero.

p.s. thanks for helping me!
Reply 10
Avatar for marcsaccount
marcsaccount
OP
5
Original post by tomtjl
Multiply out the brackets and you get what you got in part a).

You just proved this is equal to 0, so it's done.


3sin⁡x−2(1−sin⁡2x)=0[br]2sin⁡2x+3sin⁡x−23 \sin x - 2 (1 - \sin^2 x) = 0[br]2 \sin^2 x + 3 \sin x - 2

So this is the answer? How does it prove it is equal to zero though? It looks different to what I got in part A...
Reply 11
Avatar for marcsaccount
marcsaccount
OP
5
Sorry everyone! I've seen it now. For some reason I just couldn't spot it at first. Thanks for all your help.

Marc

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