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Trigonometry question? C2 ocr mei

Solve the equation tan θ = 2 sin θ for 0◦ ≤ θ ≤ 360◦.

Ive got to sin theta/cos theta = 2 sin theta but dont know where to go from here, can anyone help?
Reply 1
Avatar for davros
davros
16
Original post by liverpool2044
Solve the equation tan θ = 2 sin θ for 0◦ ≤ θ ≤ 360◦.

Ive got to sin theta/cos theta = 2 sin theta but dont know where to go from here, can anyone help?


You can rearrange to "something = 0", and factorize (common factor of sin theta) then solve as normal.
Reply 2
Avatar for ztibor
ztibor
10
Original post by liverpool2044
Solve the equation tan θ = 2 sin θ for 0◦ ≤ θ ≤ 360◦.

Ive got to sin theta/cos theta = 2 sin theta but dont know where to go from here, can anyone help?


the best method is the factorization

constraints: cos⁡θ≠0\cos \theta \neq 0

sin⁡θ(1cos⁡θ−2)=0\displaystyle \sin \theta \left (\frac{1}{\cos \theta}-2\right )=0

This product will be zero iff any of the factor wiil be zero, so we get two equations.to solve
(edited 8 years ago)
Wait, i dont really understand how you got to that, could you give me a step by step of how youve got to that, sorry if im being silly haha, appreciated
Reply 4
Avatar for ztibor
ztibor
10
Original post by liverpool2044
Wait, i dont really understand how you got to that, could you give me a step by step of how youve got to that, sorry if im being silly haha, appreciated


tan⁡θ=2⋅sin⁡θ \displaystyle \tan \theta=2 \cdot \sin \theta

sin⁡θcos⁡θ=2⋅sin⁡θ\displaystyle \frac{\sin \theta}{\cos \theta}=2 \cdot \sin \theta

subtracting 2⋅sin⁡θ2 \cdot \sin \theta from both sides

sin⁡θcos⁡θ−2⋅sin⁡θ=0\displaystyle \frac{\sin \theta}{\cos \theta}-2 \cdot \sin \theta =0

taking out the common factor

sin⁡θ⋅(1cos⁡θ−2)=0\displaystyle \sin \theta \cdot \left (\frac{1}{\cos \theta}-2 \right ) =0

Note:


sin⁡θcos⁡θ=1cos⁡θ⋅sin⁡θ=1cos⁡θ⋅11sin⁡θ=1cos⁡θsin⁡θ=1cos⁡θ1sin⁡θ=...\displaystyle \frac{\sin \theta}{\cos \theta} = \frac{1}{\cos \theta} \cdot \sin \theta =\frac{1}{\cos \theta} \cdot \frac{1}{\frac{1}{\sin \theta}}=\frac{1}{\frac{\cos \theta}{\sin \theta}}=\frac{\frac{1}{\cos \theta}}{\frac{1}{\sin \theta}}=...

and so on
(edited 8 years ago)
thats brilliant, just what i wanted, thank you!:smile:
it does say in the mark scheme its 2cos theta - 1 = 0, how do you arrange it to this?
Reply 7
Avatar for ztibor
ztibor
10
Original post by liverpool2044
it does say in the mark scheme its 2cos theta - 1 = 0, how do you arrange it to this?


well, let's answer the question

We've got

sin⁡θ⋅(1cos⁡θ−2)=0\displaystyle \sin \theta \cdot \left (\frac{1}{\cos \theta}-2 \right ) =0

As I wrote there is a statement: i.e. the product of two factors above is zero

In mathematics with no zero divider from this follows bellow

1. The first factor is zero (sin⁡θ=0 \sin \theta =0)
OR
2. The second factor is zero (1cos⁡θ−2=0\frac{1}{\cos \theta}-2=0)

THe OR is logical OR, so the two statement can be true at the same time

Arranging the 2nd: multiplying by cos⁡θ\cos \theta because it can not be zero
as i wrote

1−2cos⁡θ=0\displaystyle 1-2\cos \theta =0

Multiplying by -1

2cos⁡θ−1=0\displaystyle 2\cos \theta - 1 =0

as the mark scheme says. But this is not the final answer

Does the mark scheme say sin⁡θ=0 \displaystyle \sin \theta =0

If it does not, than there was a dividing by sin⁡θ\sin \theta which
can result loss of roots, because the value of sin⁡θ\sin \theta can be 0.

So at such dividing we have to say that sin⁡θ≠0\sin \theta \neq 0
and the end of the answer we have to examine what if sin⁡θ=0\sin \theta = 0.

This is a general rule for dividing by terms containing unknown variable.
At the other side multiplying by so terms we can get false roots.
(edited 8 years ago)
sin theta = 0 on the exam but thank you so much:smile:

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