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Trogonometrical identies question!

Help pleaase

76cos2θ=5sinθ[br] 7 - 6cos^2 \theta = 5sin\theta[br]

Thank you!


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You know an identity involving cos2θ\cos^2 \theta.
Reply 2
use cos^2+ sin^2 =1 to help :smile:
Reply 3
Original post by Libby18
use cos^2+ sin^2 =1 to help :smile:


I understand that you have to use that but I don't know how, I rearranged it so it was equal to 7


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Reply 4
Original post by Annabel_lear
I understand that you have to use that but I don't know how, I rearranged it so it was equal to 7


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erm you rearrange it to get what cos^2 is equal to, then substitute that into the equation above.



If you type your working, I can help you. I'm not allowed to give answers (we're suppose to help OP work it out).
(edited 11 years ago)
Reply 5
Original post by Libby18
erm you rearrange it to get what cos^2 is equal to, then substitute that into the equation above.



If you type your working, I can help you. I'm not allowed to give answers (we're suppose to help OP work it out).


When you make 6cos^2theta the subject you can't jut divide by 6 and theta right?


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Reply 6
Has anyone besides the OP actually tried solving it? I'm reaching a quadratic with no real solutions ...
Reply 7
Original post by Annabel_lear
When you make 6cos^2theta the subject you can't jut divide by 6 and theta right?


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nope

right first step rearrange cos^2 +sin^2 =1 ?

tell me what you get and we'll take it from that.

step at a time :smile:
Reply 8
Original post by Annabel_lear
When you make 6cos^2theta the subject you can't jut divide by 6 and theta right?


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How could you divide by θ\theta nothing is multiplied by θ\theta
Reply 9
Original post by Big-Daddy
Has anyone besides the OP actually tried solving it? I'm reaching a quadratic with no real solutions ...


Nope I haven't but following the basic rules, I would get a quadratic and then factorise that (if its possible) if not, maybe the quadratic formula? if not, then I dont know :P
Original post by Big-Daddy
Has anyone besides the OP actually tried solving it? I'm reaching a quadratic with no real solutions ...


No, you've made a mistake. The question is fine.
Reply 11
Original post by Libby18
nope

right first step rearrange cos^2 +sin^2 =1 ?

tell me what you get and we'll take it from that.

step at a time :smile:


It's not just 1 - sin^2 theta Is it?


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Original post by Mr M
No, you've made a mistake. The question is fine.


Yes you're right it's fine, I misread 5sin(x) as sin(x).

To the OP: yes it is just cos2(x)=1-sin2(x), substitute in and solve :smile:
Reply 13
Original post by Big-Daddy
Yes you're right it's fine, I misread 5sin(x) as sin(x).

To the OP: yes it is just cos2(x)=1-sin2(x), substitute in and solve :smile:


I subbed it back, got:
6sin^2 theta - 5sin theta - 1 = 0
Turned it into:
6x^2 -5x -1
Factorised:
(X-1)(6x+1)
Does that make sinx = -1/6 , 1
And then do I put that back into the
7 - 6cos^2 theta = 5sin theta ?



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Reply 14
Original post by Annabel_lear
It's not just 1 - sin^2 theta Is it?


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Exactly that :smile:
Reply 15
Original post by Annabel_lear
I subbed it back, got:
6sin^2 theta - 5sin theta - 1 = 0
Turned it into:
6x^2 -5x -1
Factorised:
(X-1)(6x+1)
Does that make sinx = -1/6 , 1
And then do I put that back into the
7 - 6cos^2 theta = 5sin theta ?



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You've made a mistake in substituting, it shouldn't be negative 1
Original post by Annabel_lear
I subbed it back, got:
6sin^2 theta - 5sin theta - 1 = 0
Turned it into:
6x^2 -5x -1
Factorised:
(X-1)(6x+1)
Does that make sinx = -1/6 , 1
And then do I put that back into the
7 - 6cos^2 theta = 5sin theta ?



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The bit in bold is where I stopped reading. :tongue: Check arithmetic!
Reply 17
Original post by Annabel_lear
I subbed it back, got:
6sin^2 theta - 5sin theta - 1 = 0

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check your algebra - I got
6sin2θ5sinθ+1=06sin^2 \theta - 5sin \theta + 1 = 0
Reply 18
Original post by davros
check your algebra - I got
6sin2θ5sinθ+1=06sin^2 \theta - 5sin \theta + 1 = 0


Yupp!
Does that make it:
(2x-1)(3x-1)

So sinx = 1/2 , 1/3



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Reply 19
Original post by Annabel_lear
Yupp!
Does that make it:
(2x-1)(3x-1)

So sinx = 1/2 , 1/3



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that's what I got :smile:

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