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# Trig Identities watch

1. Stuck on a trig identities question could someone please point me in the right direction:

2tanθ / 1 + tan^2θ = 1/2
(sorry about it not being in latex form, not too sure how to use it fully yet)

Would I start by multiplying both sides by 1 + tan^2
θ?

Thanks!
2. (Original post by Secret.)
Stuck on a trig identities question could someone please point me in the right direction:

2tanθ / 1 + tan^2θ = 1/2
(sorry about it not being in latex form, not too sure how to use it fully yet)

Would I start by multiplying both sides by 1 + tan^2
θ?

Thanks!
It may help to know that 1 + tan^2θ = sec^2θ
3. (Original post by joostan)
It may help to know that 1 + tan^2θ = sec^2θ
Sorry I should have been more clear, I'm doing C2 and so the only two trig identites we know is:

Sin θ / Cos θ = tan θ
and
Cos^θ + Sin^θ = 1
4. (Original post by Secret.)
Sorry I should have been more clear, I'm doing C2 and so the only two trig identites we know is:

Sin θ / Cos θ = tan θ
and
Cos^θ + Sin^θ = 1
Having now done the question, its not really a relevant identity anyway
Multiplying out is the way forward.
5. (Original post by Secret.)

Would I start by multiplying both sides by 1 + tan^2 [/SUP]θ?

Thanks!
Yes , no identities needed
6. (Original post by Secret.)
Stuck on a trig identities question could someone please point me in the right direction:

[B]2tanθ / 1 + tan^2[B]θ = 1/2
This is not an identity

Are you sure that you are not trying to solve an equation
7. (Original post by TenOfThem)
This is not an identity

Are you sure that you are not trying to solve an equation

Yes that's what the question says, however I thought it was because its on the same sub-chapter that the other trig identities questions are on, so do I not need trig identities at all in this question?
8. (Original post by Secret.)
Yes that's what the question says, however I thought it was because its on the same sub-chapter that the other trig identities questions are on, so do I not need trig identities at all in this question?
No - just solve the equation.
9. (Original post by joostan)
Having now done the question, its not really a relevant identity anyway
Multiplying out is the way forward.
I disagree; I think multiplying the top and the bottom of the LHS by cos^2(θ) would be the best thing to do given that he knows the double sine identity, and you can only see that if you know sec/tan.

EDIT: Especially since you'd have to end up evaluating arctan(2±sqrt(3)). I'm pretty sure he should be using a double angle identity somewhere.

Of course, if he doesn't know the double sine identity, I guess a quadratic in tangent is the next best way :/
10. (Original post by iCiaran)
Yes , no identities needed
(Original post by joostan)
Having now done the question, its not really a relevant identity anyway
Multiplying out is the way forward.

When I do that, I get a half + 1/2tan^ θ
however when I then multiply by 2 and bring it all to one side, it doesn't factorise so should I use the quadratic formula? Or does this factorise?
11. (Original post by Secret.)
When I do that, I get a half + 1/2tan^ θ
however when I then multiply by 2 and bring it all to one side, it doesn't factorise so should I use the quadratic formula? Or does this factorise?
Assuming you're right, take the half out at the front and recall the identity

12. (Original post by Secret.)
When I do that, I get a half + 1/2tan^ θ
however when I then multiply by 2 and bring it all to one side, it doesn't factorise so should I use the quadratic formula? Or does this factorise?
The quadratic formula is definitely the way to go
13. (Original post by Indeterminate)
Assuming you're right, take the half out at the front and recall the identity

The OP is in Y12 not Y13
14. (Original post by aznkid66)
I disagree; I think multiplying the top and the bottom of the LHS by cos^2(θ) would be the best thing to do given that he knows the double sine identity, and you can only see that if you know sec/tan.

EDIT: Especially since you'd have to end up evaluating arctan(2±sqrt(3)). I'm pretty sure he should be using a double angle identity somewhere.

Of course, if he doesn't know the double sine identity, I guess a quadratic in tangent is the next best way :/
Arctan (2+/- root(3)) is a nice angle.
15. (Original post by Secret.)
When I do that, I get a half + 1/2tan^ θ
however when I then multiply by 2 and bring it all to one side, it doesn't factorise so should I use the quadratic formula? Or does this factorise?
Are you sure you're not supposed to have learned the sec/tan identity yet? It merely follows from the pythagorean identity.

16. (Original post by aznkid66)
Are you sure you're not supposed to have learned the sec/tan identity yet? It merely follows from the pythagorean identity.
reciprocal trig functions come up in C3
17. (Original post by TenOfThem)
The OP is in Y12 not Y13

How could you tell? I was misled by the title of the thread.
18. (Original post by joostan)
The quadratic formula is definitely the way to go
(Original post by TenOfThem)
The OP is in Y12 not Y13

Thank you very much! Got the answer

Sorry, one more thing, a question says
1 - sin^2 θ= 3 sinθ cosθ

I started by substituting 1-cos^2 into the sin^2 part and so had

cos ^2 θ = 3sinθ cos θ
do I now have to divide by cos θ? Or do I bring it to once side and factorise like: cos θ(3sinθ). ​Thanks!
19. (Original post by Indeterminate)

How could you tell? I was misled by the title of the thread.
Because they said they're currently doing C2 up there ^
20. (Original post by Secret.)
Thank you very much! Got the answer

Sorry, one more thing, a question says
1 - sin^2 θ= 3 sinθ cosθ

I started by substituting 1-cos^2 into the sin^2 part and so had

cos ^2 θ = 3sinθ cos θ
do I now have to divide by cos θ? Or do I bring it to once side and factorise like: cos θ(3sinθ). ​Thanks!
factorise.
Cos θ may equal 0.

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