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C3 trigonometry hard ques.

By writing the following equations as quadratics in tan θ/2,solve,in the
interval 0 ≤ θ ≤ 360°:

(i) sinθ + 2cosθ = 1

So i was thinking of using this formula:
tan2θ= 2tanθ/1-2tan^2θ
which can be rewritten into:
tanθ=2tanθ/2/1-tan^2θ/2

Tanθ/2= t

so 2t/1-t^2

But when i checked the working out, it wanted:

2t/1+t^2

HELPP PLEASE!!

Thanks
Reply 1
Avatar for TeeEm
TeeEm
19
Original post by Lilly1234567890
By writing the following equations as quadratics in tan θ/2,solve,in the
interval 0 ≤ θ ≤ 360°:

(i) sinθ + 2cosθ = 1

So i was thinking of using this formula:
tan2θ= 2tanθ/1-2tan^2θ
which can be rewritten into:
tanθ=2tanθ/2/1-tan^2θ/2

Tanθ/2= t

so 2t/1-t^2

But when i checked the working out, it wanted:

2t/1+t^2

HELPP PLEASE!!

Thanks


what board are you doing please?
Original post by TeeEm
what board are you doing please?


Edexcel
Reply 3
Avatar for TeeEm
TeeEm
19
Original post by Lilly1234567890
Edexcel


this will never come up in an edexcel maths or further maths paper.
it comes from older/other boards and you require to use the "LITTLE t" identities

I will post a few examples in a few minutes for you to see how these work
Original post by TeeEm
this will never come up in an edexcel maths or further maths paper.
it comes from older/other boards and you require to use the "LITTLE t" identities

I will post a few examples in a few minutes for you to see how these work


Thanks!
the question is on pg.119 Q7
so i thought it was part of the syllabus
Reply 5
Avatar for TeeEm
TeeEm
19
Original post by Lilly1234567890
Thanks!
the question is on pg.119 Q7
so i thought it was part of the syllabus


these results are for enrichment material these days

here are 3 questions

the identities you have to prove in part (a) are common knowledge to mathematicians
PDF.pdf437.2KB
Reply 6
Avatar for Notnek
Notnek
21
Original post by Lilly1234567890
By writing the following equations as quadratics in tan θ/2,solve,in the
interval 0 ≤ θ ≤ 360°:

(i) sinθ + 2cosθ = 1

So i was thinking of using this formula:
tan2θ= 2tanθ/1-2tan^2θ
which can be rewritten into:
tanθ=2tanθ/2/1-tan^2θ/2

Tanθ/2= t

so 2t/1-t^2

But when i checked the working out, it wanted:

2t/1+t^2

HELPP PLEASE!!

Thanks


You have shown that tan⁡θ=2t1−t2\displaystyle \tan \theta = \frac{2t}{1-t^2} which is correct.

The next step is to figure out what sin⁡θ\sin \theta is.

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