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Trogonometrical identies question! watch

1. Help pleaase

Thank you!

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2. You know an identity involving .
3. use cos^2+ sin^2 =1 to help
4. (Original post by Libby18)
use cos^2+ sin^2 =1 to help
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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5. (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).
6. (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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7. Has anyone besides the OP actually tried solving it? I'm reaching a quadratic with no real solutions ...
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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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
9. (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 nothing is multiplied by
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
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.
12. (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
It's not just 1 - sin^2 theta Is it?

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13. (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
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
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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15. (Original post by Annabel_lear)
It's not just 1 - sin^2 theta Is it?

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Exactly that
16. (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
17. (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. Check arithmetic!
18. (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
19. (Original post by davros)
check your algebra - I got
Yupp!
Does that make it:
(2x-1)(3x-1)

So sinx = 1/2 , 1/3

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20. (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

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